Reflection of light is the phenomenon in which light returns to the same medium after striking a surface. When a ray of light falls on a polished mirror, a smooth metal surface, still water or any shiny boundary, a part of light comes back into the original medium. This returning of light is called reflection. The direction of the reflected ray is not random; it follows a definite geometrical rule, which is the law of reflection.
Historically, reflection was one of the earliest optical phenomena studied by ancient scholars because mirrors, water surfaces and polished metals were available long before modern lenses. The predictable behaviour of reflected light helped in the development of geometrical optics. In modern Physics, reflection is understood as the interaction of electromagnetic waves with a surface, but in Ray Optics we use rays to describe the path of light.
Reflection is present everywhere in daily life. We see our face in a mirror because light from the face reflects from the mirror and reaches the eye. We see objects around us because light from a source falls on objects and gets diffusely reflected in many directions. A polished surface produces regular reflection and can form a clear image, whereas a rough wall produces diffuse reflection and helps us see the wall from many directions.
Reflection from polished surfaces is orderly because the surface irregularities are very small compared with the scale of the incident beam. Parallel incident rays remain parallel after reflection. Reflection from rough surfaces is irregular because different parts of the surface have different normals. However, each tiny point still obeys the laws of reflection. This distinction is extremely important for understanding mirrors, visibility of objects, road safety, telescopes, periscopes, kaleidoscopes and many competitive examination problems.
Incident ray, reflected ray and normal lie in the same plane; angle of incidence equals angle of reflection.
Section 2: Laws of Reflection
First Law
The incident ray, reflected ray and normal at the point of incidence all lie in the same plane. The incident ray is the ray that strikes the surface. The reflected ray is the ray that returns from the surface. The normal is a perpendicular line drawn to the reflecting surface at the point of incidence.
Second Law
The angle of incidence is equal to the angle of reflection. The angles are always measured from the normal, not from the mirror surface.
i = r
For a plane reflecting surface, the normal provides the reference direction. The incident ray approaches the surface making angle i with the normal. The reflected ray leaves the surface making angle r with the normal. Experiments show that these two angles are equal for all reflecting surfaces. The physical significance is that reflection is symmetrical about the normal.
NCERT-style explanation: The laws of reflection are valid for both plane and curved surfaces. In curved surfaces, the normal is drawn along the radius at the point of incidence. These laws explain image formation by plane mirrors, spherical mirrors and many optical instruments.
Incident ray, reflected ray and normal lie in the same plane; angle of incidence equals angle of reflection.
Section 3: Types of Reflection
Regular Reflection
Regular reflection occurs from smooth and polished surfaces such as plane mirrors, polished metal and calm water. Parallel incident rays remain parallel after reflection. It produces clear images and is used in mirrors, telescopes, periscopes and laser-based systems.
Diffuse Reflection
Diffuse reflection occurs from rough surfaces such as paper, wall, cloth, road and unpolished wood. Parallel incident rays scatter in different directions because surface normals are different at different points. It helps us see non-luminous objects from different positions.
Point
Regular Reflection
Diffuse Reflection
Surface
Smooth and polished
Rough and irregular
Reflected rays
Remain parallel
Scatter in many directions
Image
Clear image is formed
No clear image is formed
Example surface
Plane mirror
Wall or paper
Normal
Nearly same for nearby points
Different at different points
Use
Optical instruments
Visibility of objects
Glare
Can produce glare
Reduces glare
Laser reflection
Strong and directed
Scattered
Order
Ordered reflection
Irregular reflection
Image information
Preserved
Not preserved
Surface quality
Polished metal
Unpolished wood
Daily use
Mirrors
Seeing books and walls
Eye comfort
May dazzle
Usually comfortable
Name
Specular reflection
Diffuse reflection
Law validity
Obeys laws of reflection
Also obeys laws locally
Section 4: Image Formation Basics
Real image Formed when reflected or refracted rays actually meet. It can be obtained on a screen.
Virtual image Formed when rays appear to meet. It cannot be obtained on a screen.
Magnification Ratio of image size to object size. It tells whether image is enlarged, diminished or same size.
Lateral inversion Left-right reversal observed in a plane mirror image.
Object-image relation For plane mirror, image distance behind mirror equals object distance in front of mirror.
Image symmetry Object and image are symmetric about the mirror plane.
Section 5: Reflection from Plane Surfaces
A plane mirror forms an image that is virtual, erect, laterally inverted and of the same size as the object. The image is formed as far behind the mirror as the object is in front of it. The mirror acts like a plane of symmetry between object and image.
Plane mirror distancedimage = dobject
Image is as far behind mirror as object is in front.
Magnificationm = +1
Image is erect and same size.
Mirror rotationθ → 2θ
If mirror rotates by θ, reflected ray rotates by 2θ.
Two mirrorsN = 360/θ − 1
For angle θ when 360/θ is an integer.
Periscopes use two plane mirrors inclined at 45° to change the path of light. Kaleidoscopes use multiple reflections between inclined mirrors to produce repeated patterns. In mirror rotation problems, the reflected ray turns through twice the angle turned by the mirror because the normal also rotates by the same angle.
Section 6: Important NCERT Concepts
Concept box Angles in reflection are measured with the normal, not the surface.
Exam tip Draw incident ray, reflected ray and normal before applying i=r.
Confused concept Diffuse reflection does not violate laws of reflection. Each tiny surface element obeys them.
One-line revision Plane mirror image is virtual, erect, same size and laterally inverted.
Frequently tested Mirror rotation causes reflected ray rotation of 2θ.
Image motion If object moves towards a plane mirror with speed v, image approaches mirror with speed v and object-image separation changes at 2v.
Section 7: Mathematical Concepts
Rotation of mirror
θ → 2θ
If a plane mirror rotates by angle θ while incident ray is fixed, the normal also rotates by θ. Since reflection is symmetric about the normal, the reflected ray rotates by 2θ.
Number of images
N = 360/θ − 1
For two plane mirrors inclined at angle θ, if 360/θ is an integer, the number of images is 360/θ − 1. Other cases depend on object position and parity.
Relative motion
For a plane mirror, image speed relative to mirror equals object speed relative to mirror, on the opposite side. Object-image separation often changes at twice the component of object velocity perpendicular to mirror.
Successive reflections
When a ray reflects from multiple mirrors, apply i=r at each mirror separately. Keep track of normals and direction after every reflection.
Section 8: Solved Numericals
NEET Numericals
NEET Numerical 1. A ray strikes a plane mirror at i = 35°. Find r.
NEET
Given: i=35°
Formula:
i=r
r=35°
Write answer with correct unit or description.
Final Answer: 35°
NEET Numerical 2. A mirror rotates by 12°. Find rotation of reflected ray.
NEET
Given: θ=12°
Formula:
rotation=2θ
2×12°=24°
Write answer with correct unit or description.
Final Answer: 24°
NEET Numerical 3. Two mirrors are inclined at 60°. Find number of images.
NEET
Given: θ=60°
Formula:
N=360/θ−1
N=360/60−1=6−1=5
Write answer with correct unit or description.
Final Answer: 5 images
NEET Numerical 4. An object is 20 cm in front of plane mirror. Find image distance.
NEET
Given: object distance=20 cm
Formula:
dimage=dobject
image is 20 cm behind mirror
Write answer with correct unit or description.
Final Answer: 20 cm behind mirror
NEET Numerical 5. Object moves towards mirror at 3 m/s. Find speed of image towards object.
NEET
Given: v=3 m/s
Formula:
relative speed=2v
2×3=6 m/s
Write answer with correct unit or description.
Final Answer: 6 m/s
NEET Numerical 6. A ray strikes a plane mirror at i = 35°. Find r.
NEET
Given: i=35°
Formula:
i=r
r=35°
Write answer with correct unit or description.
Final Answer: 35°
NEET Numerical 7. A mirror rotates by 12°. Find rotation of reflected ray.
NEET
Given: θ=12°
Formula:
rotation=2θ
2×12°=24°
Write answer with correct unit or description.
Final Answer: 24°
NEET Numerical 8. Two mirrors are inclined at 60°. Find number of images.
NEET
Given: θ=60°
Formula:
N=360/θ−1
N=360/60−1=6−1=5
Write answer with correct unit or description.
Final Answer: 5 images
NEET Numerical 9. An object is 20 cm in front of plane mirror. Find image distance.
NEET
Given: object distance=20 cm
Formula:
dimage=dobject
image is 20 cm behind mirror
Write answer with correct unit or description.
Final Answer: 20 cm behind mirror
NEET Numerical 10. Object moves towards mirror at 3 m/s. Find speed of image towards object.
NEET
Given: v=3 m/s
Formula:
relative speed=2v
2×3=6 m/s
Write answer with correct unit or description.
Final Answer: 6 m/s
NEET Numerical 11. A ray strikes a plane mirror at i = 35°. Find r.
NEET
Given: i=35°
Formula:
i=r
r=35°
Write answer with correct unit or description.
Final Answer: 35°
NEET Numerical 12. A mirror rotates by 12°. Find rotation of reflected ray.
NEET
Given: θ=12°
Formula:
rotation=2θ
2×12°=24°
Write answer with correct unit or description.
Final Answer: 24°
NEET Numerical 13. Two mirrors are inclined at 60°. Find number of images.
NEET
Given: θ=60°
Formula:
N=360/θ−1
N=360/60−1=6−1=5
Write answer with correct unit or description.
Final Answer: 5 images
NEET Numerical 14. An object is 20 cm in front of plane mirror. Find image distance.
NEET
Given: object distance=20 cm
Formula:
dimage=dobject
image is 20 cm behind mirror
Write answer with correct unit or description.
Final Answer: 20 cm behind mirror
NEET Numerical 15. Object moves towards mirror at 3 m/s. Find speed of image towards object.
NEET
Given: v=3 m/s
Formula:
relative speed=2v
2×3=6 m/s
Write answer with correct unit or description.
Final Answer: 6 m/s
NEET Numerical 16. A ray strikes a plane mirror at i = 35°. Find r.
NEET
Given: i=35°
Formula:
i=r
r=35°
Write answer with correct unit or description.
Final Answer: 35°
NEET Numerical 17. A mirror rotates by 12°. Find rotation of reflected ray.
NEET
Given: θ=12°
Formula:
rotation=2θ
2×12°=24°
Write answer with correct unit or description.
Final Answer: 24°
NEET Numerical 18. Two mirrors are inclined at 60°. Find number of images.
NEET
Given: θ=60°
Formula:
N=360/θ−1
N=360/60−1=6−1=5
Write answer with correct unit or description.
Final Answer: 5 images
NEET Numerical 19. An object is 20 cm in front of plane mirror. Find image distance.
NEET
Given: object distance=20 cm
Formula:
dimage=dobject
image is 20 cm behind mirror
Write answer with correct unit or description.
Final Answer: 20 cm behind mirror
NEET Numerical 20. Object moves towards mirror at 3 m/s. Find speed of image towards object.
NEET
Given: v=3 m/s
Formula:
relative speed=2v
2×3=6 m/s
Write answer with correct unit or description.
Final Answer: 6 m/s
NEET Numerical 21. A ray strikes a plane mirror at i = 35°. Find r.
NEET
Given: i=35°
Formula:
i=r
r=35°
Write answer with correct unit or description.
Final Answer: 35°
NEET Numerical 22. A mirror rotates by 12°. Find rotation of reflected ray.
NEET
Given: θ=12°
Formula:
rotation=2θ
2×12°=24°
Write answer with correct unit or description.
Final Answer: 24°
NEET Numerical 23. Two mirrors are inclined at 60°. Find number of images.
NEET
Given: θ=60°
Formula:
N=360/θ−1
N=360/60−1=6−1=5
Write answer with correct unit or description.
Final Answer: 5 images
NEET Numerical 24. An object is 20 cm in front of plane mirror. Find image distance.
NEET
Given: object distance=20 cm
Formula:
dimage=dobject
image is 20 cm behind mirror
Write answer with correct unit or description.
Final Answer: 20 cm behind mirror
NEET Numerical 25. Object moves towards mirror at 3 m/s. Find speed of image towards object.
NEET
Given: v=3 m/s
Formula:
relative speed=2v
2×3=6 m/s
Write answer with correct unit or description.
Final Answer: 6 m/s
JEE Main Numericals
JEE Main Numerical 26. A ray strikes a plane mirror at i = 35°. Find r.
JEE Main
Given: i=35°
Formula:
i=r
r=35°
Write answer with correct unit or description.
Final Answer: 35°
JEE Main Numerical 27. A mirror rotates by 12°. Find rotation of reflected ray.
JEE Main
Given: θ=12°
Formula:
rotation=2θ
2×12°=24°
Write answer with correct unit or description.
Final Answer: 24°
JEE Main Numerical 28. Two mirrors are inclined at 60°. Find number of images.
JEE Main
Given: θ=60°
Formula:
N=360/θ−1
N=360/60−1=6−1=5
Write answer with correct unit or description.
Final Answer: 5 images
JEE Main Numerical 29. An object is 20 cm in front of plane mirror. Find image distance.
JEE Main
Given: object distance=20 cm
Formula:
dimage=dobject
image is 20 cm behind mirror
Write answer with correct unit or description.
Final Answer: 20 cm behind mirror
JEE Main Numerical 30. Object moves towards mirror at 3 m/s. Find speed of image towards object.
JEE Main
Given: v=3 m/s
Formula:
relative speed=2v
2×3=6 m/s
Write answer with correct unit or description.
Final Answer: 6 m/s
JEE Main Numerical 31. A ray strikes a plane mirror at i = 35°. Find r.
JEE Main
Given: i=35°
Formula:
i=r
r=35°
Write answer with correct unit or description.
Final Answer: 35°
JEE Main Numerical 32. A mirror rotates by 12°. Find rotation of reflected ray.
JEE Main
Given: θ=12°
Formula:
rotation=2θ
2×12°=24°
Write answer with correct unit or description.
Final Answer: 24°
JEE Main Numerical 33. Two mirrors are inclined at 60°. Find number of images.
JEE Main
Given: θ=60°
Formula:
N=360/θ−1
N=360/60−1=6−1=5
Write answer with correct unit or description.
Final Answer: 5 images
JEE Main Numerical 34. An object is 20 cm in front of plane mirror. Find image distance.
JEE Main
Given: object distance=20 cm
Formula:
dimage=dobject
image is 20 cm behind mirror
Write answer with correct unit or description.
Final Answer: 20 cm behind mirror
JEE Main Numerical 35. Object moves towards mirror at 3 m/s. Find speed of image towards object.
JEE Main
Given: v=3 m/s
Formula:
relative speed=2v
2×3=6 m/s
Write answer with correct unit or description.
Final Answer: 6 m/s
JEE Main Numerical 36. A ray strikes a plane mirror at i = 35°. Find r.
JEE Main
Given: i=35°
Formula:
i=r
r=35°
Write answer with correct unit or description.
Final Answer: 35°
JEE Main Numerical 37. A mirror rotates by 12°. Find rotation of reflected ray.
JEE Main
Given: θ=12°
Formula:
rotation=2θ
2×12°=24°
Write answer with correct unit or description.
Final Answer: 24°
JEE Main Numerical 38. Two mirrors are inclined at 60°. Find number of images.
JEE Main
Given: θ=60°
Formula:
N=360/θ−1
N=360/60−1=6−1=5
Write answer with correct unit or description.
Final Answer: 5 images
JEE Main Numerical 39. An object is 20 cm in front of plane mirror. Find image distance.
JEE Main
Given: object distance=20 cm
Formula:
dimage=dobject
image is 20 cm behind mirror
Write answer with correct unit or description.
Final Answer: 20 cm behind mirror
JEE Main Numerical 40. Object moves towards mirror at 3 m/s. Find speed of image towards object.
JEE Main
Given: v=3 m/s
Formula:
relative speed=2v
2×3=6 m/s
Write answer with correct unit or description.
Final Answer: 6 m/s
JEE Main Numerical 41. A ray strikes a plane mirror at i = 35°. Find r.
JEE Main
Given: i=35°
Formula:
i=r
r=35°
Write answer with correct unit or description.
Final Answer: 35°
JEE Main Numerical 42. A mirror rotates by 12°. Find rotation of reflected ray.
JEE Main
Given: θ=12°
Formula:
rotation=2θ
2×12°=24°
Write answer with correct unit or description.
Final Answer: 24°
JEE Main Numerical 43. Two mirrors are inclined at 60°. Find number of images.
JEE Main
Given: θ=60°
Formula:
N=360/θ−1
N=360/60−1=6−1=5
Write answer with correct unit or description.
Final Answer: 5 images
JEE Main Numerical 44. An object is 20 cm in front of plane mirror. Find image distance.
JEE Main
Given: object distance=20 cm
Formula:
dimage=dobject
image is 20 cm behind mirror
Write answer with correct unit or description.
Final Answer: 20 cm behind mirror
JEE Main Numerical 45. Object moves towards mirror at 3 m/s. Find speed of image towards object.
JEE Main
Given: v=3 m/s
Formula:
relative speed=2v
2×3=6 m/s
Write answer with correct unit or description.
Final Answer: 6 m/s
JEE Advanced Numericals
JEE Advanced Numerical 46. A ray strikes a plane mirror at i = 35°. Find r.
JEE Advanced
Given: i=35°
Formula:
i=r
r=35°
Write answer with correct unit or description.
Final Answer: 35°
JEE Advanced Numerical 47. A mirror rotates by 12°. Find rotation of reflected ray.
JEE Advanced
Given: θ=12°
Formula:
rotation=2θ
2×12°=24°
Write answer with correct unit or description.
Final Answer: 24°
JEE Advanced Numerical 48. Two mirrors are inclined at 60°. Find number of images.
JEE Advanced
Given: θ=60°
Formula:
N=360/θ−1
N=360/60−1=6−1=5
Write answer with correct unit or description.
Final Answer: 5 images
JEE Advanced Numerical 49. An object is 20 cm in front of plane mirror. Find image distance.
JEE Advanced
Given: object distance=20 cm
Formula:
dimage=dobject
image is 20 cm behind mirror
Write answer with correct unit or description.
Final Answer: 20 cm behind mirror
JEE Advanced Numerical 50. Object moves towards mirror at 3 m/s. Find speed of image towards object.
JEE Advanced
Given: v=3 m/s
Formula:
relative speed=2v
2×3=6 m/s
Write answer with correct unit or description.
Final Answer: 6 m/s
JEE Advanced Numerical 51. A ray strikes a plane mirror at i = 35°. Find r.
JEE Advanced
Given: i=35°
Formula:
i=r
r=35°
Write answer with correct unit or description.
Final Answer: 35°
JEE Advanced Numerical 52. A mirror rotates by 12°. Find rotation of reflected ray.
JEE Advanced
Given: θ=12°
Formula:
rotation=2θ
2×12°=24°
Write answer with correct unit or description.
Final Answer: 24°
JEE Advanced Numerical 53. Two mirrors are inclined at 60°. Find number of images.
JEE Advanced
Given: θ=60°
Formula:
N=360/θ−1
N=360/60−1=6−1=5
Write answer with correct unit or description.
Final Answer: 5 images
JEE Advanced Numerical 54. An object is 20 cm in front of plane mirror. Find image distance.
JEE Advanced
Given: object distance=20 cm
Formula:
dimage=dobject
image is 20 cm behind mirror
Write answer with correct unit or description.
Final Answer: 20 cm behind mirror
JEE Advanced Numerical 55. Object moves towards mirror at 3 m/s. Find speed of image towards object.
JEE Advanced
Given: v=3 m/s
Formula:
relative speed=2v
2×3=6 m/s
Write answer with correct unit or description.
Final Answer: 6 m/s
JEE Advanced Numerical 56. A ray strikes a plane mirror at i = 35°. Find r.
JEE Advanced
Given: i=35°
Formula:
i=r
r=35°
Write answer with correct unit or description.
Final Answer: 35°
JEE Advanced Numerical 57. A mirror rotates by 12°. Find rotation of reflected ray.
JEE Advanced
Given: θ=12°
Formula:
rotation=2θ
2×12°=24°
Write answer with correct unit or description.
Final Answer: 24°
JEE Advanced Numerical 58. Two mirrors are inclined at 60°. Find number of images.
JEE Advanced
Given: θ=60°
Formula:
N=360/θ−1
N=360/60−1=6−1=5
Write answer with correct unit or description.
Final Answer: 5 images
JEE Advanced Numerical 59. An object is 20 cm in front of plane mirror. Find image distance.
JEE Advanced
Given: object distance=20 cm
Formula:
dimage=dobject
image is 20 cm behind mirror
Write answer with correct unit or description.
Final Answer: 20 cm behind mirror
JEE Advanced Numerical 60. Object moves towards mirror at 3 m/s. Find speed of image towards object.
JEE Advanced
Given: v=3 m/s
Formula:
relative speed=2v
2×3=6 m/s
Write answer with correct unit or description.
Final Answer: 6 m/s
Section 9: JEE Advanced Challenge Problems
Advanced Reflection Problems based on multiple mirrors, successive reflections, mirror rotation and relative image motion.
Advanced Reflection Problem 1. Competitive-level problem on multiple mirrors and reflected ray direction.
JEE Advanced
Approach: Draw each mirror, mark normal, apply i=r at every reflection and track image position using symmetry.
Identify mirror geometry and angle between mirrors.
Use N=360/θ−1 when applicable.
For rotation, use reflected ray rotation = 2θ.
For relative motion, resolve velocity perpendicular to mirror.
Solved using multiple reflection geometry.
Advanced Reflection Problem 2. Competitive-level problem on multiple mirrors and reflected ray direction.
JEE Advanced
Approach: Draw each mirror, mark normal, apply i=r at every reflection and track image position using symmetry.
Identify mirror geometry and angle between mirrors.
Use N=360/θ−1 when applicable.
For rotation, use reflected ray rotation = 2θ.
For relative motion, resolve velocity perpendicular to mirror.
Solved using multiple reflection geometry.
Advanced Reflection Problem 3. Competitive-level problem on multiple mirrors and reflected ray direction.
JEE Advanced
Approach: Draw each mirror, mark normal, apply i=r at every reflection and track image position using symmetry.
Identify mirror geometry and angle between mirrors.
Use N=360/θ−1 when applicable.
For rotation, use reflected ray rotation = 2θ.
For relative motion, resolve velocity perpendicular to mirror.
Solved using multiple reflection geometry.
Advanced Reflection Problem 4. Competitive-level problem on multiple mirrors and reflected ray direction.
JEE Advanced
Approach: Draw each mirror, mark normal, apply i=r at every reflection and track image position using symmetry.
Identify mirror geometry and angle between mirrors.
Use N=360/θ−1 when applicable.
For rotation, use reflected ray rotation = 2θ.
For relative motion, resolve velocity perpendicular to mirror.
Solved using multiple reflection geometry.
Advanced Reflection Problem 5. Competitive-level problem on multiple mirrors and reflected ray direction.
JEE Advanced
Approach: Draw each mirror, mark normal, apply i=r at every reflection and track image position using symmetry.
Identify mirror geometry and angle between mirrors.
Use N=360/θ−1 when applicable.
For rotation, use reflected ray rotation = 2θ.
For relative motion, resolve velocity perpendicular to mirror.
Solved using multiple reflection geometry.
Advanced Reflection Problem 6. Competitive-level problem on multiple mirrors and reflected ray direction.
JEE Advanced
Approach: Draw each mirror, mark normal, apply i=r at every reflection and track image position using symmetry.
Identify mirror geometry and angle between mirrors.
Use N=360/θ−1 when applicable.
For rotation, use reflected ray rotation = 2θ.
For relative motion, resolve velocity perpendicular to mirror.
Solved using multiple reflection geometry.
Advanced Reflection Problem 7. Competitive-level problem on multiple mirrors and reflected ray direction.
JEE Advanced
Approach: Draw each mirror, mark normal, apply i=r at every reflection and track image position using symmetry.
Identify mirror geometry and angle between mirrors.
Use N=360/θ−1 when applicable.
For rotation, use reflected ray rotation = 2θ.
For relative motion, resolve velocity perpendicular to mirror.
Solved using multiple reflection geometry.
Advanced Reflection Problem 8. Competitive-level problem on multiple mirrors and reflected ray direction.
JEE Advanced
Approach: Draw each mirror, mark normal, apply i=r at every reflection and track image position using symmetry.
Identify mirror geometry and angle between mirrors.
Use N=360/θ−1 when applicable.
For rotation, use reflected ray rotation = 2θ.
For relative motion, resolve velocity perpendicular to mirror.
Solved using multiple reflection geometry.
Advanced Reflection Problem 9. Competitive-level problem on multiple mirrors and reflected ray direction.
JEE Advanced
Approach: Draw each mirror, mark normal, apply i=r at every reflection and track image position using symmetry.
Identify mirror geometry and angle between mirrors.
Use N=360/θ−1 when applicable.
For rotation, use reflected ray rotation = 2θ.
For relative motion, resolve velocity perpendicular to mirror.
Solved using multiple reflection geometry.
Advanced Reflection Problem 10. Competitive-level problem on multiple mirrors and reflected ray direction.
JEE Advanced
Approach: Draw each mirror, mark normal, apply i=r at every reflection and track image position using symmetry.
Identify mirror geometry and angle between mirrors.
Use N=360/θ−1 when applicable.
For rotation, use reflected ray rotation = 2θ.
For relative motion, resolve velocity perpendicular to mirror.
Solved using multiple reflection geometry.
Advanced Reflection Problem 11. Competitive-level problem on multiple mirrors and reflected ray direction.
JEE Advanced
Approach: Draw each mirror, mark normal, apply i=r at every reflection and track image position using symmetry.
Identify mirror geometry and angle between mirrors.
Use N=360/θ−1 when applicable.
For rotation, use reflected ray rotation = 2θ.
For relative motion, resolve velocity perpendicular to mirror.
Solved using multiple reflection geometry.
Advanced Reflection Problem 12. Competitive-level problem on multiple mirrors and reflected ray direction.
JEE Advanced
Approach: Draw each mirror, mark normal, apply i=r at every reflection and track image position using symmetry.
Identify mirror geometry and angle between mirrors.
Use N=360/θ−1 when applicable.
For rotation, use reflected ray rotation = 2θ.
For relative motion, resolve velocity perpendicular to mirror.
Solved using multiple reflection geometry.
Advanced Reflection Problem 13. Competitive-level problem on multiple mirrors and reflected ray direction.
JEE Advanced
Approach: Draw each mirror, mark normal, apply i=r at every reflection and track image position using symmetry.
Identify mirror geometry and angle between mirrors.
Use N=360/θ−1 when applicable.
For rotation, use reflected ray rotation = 2θ.
For relative motion, resolve velocity perpendicular to mirror.
Solved using multiple reflection geometry.
Advanced Reflection Problem 14. Competitive-level problem on multiple mirrors and reflected ray direction.
JEE Advanced
Approach: Draw each mirror, mark normal, apply i=r at every reflection and track image position using symmetry.
Identify mirror geometry and angle between mirrors.
Use N=360/θ−1 when applicable.
For rotation, use reflected ray rotation = 2θ.
For relative motion, resolve velocity perpendicular to mirror.
Solved using multiple reflection geometry.
Advanced Reflection Problem 15. Competitive-level problem on multiple mirrors and reflected ray direction.
JEE Advanced
Approach: Draw each mirror, mark normal, apply i=r at every reflection and track image position using symmetry.
Identify mirror geometry and angle between mirrors.
Use N=360/θ−1 when applicable.
For rotation, use reflected ray rotation = 2θ.
For relative motion, resolve velocity perpendicular to mirror.
Solved using multiple reflection geometry.
Advanced Reflection Problem 16. Competitive-level problem on multiple mirrors and reflected ray direction.
JEE Advanced
Approach: Draw each mirror, mark normal, apply i=r at every reflection and track image position using symmetry.
Identify mirror geometry and angle between mirrors.
Use N=360/θ−1 when applicable.
For rotation, use reflected ray rotation = 2θ.
For relative motion, resolve velocity perpendicular to mirror.
Solved using multiple reflection geometry.
Advanced Reflection Problem 17. Competitive-level problem on multiple mirrors and reflected ray direction.
JEE Advanced
Approach: Draw each mirror, mark normal, apply i=r at every reflection and track image position using symmetry.
Identify mirror geometry and angle between mirrors.
Use N=360/θ−1 when applicable.
For rotation, use reflected ray rotation = 2θ.
For relative motion, resolve velocity perpendicular to mirror.
Solved using multiple reflection geometry.
Advanced Reflection Problem 18. Competitive-level problem on multiple mirrors and reflected ray direction.
JEE Advanced
Approach: Draw each mirror, mark normal, apply i=r at every reflection and track image position using symmetry.
Identify mirror geometry and angle between mirrors.
Use N=360/θ−1 when applicable.
For rotation, use reflected ray rotation = 2θ.
For relative motion, resolve velocity perpendicular to mirror.
Solved using multiple reflection geometry.
Advanced Reflection Problem 19. Competitive-level problem on multiple mirrors and reflected ray direction.
JEE Advanced
Approach: Draw each mirror, mark normal, apply i=r at every reflection and track image position using symmetry.
Identify mirror geometry and angle between mirrors.
Use N=360/θ−1 when applicable.
For rotation, use reflected ray rotation = 2θ.
For relative motion, resolve velocity perpendicular to mirror.
Solved using multiple reflection geometry.
Advanced Reflection Problem 20. Competitive-level problem on multiple mirrors and reflected ray direction.
JEE Advanced
Approach: Draw each mirror, mark normal, apply i=r at every reflection and track image position using symmetry.
Identify mirror geometry and angle between mirrors.
Use N=360/θ−1 when applicable.
For rotation, use reflected ray rotation = 2θ.
For relative motion, resolve velocity perpendicular to mirror.
Solved using multiple reflection geometry.
Section 10: Previous Year Questions
CBSE PYQs
CBSE PYQs 1. State laws of reflection.
CBSE
Incident ray, reflected ray and normal lie in same plane, and angle of incidence equals angle of reflection.
Answer: Incident ray, reflected ray and normal lie in same plane, and angle of incidence equals angle of reflection.
CBSE PYQs 2. Why diffuse reflection helps us see objects?
CBSE
It scatters light in many directions so light can enter the eye from many positions.
Answer: It scatters light in many directions so light can enter the eye from many positions.
CBSE PYQs 3. What is lateral inversion?
CBSE
It is left-right reversal in plane mirror image.
Answer: It is left-right reversal in plane mirror image.
CBSE PYQs 4. Why image in plane mirror is virtual?
CBSE
Reflected rays do not actually meet; they appear to come from behind the mirror.
Answer: Reflected rays do not actually meet; they appear to come from behind the mirror.
CBSE PYQs 5. What happens to reflected ray if mirror rotates by θ?
CBSE
The reflected ray rotates by 2θ.
Answer: The reflected ray rotates by 2θ.
CBSE PYQs 6. Find number of images for two mirrors at 90°.
CBSE
N=360/90−1=3.
Answer: N=360/90−1=3.
CBSE PYQs 7. What is regular reflection?
CBSE
Reflection from smooth surface where parallel rays remain parallel.
Answer: Reflection from smooth surface where parallel rays remain parallel.
CBSE PYQs 8. Do laws of reflection hold for rough surfaces?
CBSE
Yes, at each small surface element the laws are obeyed.
Answer: Yes, at each small surface element the laws are obeyed.
CBSE PYQs 9. Why are mirrors used in telescopes?
CBSE
They can collect and reflect light to form images without chromatic aberration.
Answer: They can collect and reflect light to form images without chromatic aberration.
CBSE PYQs 10. What is magnification of plane mirror?
CBSE
m=+1.
Answer: m=+1.
CBSE PYQs 11. State laws of reflection.
CBSE
Incident ray, reflected ray and normal lie in same plane, and angle of incidence equals angle of reflection.
Answer: Incident ray, reflected ray and normal lie in same plane, and angle of incidence equals angle of reflection.
CBSE PYQs 12. Why diffuse reflection helps us see objects?
CBSE
It scatters light in many directions so light can enter the eye from many positions.
Answer: It scatters light in many directions so light can enter the eye from many positions.
CBSE PYQs 13. What is lateral inversion?
CBSE
It is left-right reversal in plane mirror image.
Answer: It is left-right reversal in plane mirror image.
CBSE PYQs 14. Why image in plane mirror is virtual?
CBSE
Reflected rays do not actually meet; they appear to come from behind the mirror.
Answer: Reflected rays do not actually meet; they appear to come from behind the mirror.
CBSE PYQs 15. What happens to reflected ray if mirror rotates by θ?
CBSE
The reflected ray rotates by 2θ.
Answer: The reflected ray rotates by 2θ.
NEET PYQs
NEET PYQs 1. State laws of reflection.
NEET
Incident ray, reflected ray and normal lie in same plane, and angle of incidence equals angle of reflection.
Answer: Incident ray, reflected ray and normal lie in same plane, and angle of incidence equals angle of reflection.
NEET PYQs 2. Why diffuse reflection helps us see objects?
NEET
It scatters light in many directions so light can enter the eye from many positions.
Answer: It scatters light in many directions so light can enter the eye from many positions.
NEET PYQs 3. What is lateral inversion?
NEET
It is left-right reversal in plane mirror image.
Answer: It is left-right reversal in plane mirror image.
NEET PYQs 4. Why image in plane mirror is virtual?
NEET
Reflected rays do not actually meet; they appear to come from behind the mirror.
Answer: Reflected rays do not actually meet; they appear to come from behind the mirror.
NEET PYQs 5. What happens to reflected ray if mirror rotates by θ?
NEET
The reflected ray rotates by 2θ.
Answer: The reflected ray rotates by 2θ.
NEET PYQs 6. Find number of images for two mirrors at 90°.
NEET
N=360/90−1=3.
Answer: N=360/90−1=3.
NEET PYQs 7. What is regular reflection?
NEET
Reflection from smooth surface where parallel rays remain parallel.
Answer: Reflection from smooth surface where parallel rays remain parallel.
NEET PYQs 8. Do laws of reflection hold for rough surfaces?
NEET
Yes, at each small surface element the laws are obeyed.
Answer: Yes, at each small surface element the laws are obeyed.
NEET PYQs 9. Why are mirrors used in telescopes?
NEET
They can collect and reflect light to form images without chromatic aberration.
Answer: They can collect and reflect light to form images without chromatic aberration.
NEET PYQs 10. What is magnification of plane mirror?
NEET
m=+1.
Answer: m=+1.
NEET PYQs 11. State laws of reflection.
NEET
Incident ray, reflected ray and normal lie in same plane, and angle of incidence equals angle of reflection.
Answer: Incident ray, reflected ray and normal lie in same plane, and angle of incidence equals angle of reflection.
NEET PYQs 12. Why diffuse reflection helps us see objects?
NEET
It scatters light in many directions so light can enter the eye from many positions.
Answer: It scatters light in many directions so light can enter the eye from many positions.
NEET PYQs 13. What is lateral inversion?
NEET
It is left-right reversal in plane mirror image.
Answer: It is left-right reversal in plane mirror image.
NEET PYQs 14. Why image in plane mirror is virtual?
NEET
Reflected rays do not actually meet; they appear to come from behind the mirror.
Answer: Reflected rays do not actually meet; they appear to come from behind the mirror.
NEET PYQs 15. What happens to reflected ray if mirror rotates by θ?
NEET
The reflected ray rotates by 2θ.
Answer: The reflected ray rotates by 2θ.
NEET PYQs 16. Find number of images for two mirrors at 90°.
NEET
N=360/90−1=3.
Answer: N=360/90−1=3.
NEET PYQs 17. What is regular reflection?
NEET
Reflection from smooth surface where parallel rays remain parallel.
Answer: Reflection from smooth surface where parallel rays remain parallel.
NEET PYQs 18. Do laws of reflection hold for rough surfaces?
NEET
Yes, at each small surface element the laws are obeyed.
Answer: Yes, at each small surface element the laws are obeyed.
NEET PYQs 19. Why are mirrors used in telescopes?
NEET
They can collect and reflect light to form images without chromatic aberration.
Answer: They can collect and reflect light to form images without chromatic aberration.
NEET PYQs 20. What is magnification of plane mirror?
NEET
m=+1.
Answer: m=+1.
JEE Main PYQs
JEE Main PYQs 1. State laws of reflection.
JEE Main
Incident ray, reflected ray and normal lie in same plane, and angle of incidence equals angle of reflection.
Answer: Incident ray, reflected ray and normal lie in same plane, and angle of incidence equals angle of reflection.
JEE Main PYQs 2. Why diffuse reflection helps us see objects?
JEE Main
It scatters light in many directions so light can enter the eye from many positions.
Answer: It scatters light in many directions so light can enter the eye from many positions.
JEE Main PYQs 3. What is lateral inversion?
JEE Main
It is left-right reversal in plane mirror image.
Answer: It is left-right reversal in plane mirror image.
JEE Main PYQs 4. Why image in plane mirror is virtual?
JEE Main
Reflected rays do not actually meet; they appear to come from behind the mirror.
Answer: Reflected rays do not actually meet; they appear to come from behind the mirror.
JEE Main PYQs 5. What happens to reflected ray if mirror rotates by θ?
JEE Main
The reflected ray rotates by 2θ.
Answer: The reflected ray rotates by 2θ.
JEE Main PYQs 6. Find number of images for two mirrors at 90°.
JEE Main
N=360/90−1=3.
Answer: N=360/90−1=3.
JEE Main PYQs 7. What is regular reflection?
JEE Main
Reflection from smooth surface where parallel rays remain parallel.
Answer: Reflection from smooth surface where parallel rays remain parallel.
JEE Main PYQs 8. Do laws of reflection hold for rough surfaces?
JEE Main
Yes, at each small surface element the laws are obeyed.
Answer: Yes, at each small surface element the laws are obeyed.
JEE Main PYQs 9. Why are mirrors used in telescopes?
JEE Main
They can collect and reflect light to form images without chromatic aberration.
Answer: They can collect and reflect light to form images without chromatic aberration.
JEE Main PYQs 10. What is magnification of plane mirror?
JEE Main
m=+1.
Answer: m=+1.
JEE Main PYQs 11. State laws of reflection.
JEE Main
Incident ray, reflected ray and normal lie in same plane, and angle of incidence equals angle of reflection.
Answer: Incident ray, reflected ray and normal lie in same plane, and angle of incidence equals angle of reflection.
JEE Main PYQs 12. Why diffuse reflection helps us see objects?
JEE Main
It scatters light in many directions so light can enter the eye from many positions.
Answer: It scatters light in many directions so light can enter the eye from many positions.
JEE Main PYQs 13. What is lateral inversion?
JEE Main
It is left-right reversal in plane mirror image.
Answer: It is left-right reversal in plane mirror image.
JEE Main PYQs 14. Why image in plane mirror is virtual?
JEE Main
Reflected rays do not actually meet; they appear to come from behind the mirror.
Answer: Reflected rays do not actually meet; they appear to come from behind the mirror.
JEE Main PYQs 15. What happens to reflected ray if mirror rotates by θ?
JEE Main
The reflected ray rotates by 2θ.
Answer: The reflected ray rotates by 2θ.
JEE Main PYQs 16. Find number of images for two mirrors at 90°.
JEE Main
N=360/90−1=3.
Answer: N=360/90−1=3.
JEE Main PYQs 17. What is regular reflection?
JEE Main
Reflection from smooth surface where parallel rays remain parallel.
Answer: Reflection from smooth surface where parallel rays remain parallel.
JEE Main PYQs 18. Do laws of reflection hold for rough surfaces?
JEE Main
Yes, at each small surface element the laws are obeyed.
Answer: Yes, at each small surface element the laws are obeyed.
JEE Main PYQs 19. Why are mirrors used in telescopes?
JEE Main
They can collect and reflect light to form images without chromatic aberration.
Answer: They can collect and reflect light to form images without chromatic aberration.
JEE Main PYQs 20. What is magnification of plane mirror?
JEE Main
m=+1.
Answer: m=+1.
JEE Advanced PYQs
JEE Advanced PYQs 1. State laws of reflection.
JEE Advanced
Incident ray, reflected ray and normal lie in same plane, and angle of incidence equals angle of reflection.
Answer: Incident ray, reflected ray and normal lie in same plane, and angle of incidence equals angle of reflection.
JEE Advanced PYQs 2. Why diffuse reflection helps us see objects?
JEE Advanced
It scatters light in many directions so light can enter the eye from many positions.
Answer: It scatters light in many directions so light can enter the eye from many positions.
JEE Advanced PYQs 3. What is lateral inversion?
JEE Advanced
It is left-right reversal in plane mirror image.
Answer: It is left-right reversal in plane mirror image.
JEE Advanced PYQs 4. Why image in plane mirror is virtual?
JEE Advanced
Reflected rays do not actually meet; they appear to come from behind the mirror.
Answer: Reflected rays do not actually meet; they appear to come from behind the mirror.
JEE Advanced PYQs 5. What happens to reflected ray if mirror rotates by θ?
JEE Advanced
The reflected ray rotates by 2θ.
Answer: The reflected ray rotates by 2θ.
JEE Advanced PYQs 6. Find number of images for two mirrors at 90°.
JEE Advanced
N=360/90−1=3.
Answer: N=360/90−1=3.
JEE Advanced PYQs 7. What is regular reflection?
JEE Advanced
Reflection from smooth surface where parallel rays remain parallel.
Answer: Reflection from smooth surface where parallel rays remain parallel.
JEE Advanced PYQs 8. Do laws of reflection hold for rough surfaces?
JEE Advanced
Yes, at each small surface element the laws are obeyed.
Answer: Yes, at each small surface element the laws are obeyed.
JEE Advanced PYQs 9. Why are mirrors used in telescopes?
JEE Advanced
They can collect and reflect light to form images without chromatic aberration.
Answer: They can collect and reflect light to form images without chromatic aberration.
JEE Advanced PYQs 10. What is magnification of plane mirror?
JEE Advanced
m=+1.
Answer: m=+1.
JEE Advanced PYQs 11. State laws of reflection.
JEE Advanced
Incident ray, reflected ray and normal lie in same plane, and angle of incidence equals angle of reflection.
Answer: Incident ray, reflected ray and normal lie in same plane, and angle of incidence equals angle of reflection.
JEE Advanced PYQs 12. Why diffuse reflection helps us see objects?
JEE Advanced
It scatters light in many directions so light can enter the eye from many positions.
Answer: It scatters light in many directions so light can enter the eye from many positions.
JEE Advanced PYQs 13. What is lateral inversion?
JEE Advanced
It is left-right reversal in plane mirror image.
Answer: It is left-right reversal in plane mirror image.
JEE Advanced PYQs 14. Why image in plane mirror is virtual?
JEE Advanced
Reflected rays do not actually meet; they appear to come from behind the mirror.
Answer: Reflected rays do not actually meet; they appear to come from behind the mirror.
JEE Advanced PYQs 15. What happens to reflected ray if mirror rotates by θ?
JEE Advanced
The reflected ray rotates by 2θ.
Answer: The reflected ray rotates by 2θ.
JEE Advanced PYQs 16. Find number of images for two mirrors at 90°.
JEE Advanced
N=360/90−1=3.
Answer: N=360/90−1=3.
JEE Advanced PYQs 17. What is regular reflection?
JEE Advanced
Reflection from smooth surface where parallel rays remain parallel.
Answer: Reflection from smooth surface where parallel rays remain parallel.
JEE Advanced PYQs 18. Do laws of reflection hold for rough surfaces?
JEE Advanced
Yes, at each small surface element the laws are obeyed.
Answer: Yes, at each small surface element the laws are obeyed.
JEE Advanced PYQs 19. Why are mirrors used in telescopes?
JEE Advanced
They can collect and reflect light to form images without chromatic aberration.
Answer: They can collect and reflect light to form images without chromatic aberration.
JEE Advanced PYQs 20. What is magnification of plane mirror?
JEE Advanced
m=+1.
Answer: m=+1.
IB Physics Questions
IB Physics Questions 1. State laws of reflection.
IB Physics Questions
Incident ray, reflected ray and normal lie in same plane, and angle of incidence equals angle of reflection.
Answer: Incident ray, reflected ray and normal lie in same plane, and angle of incidence equals angle of reflection.
IB Physics Questions 2. Why diffuse reflection helps us see objects?
IB Physics Questions
It scatters light in many directions so light can enter the eye from many positions.
Answer: It scatters light in many directions so light can enter the eye from many positions.
IB Physics Questions 3. What is lateral inversion?
IB Physics Questions
It is left-right reversal in plane mirror image.
Answer: It is left-right reversal in plane mirror image.
IB Physics Questions 4. Why image in plane mirror is virtual?
IB Physics Questions
Reflected rays do not actually meet; they appear to come from behind the mirror.
Answer: Reflected rays do not actually meet; they appear to come from behind the mirror.
IB Physics Questions 5. What happens to reflected ray if mirror rotates by θ?
IB Physics Questions
The reflected ray rotates by 2θ.
Answer: The reflected ray rotates by 2θ.
IB Physics Questions 6. Find number of images for two mirrors at 90°.
IB Physics Questions
N=360/90−1=3.
Answer: N=360/90−1=3.
IB Physics Questions 7. What is regular reflection?
IB Physics Questions
Reflection from smooth surface where parallel rays remain parallel.
Answer: Reflection from smooth surface where parallel rays remain parallel.
IB Physics Questions 8. Do laws of reflection hold for rough surfaces?
IB Physics Questions
Yes, at each small surface element the laws are obeyed.
Answer: Yes, at each small surface element the laws are obeyed.
IB Physics Questions 9. Why are mirrors used in telescopes?
IB Physics Questions
They can collect and reflect light to form images without chromatic aberration.
Answer: They can collect and reflect light to form images without chromatic aberration.
IB Physics Questions 10. What is magnification of plane mirror?
IB Physics Questions
m=+1.
Answer: m=+1.
IB Physics Questions 11. State laws of reflection.
IB Physics Questions
Incident ray, reflected ray and normal lie in same plane, and angle of incidence equals angle of reflection.
Answer: Incident ray, reflected ray and normal lie in same plane, and angle of incidence equals angle of reflection.
IB Physics Questions 12. Why diffuse reflection helps us see objects?
IB Physics Questions
It scatters light in many directions so light can enter the eye from many positions.
Answer: It scatters light in many directions so light can enter the eye from many positions.
IB Physics Questions 13. What is lateral inversion?
IB Physics Questions
It is left-right reversal in plane mirror image.
Answer: It is left-right reversal in plane mirror image.
IB Physics Questions 14. Why image in plane mirror is virtual?
IB Physics Questions
Reflected rays do not actually meet; they appear to come from behind the mirror.
Answer: Reflected rays do not actually meet; they appear to come from behind the mirror.
IB Physics Questions 15. What happens to reflected ray if mirror rotates by θ?
IB Physics Questions
The reflected ray rotates by 2θ.
Answer: The reflected ray rotates by 2θ.
IGCSE Physics Questions
IGCSE Physics Questions 1. State laws of reflection.
IGCSE Physics Questions
Incident ray, reflected ray and normal lie in same plane, and angle of incidence equals angle of reflection.
Answer: Incident ray, reflected ray and normal lie in same plane, and angle of incidence equals angle of reflection.
IGCSE Physics Questions 2. Why diffuse reflection helps us see objects?
IGCSE Physics Questions
It scatters light in many directions so light can enter the eye from many positions.
Answer: It scatters light in many directions so light can enter the eye from many positions.
IGCSE Physics Questions 3. What is lateral inversion?
IGCSE Physics Questions
It is left-right reversal in plane mirror image.
Answer: It is left-right reversal in plane mirror image.
IGCSE Physics Questions 4. Why image in plane mirror is virtual?
IGCSE Physics Questions
Reflected rays do not actually meet; they appear to come from behind the mirror.
Answer: Reflected rays do not actually meet; they appear to come from behind the mirror.
IGCSE Physics Questions 5. What happens to reflected ray if mirror rotates by θ?
IGCSE Physics Questions
The reflected ray rotates by 2θ.
Answer: The reflected ray rotates by 2θ.
IGCSE Physics Questions 6. Find number of images for two mirrors at 90°.
IGCSE Physics Questions
N=360/90−1=3.
Answer: N=360/90−1=3.
IGCSE Physics Questions 7. What is regular reflection?
IGCSE Physics Questions
Reflection from smooth surface where parallel rays remain parallel.
Answer: Reflection from smooth surface where parallel rays remain parallel.
IGCSE Physics Questions 8. Do laws of reflection hold for rough surfaces?
IGCSE Physics Questions
Yes, at each small surface element the laws are obeyed.
Answer: Yes, at each small surface element the laws are obeyed.
IGCSE Physics Questions 9. Why are mirrors used in telescopes?
IGCSE Physics Questions
They can collect and reflect light to form images without chromatic aberration.
Answer: They can collect and reflect light to form images without chromatic aberration.
IGCSE Physics Questions 10. What is magnification of plane mirror?
IGCSE Physics Questions
m=+1.
Answer: m=+1.
IGCSE Physics Questions 11. State laws of reflection.
IGCSE Physics Questions
Incident ray, reflected ray and normal lie in same plane, and angle of incidence equals angle of reflection.
Answer: Incident ray, reflected ray and normal lie in same plane, and angle of incidence equals angle of reflection.
IGCSE Physics Questions 12. Why diffuse reflection helps us see objects?
IGCSE Physics Questions
It scatters light in many directions so light can enter the eye from many positions.
Answer: It scatters light in many directions so light can enter the eye from many positions.
IGCSE Physics Questions 13. What is lateral inversion?
IGCSE Physics Questions
It is left-right reversal in plane mirror image.
Answer: It is left-right reversal in plane mirror image.
IGCSE Physics Questions 14. Why image in plane mirror is virtual?
IGCSE Physics Questions
Reflected rays do not actually meet; they appear to come from behind the mirror.
Answer: Reflected rays do not actually meet; they appear to come from behind the mirror.
IGCSE Physics Questions 15. What happens to reflected ray if mirror rotates by θ?
IGCSE Physics Questions
The reflected ray rotates by 2θ.
Answer: The reflected ray rotates by 2θ.
A-Level Physics Questions
A-Level Physics Questions 1. State laws of reflection.
A-Level Physics Questions
Incident ray, reflected ray and normal lie in same plane, and angle of incidence equals angle of reflection.
Answer: Incident ray, reflected ray and normal lie in same plane, and angle of incidence equals angle of reflection.
A-Level Physics Questions 2. Why diffuse reflection helps us see objects?
A-Level Physics Questions
It scatters light in many directions so light can enter the eye from many positions.
Answer: It scatters light in many directions so light can enter the eye from many positions.
A-Level Physics Questions 3. What is lateral inversion?
A-Level Physics Questions
It is left-right reversal in plane mirror image.
Answer: It is left-right reversal in plane mirror image.
A-Level Physics Questions 4. Why image in plane mirror is virtual?
A-Level Physics Questions
Reflected rays do not actually meet; they appear to come from behind the mirror.
Answer: Reflected rays do not actually meet; they appear to come from behind the mirror.
A-Level Physics Questions 5. What happens to reflected ray if mirror rotates by θ?
A-Level Physics Questions
The reflected ray rotates by 2θ.
Answer: The reflected ray rotates by 2θ.
A-Level Physics Questions 6. Find number of images for two mirrors at 90°.
A-Level Physics Questions
N=360/90−1=3.
Answer: N=360/90−1=3.
A-Level Physics Questions 7. What is regular reflection?
A-Level Physics Questions
Reflection from smooth surface where parallel rays remain parallel.
Answer: Reflection from smooth surface where parallel rays remain parallel.
A-Level Physics Questions 8. Do laws of reflection hold for rough surfaces?
A-Level Physics Questions
Yes, at each small surface element the laws are obeyed.
Answer: Yes, at each small surface element the laws are obeyed.
A-Level Physics Questions 9. Why are mirrors used in telescopes?
A-Level Physics Questions
They can collect and reflect light to form images without chromatic aberration.
Answer: They can collect and reflect light to form images without chromatic aberration.
A-Level Physics Questions 10. What is magnification of plane mirror?
A-Level Physics Questions
m=+1.
Answer: m=+1.
A-Level Physics Questions 11. State laws of reflection.
A-Level Physics Questions
Incident ray, reflected ray and normal lie in same plane, and angle of incidence equals angle of reflection.
Answer: Incident ray, reflected ray and normal lie in same plane, and angle of incidence equals angle of reflection.
A-Level Physics Questions 12. Why diffuse reflection helps us see objects?
A-Level Physics Questions
It scatters light in many directions so light can enter the eye from many positions.
Answer: It scatters light in many directions so light can enter the eye from many positions.
A-Level Physics Questions 13. What is lateral inversion?
A-Level Physics Questions
It is left-right reversal in plane mirror image.
Answer: It is left-right reversal in plane mirror image.
A-Level Physics Questions 14. Why image in plane mirror is virtual?
A-Level Physics Questions
Reflected rays do not actually meet; they appear to come from behind the mirror.
Answer: Reflected rays do not actually meet; they appear to come from behind the mirror.
A-Level Physics Questions 15. What happens to reflected ray if mirror rotates by θ?
A-Level Physics Questions
The reflected ray rotates by 2θ.
Answer: The reflected ray rotates by 2θ.
Section 11: Conceptual Questions
Conceptual Question 1. Why can we see objects?
Conceptual
The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Answer: The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Conceptual Question 2. Why does diffuse reflection occur?
Conceptual
The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Answer: The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Conceptual Question 3. Why is image laterally inverted?
Conceptual
The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Answer: The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Conceptual Question 4. Why does mirror obey reflection laws?
Conceptual
The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Answer: The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Conceptual Question 5. Why are mirrors used in telescopes?
Conceptual
The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Answer: The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Conceptual Question 6. Why is plane mirror image virtual?
Conceptual
The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Answer: The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Conceptual Question 7. Why are angles measured from normal?
Conceptual
The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Answer: The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Conceptual Question 8. Why polished surfaces form clear images?
Conceptual
The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Answer: The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Conceptual Question 9. Why rough surfaces do not form clear images?
Conceptual
The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Answer: The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Conceptual Question 10. Why reflected ray rotates by twice mirror rotation?
Conceptual
The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Answer: The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Conceptual Question 11. Why can we see objects?
Conceptual
The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Answer: The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Conceptual Question 12. Why does diffuse reflection occur?
Conceptual
The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Answer: The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Conceptual Question 13. Why is image laterally inverted?
Conceptual
The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Answer: The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Conceptual Question 14. Why does mirror obey reflection laws?
Conceptual
The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Answer: The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Conceptual Question 15. Why are mirrors used in telescopes?
Conceptual
The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Answer: The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Conceptual Question 16. Why is plane mirror image virtual?
Conceptual
The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Answer: The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Conceptual Question 17. Why are angles measured from normal?
Conceptual
The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Answer: The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Conceptual Question 18. Why polished surfaces form clear images?
Conceptual
The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Answer: The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Conceptual Question 19. Why rough surfaces do not form clear images?
Conceptual
The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Answer: The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Conceptual Question 20. Why reflected ray rotates by twice mirror rotation?
Conceptual
The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Answer: The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Conceptual Question 21. Why can we see objects?
Conceptual
The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Answer: The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Conceptual Question 22. Why does diffuse reflection occur?
Conceptual
The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Answer: The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Conceptual Question 23. Why is image laterally inverted?
Conceptual
The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Answer: The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Conceptual Question 24. Why does mirror obey reflection laws?
Conceptual
The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Answer: The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Conceptual Question 25. Why are mirrors used in telescopes?
Conceptual
The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Answer: The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Conceptual Question 26. Why is plane mirror image virtual?
Conceptual
The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Answer: The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Conceptual Question 27. Why are angles measured from normal?
Conceptual
The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Answer: The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Conceptual Question 28. Why polished surfaces form clear images?
Conceptual
The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Answer: The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Conceptual Question 29. Why rough surfaces do not form clear images?
Conceptual
The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Answer: The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Conceptual Question 30. Why reflected ray rotates by twice mirror rotation?
Conceptual
The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Answer: The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Conceptual Question 31. Why can we see objects?
Conceptual
The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Answer: The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Conceptual Question 32. Why does diffuse reflection occur?
Conceptual
The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Answer: The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Conceptual Question 33. Why is image laterally inverted?
Conceptual
The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Answer: The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Conceptual Question 34. Why does mirror obey reflection laws?
Conceptual
The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Answer: The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Conceptual Question 35. Why are mirrors used in telescopes?
Conceptual
The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Answer: The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Conceptual Question 36. Why is plane mirror image virtual?
Conceptual
The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Answer: The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Conceptual Question 37. Why are angles measured from normal?
Conceptual
The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Answer: The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Conceptual Question 38. Why polished surfaces form clear images?
Conceptual
The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Answer: The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Conceptual Question 39. Why rough surfaces do not form clear images?
Conceptual
The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Answer: The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Conceptual Question 40. Why reflected ray rotates by twice mirror rotation?
Conceptual
The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Answer: The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Conceptual Question 41. Why can we see objects?
Conceptual
The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Answer: The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Conceptual Question 42. Why does diffuse reflection occur?
Conceptual
The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Answer: The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Conceptual Question 43. Why is image laterally inverted?
Conceptual
The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Answer: The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Conceptual Question 44. Why does mirror obey reflection laws?
Conceptual
The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Answer: The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Conceptual Question 45. Why are mirrors used in telescopes?
Conceptual
The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Answer: The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Conceptual Question 46. Why is plane mirror image virtual?
Conceptual
The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Answer: The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Conceptual Question 47. Why are angles measured from normal?
Conceptual
The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Answer: The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Conceptual Question 48. Why polished surfaces form clear images?
Conceptual
The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Answer: The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Conceptual Question 49. Why rough surfaces do not form clear images?
Conceptual
The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Answer: The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Conceptual Question 50. Why reflected ray rotates by twice mirror rotation?
Conceptual
The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Answer: The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Conceptual Question 51. Why can we see objects?
Conceptual
The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Answer: The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Conceptual Question 52. Why does diffuse reflection occur?
Conceptual
The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Answer: The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Conceptual Question 53. Why is image laterally inverted?
Conceptual
The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Answer: The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Conceptual Question 54. Why does mirror obey reflection laws?
Conceptual
The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Answer: The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Conceptual Question 55. Why are mirrors used in telescopes?
Conceptual
The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Answer: The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Conceptual Question 56. Why is plane mirror image virtual?
Conceptual
The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Answer: The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Conceptual Question 57. Why are angles measured from normal?
Conceptual
The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Answer: The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Conceptual Question 58. Why polished surfaces form clear images?
Conceptual
The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Answer: The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Conceptual Question 59. Why rough surfaces do not form clear images?
Conceptual
The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Answer: The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Conceptual Question 60. Why reflected ray rotates by twice mirror rotation?
Conceptual
The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Answer: The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Conceptual Question 61. Why can we see objects?
Conceptual
The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Answer: The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Conceptual Question 62. Why does diffuse reflection occur?
Conceptual
The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Answer: The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Conceptual Question 63. Why is image laterally inverted?
Conceptual
The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Answer: The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Conceptual Question 64. Why does mirror obey reflection laws?
Conceptual
The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Answer: The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Conceptual Question 65. Why are mirrors used in telescopes?
Conceptual
The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Answer: The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Conceptual Question 66. Why is plane mirror image virtual?
Conceptual
The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Answer: The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Conceptual Question 67. Why are angles measured from normal?
Conceptual
The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Answer: The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Conceptual Question 68. Why polished surfaces form clear images?
Conceptual
The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Answer: The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Conceptual Question 69. Why rough surfaces do not form clear images?
Conceptual
The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Answer: The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Conceptual Question 70. Why reflected ray rotates by twice mirror rotation?
Conceptual
The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Answer: The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Conceptual Question 71. Why can we see objects?
Conceptual
The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Answer: The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Conceptual Question 72. Why does diffuse reflection occur?
Conceptual
The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Answer: The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Conceptual Question 73. Why is image laterally inverted?
Conceptual
The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Answer: The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Conceptual Question 74. Why does mirror obey reflection laws?
Conceptual
The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Answer: The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Conceptual Question 75. Why are mirrors used in telescopes?
Conceptual
The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Answer: The concept is explained using the laws of reflection, surface normal, ray geometry and the difference between regular and diffuse reflection. In exams, draw the normal and apply i=r clearly.
Section 12: Assertion Reason Questions
Assertion-Reason 1. Assertion: Angle of incidence equals angle of reflection. Reason: Reflection is symmetric about the normal.
Assertion-Reason
Answer: Both are true and Reason correctly explains Assertion.
Explanation: Check truth of both statements, then test whether reason explains assertion through reflection geometry.
Both are true and Reason correctly explains Assertion.
Assertion-Reason 2. Assertion: Diffuse reflection does not obey reflection laws. Reason: Rough surfaces have many normals.
Assertion-Reason
Answer: Assertion is false; laws are obeyed locally at every small surface element.
Explanation: Check truth of both statements, then test whether reason explains assertion through reflection geometry.
Assertion is false; laws are obeyed locally at every small surface element.
Assertion-Reason 3. Assertion: Plane mirror image is virtual. Reason: Reflected rays appear to meet behind the mirror.
Assertion-Reason
Answer: Both are true and Reason correctly explains Assertion.
Explanation: Check truth of both statements, then test whether reason explains assertion through reflection geometry.
Both are true and Reason correctly explains Assertion.
Assertion-Reason 4. Assertion: Rotating mirror by θ rotates reflected ray by 2θ. Reason: Normal rotates by θ and reflection is symmetric.
Assertion-Reason
Answer: Both are true and Reason correctly explains Assertion.
Explanation: Check truth of both statements, then test whether reason explains assertion through reflection geometry.
Both are true and Reason correctly explains Assertion.
Assertion-Reason 5. Assertion: Plane mirror magnification is +1. Reason: Image is erect and same size as object.
Assertion-Reason
Answer: Both are true and Reason correctly explains Assertion.
Explanation: Check truth of both statements, then test whether reason explains assertion through reflection geometry.
Both are true and Reason correctly explains Assertion.
Assertion-Reason 6. Assertion: Angle of incidence equals angle of reflection. Reason: Reflection is symmetric about the normal.
Assertion-Reason
Answer: Both are true and Reason correctly explains Assertion.
Explanation: Check truth of both statements, then test whether reason explains assertion through reflection geometry.
Both are true and Reason correctly explains Assertion.
Assertion-Reason 7. Assertion: Diffuse reflection does not obey reflection laws. Reason: Rough surfaces have many normals.
Assertion-Reason
Answer: Assertion is false; laws are obeyed locally at every small surface element.
Explanation: Check truth of both statements, then test whether reason explains assertion through reflection geometry.
Assertion is false; laws are obeyed locally at every small surface element.
Assertion-Reason 8. Assertion: Plane mirror image is virtual. Reason: Reflected rays appear to meet behind the mirror.
Assertion-Reason
Answer: Both are true and Reason correctly explains Assertion.
Explanation: Check truth of both statements, then test whether reason explains assertion through reflection geometry.
Both are true and Reason correctly explains Assertion.
Assertion-Reason 9. Assertion: Rotating mirror by θ rotates reflected ray by 2θ. Reason: Normal rotates by θ and reflection is symmetric.
Assertion-Reason
Answer: Both are true and Reason correctly explains Assertion.
Explanation: Check truth of both statements, then test whether reason explains assertion through reflection geometry.
Both are true and Reason correctly explains Assertion.
Assertion-Reason 10. Assertion: Plane mirror magnification is +1. Reason: Image is erect and same size as object.
Assertion-Reason
Answer: Both are true and Reason correctly explains Assertion.
Explanation: Check truth of both statements, then test whether reason explains assertion through reflection geometry.
Both are true and Reason correctly explains Assertion.
Assertion-Reason 11. Assertion: Angle of incidence equals angle of reflection. Reason: Reflection is symmetric about the normal.
Assertion-Reason
Answer: Both are true and Reason correctly explains Assertion.
Explanation: Check truth of both statements, then test whether reason explains assertion through reflection geometry.
Both are true and Reason correctly explains Assertion.
Assertion-Reason 12. Assertion: Diffuse reflection does not obey reflection laws. Reason: Rough surfaces have many normals.
Assertion-Reason
Answer: Assertion is false; laws are obeyed locally at every small surface element.
Explanation: Check truth of both statements, then test whether reason explains assertion through reflection geometry.
Assertion is false; laws are obeyed locally at every small surface element.
Assertion-Reason 13. Assertion: Plane mirror image is virtual. Reason: Reflected rays appear to meet behind the mirror.
Assertion-Reason
Answer: Both are true and Reason correctly explains Assertion.
Explanation: Check truth of both statements, then test whether reason explains assertion through reflection geometry.
Both are true and Reason correctly explains Assertion.
Assertion-Reason 14. Assertion: Rotating mirror by θ rotates reflected ray by 2θ. Reason: Normal rotates by θ and reflection is symmetric.
Assertion-Reason
Answer: Both are true and Reason correctly explains Assertion.
Explanation: Check truth of both statements, then test whether reason explains assertion through reflection geometry.
Both are true and Reason correctly explains Assertion.
Assertion-Reason 15. Assertion: Plane mirror magnification is +1. Reason: Image is erect and same size as object.
Assertion-Reason
Answer: Both are true and Reason correctly explains Assertion.
Explanation: Check truth of both statements, then test whether reason explains assertion through reflection geometry.
Both are true and Reason correctly explains Assertion.
Assertion-Reason 16. Assertion: Angle of incidence equals angle of reflection. Reason: Reflection is symmetric about the normal.
Assertion-Reason
Answer: Both are true and Reason correctly explains Assertion.
Explanation: Check truth of both statements, then test whether reason explains assertion through reflection geometry.
Both are true and Reason correctly explains Assertion.
Assertion-Reason 17. Assertion: Diffuse reflection does not obey reflection laws. Reason: Rough surfaces have many normals.
Assertion-Reason
Answer: Assertion is false; laws are obeyed locally at every small surface element.
Explanation: Check truth of both statements, then test whether reason explains assertion through reflection geometry.
Assertion is false; laws are obeyed locally at every small surface element.
Assertion-Reason 18. Assertion: Plane mirror image is virtual. Reason: Reflected rays appear to meet behind the mirror.
Assertion-Reason
Answer: Both are true and Reason correctly explains Assertion.
Explanation: Check truth of both statements, then test whether reason explains assertion through reflection geometry.
Both are true and Reason correctly explains Assertion.
Assertion-Reason 19. Assertion: Rotating mirror by θ rotates reflected ray by 2θ. Reason: Normal rotates by θ and reflection is symmetric.
Assertion-Reason
Answer: Both are true and Reason correctly explains Assertion.
Explanation: Check truth of both statements, then test whether reason explains assertion through reflection geometry.
Both are true and Reason correctly explains Assertion.
Assertion-Reason 20. Assertion: Plane mirror magnification is +1. Reason: Image is erect and same size as object.
Assertion-Reason
Answer: Both are true and Reason correctly explains Assertion.
Explanation: Check truth of both statements, then test whether reason explains assertion through reflection geometry.
Both are true and Reason correctly explains Assertion.
Assertion-Reason 21. Assertion: Angle of incidence equals angle of reflection. Reason: Reflection is symmetric about the normal.
Assertion-Reason
Answer: Both are true and Reason correctly explains Assertion.
Explanation: Check truth of both statements, then test whether reason explains assertion through reflection geometry.
Both are true and Reason correctly explains Assertion.
Assertion-Reason 22. Assertion: Diffuse reflection does not obey reflection laws. Reason: Rough surfaces have many normals.
Assertion-Reason
Answer: Assertion is false; laws are obeyed locally at every small surface element.
Explanation: Check truth of both statements, then test whether reason explains assertion through reflection geometry.
Assertion is false; laws are obeyed locally at every small surface element.
Assertion-Reason 23. Assertion: Plane mirror image is virtual. Reason: Reflected rays appear to meet behind the mirror.
Assertion-Reason
Answer: Both are true and Reason correctly explains Assertion.
Explanation: Check truth of both statements, then test whether reason explains assertion through reflection geometry.
Both are true and Reason correctly explains Assertion.
Assertion-Reason 24. Assertion: Rotating mirror by θ rotates reflected ray by 2θ. Reason: Normal rotates by θ and reflection is symmetric.
Assertion-Reason
Answer: Both are true and Reason correctly explains Assertion.
Explanation: Check truth of both statements, then test whether reason explains assertion through reflection geometry.
Both are true and Reason correctly explains Assertion.
Assertion-Reason 25. Assertion: Plane mirror magnification is +1. Reason: Image is erect and same size as object.
Assertion-Reason
Answer: Both are true and Reason correctly explains Assertion.
Explanation: Check truth of both statements, then test whether reason explains assertion through reflection geometry.
Both are true and Reason correctly explains Assertion.
Section 13: Case Study Questions
Case Study 1: Rear-view mirror
Scenario: Rear-view mirror uses reflection of light. The observed behaviour depends on whether reflection is regular or diffuse, how the mirror is placed and how rays obey the law i=r.
Representative reflection diagram for Rear-view mirror.
Case 1.1. Explain reflection point 1 for Rear-view mirror.
Case Study
Rear-view mirror can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Answer: Rear-view mirror can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Case 1.2. Explain reflection point 2 for Rear-view mirror.
Case Study
Rear-view mirror can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Answer: Rear-view mirror can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Case 1.3. Explain reflection point 3 for Rear-view mirror.
Case Study
Rear-view mirror can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Answer: Rear-view mirror can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Case 1.4. Explain reflection point 4 for Rear-view mirror.
Case Study
Rear-view mirror can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Answer: Rear-view mirror can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Case 1.5. Explain reflection point 5 for Rear-view mirror.
Case Study
Rear-view mirror can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Answer: Rear-view mirror can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Case Study 2: Periscope
Scenario: Periscope uses reflection of light. The observed behaviour depends on whether reflection is regular or diffuse, how the mirror is placed and how rays obey the law i=r.
Representative reflection diagram for Periscope.
Case 2.1. Explain reflection point 1 for Periscope.
Case Study
Periscope can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Answer: Periscope can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Case 2.2. Explain reflection point 2 for Periscope.
Case Study
Periscope can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Answer: Periscope can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Case 2.3. Explain reflection point 3 for Periscope.
Case Study
Periscope can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Answer: Periscope can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Case 2.4. Explain reflection point 4 for Periscope.
Case Study
Periscope can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Answer: Periscope can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Case 2.5. Explain reflection point 5 for Periscope.
Case Study
Periscope can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Answer: Periscope can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Case Study 3: Kaleidoscope
Scenario: Kaleidoscope uses reflection of light. The observed behaviour depends on whether reflection is regular or diffuse, how the mirror is placed and how rays obey the law i=r.
Representative reflection diagram for Kaleidoscope.
Case 3.1. Explain reflection point 1 for Kaleidoscope.
Case Study
Kaleidoscope can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Answer: Kaleidoscope can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Case 3.2. Explain reflection point 2 for Kaleidoscope.
Case Study
Kaleidoscope can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Answer: Kaleidoscope can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Case 3.3. Explain reflection point 3 for Kaleidoscope.
Case Study
Kaleidoscope can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Answer: Kaleidoscope can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Case 3.4. Explain reflection point 4 for Kaleidoscope.
Case Study
Kaleidoscope can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Answer: Kaleidoscope can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Case 3.5. Explain reflection point 5 for Kaleidoscope.
Case Study
Kaleidoscope can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Answer: Kaleidoscope can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Case Study 4: Dressing mirror
Scenario: Dressing mirror uses reflection of light. The observed behaviour depends on whether reflection is regular or diffuse, how the mirror is placed and how rays obey the law i=r.
Representative reflection diagram for Dressing mirror.
Case 4.1. Explain reflection point 1 for Dressing mirror.
Case Study
Dressing mirror can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Answer: Dressing mirror can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Case 4.2. Explain reflection point 2 for Dressing mirror.
Case Study
Dressing mirror can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Answer: Dressing mirror can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Case 4.3. Explain reflection point 3 for Dressing mirror.
Case Study
Dressing mirror can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Answer: Dressing mirror can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Case 4.4. Explain reflection point 4 for Dressing mirror.
Case Study
Dressing mirror can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Answer: Dressing mirror can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Case 4.5. Explain reflection point 5 for Dressing mirror.
Case Study
Dressing mirror can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Answer: Dressing mirror can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Case Study 5: Road safety
Scenario: Road safety uses reflection of light. The observed behaviour depends on whether reflection is regular or diffuse, how the mirror is placed and how rays obey the law i=r.
Representative reflection diagram for Road safety.
Case 5.1. Explain reflection point 1 for Road safety.
Case Study
Road safety can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Answer: Road safety can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Case 5.2. Explain reflection point 2 for Road safety.
Case Study
Road safety can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Answer: Road safety can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Case 5.3. Explain reflection point 3 for Road safety.
Case Study
Road safety can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Answer: Road safety can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Case 5.4. Explain reflection point 4 for Road safety.
Case Study
Road safety can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Answer: Road safety can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Case 5.5. Explain reflection point 5 for Road safety.
Case Study
Road safety can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Answer: Road safety can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Case Study 6: Laser reflection
Scenario: Laser reflection uses reflection of light. The observed behaviour depends on whether reflection is regular or diffuse, how the mirror is placed and how rays obey the law i=r.
Representative reflection diagram for Laser reflection.
Case 6.1. Explain reflection point 1 for Laser reflection.
Case Study
Laser reflection can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Answer: Laser reflection can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Case 6.2. Explain reflection point 2 for Laser reflection.
Case Study
Laser reflection can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Answer: Laser reflection can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Case 6.3. Explain reflection point 3 for Laser reflection.
Case Study
Laser reflection can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Answer: Laser reflection can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Case 6.4. Explain reflection point 4 for Laser reflection.
Case Study
Laser reflection can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Answer: Laser reflection can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Case 6.5. Explain reflection point 5 for Laser reflection.
Case Study
Laser reflection can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Answer: Laser reflection can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Case Study 7: Solar concentrators
Scenario: Solar concentrators uses reflection of light. The observed behaviour depends on whether reflection is regular or diffuse, how the mirror is placed and how rays obey the law i=r.
Representative reflection diagram for Solar concentrators.
Case 7.1. Explain reflection point 1 for Solar concentrators.
Case Study
Solar concentrators can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Answer: Solar concentrators can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Case 7.2. Explain reflection point 2 for Solar concentrators.
Case Study
Solar concentrators can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Answer: Solar concentrators can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Case 7.3. Explain reflection point 3 for Solar concentrators.
Case Study
Solar concentrators can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Answer: Solar concentrators can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Case 7.4. Explain reflection point 4 for Solar concentrators.
Case Study
Solar concentrators can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Answer: Solar concentrators can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Case 7.5. Explain reflection point 5 for Solar concentrators.
Case Study
Solar concentrators can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Answer: Solar concentrators can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Case Study 8: Security mirrors
Scenario: Security mirrors uses reflection of light. The observed behaviour depends on whether reflection is regular or diffuse, how the mirror is placed and how rays obey the law i=r.
Representative reflection diagram for Security mirrors.
Case 8.1. Explain reflection point 1 for Security mirrors.
Case Study
Security mirrors can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Answer: Security mirrors can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Case 8.2. Explain reflection point 2 for Security mirrors.
Case Study
Security mirrors can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Answer: Security mirrors can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Case 8.3. Explain reflection point 3 for Security mirrors.
Case Study
Security mirrors can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Answer: Security mirrors can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Case 8.4. Explain reflection point 4 for Security mirrors.
Case Study
Security mirrors can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Answer: Security mirrors can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Case 8.5. Explain reflection point 5 for Security mirrors.
Case Study
Security mirrors can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Answer: Security mirrors can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Case Study 9: Optical communication
Scenario: Optical communication uses reflection of light. The observed behaviour depends on whether reflection is regular or diffuse, how the mirror is placed and how rays obey the law i=r.
Representative reflection diagram for Optical communication.
Case 9.1. Explain reflection point 1 for Optical communication.
Case Study
Optical communication can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Answer: Optical communication can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Case 9.2. Explain reflection point 2 for Optical communication.
Case Study
Optical communication can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Answer: Optical communication can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Case 9.3. Explain reflection point 3 for Optical communication.
Case Study
Optical communication can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Answer: Optical communication can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Case 9.4. Explain reflection point 4 for Optical communication.
Case Study
Optical communication can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Answer: Optical communication can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Case 9.5. Explain reflection point 5 for Optical communication.
Case Study
Optical communication can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Answer: Optical communication can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Case Study 10: Plane mirror in salon
Scenario: Plane mirror in salon uses reflection of light. The observed behaviour depends on whether reflection is regular or diffuse, how the mirror is placed and how rays obey the law i=r.
Representative reflection diagram for Plane mirror in salon.
Case 10.1. Explain reflection point 1 for Plane mirror in salon.
Case Study
Plane mirror in salon can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Answer: Plane mirror in salon can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Case 10.2. Explain reflection point 2 for Plane mirror in salon.
Case Study
Plane mirror in salon can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Answer: Plane mirror in salon can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Case 10.3. Explain reflection point 3 for Plane mirror in salon.
Case Study
Plane mirror in salon can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Answer: Plane mirror in salon can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Case 10.4. Explain reflection point 4 for Plane mirror in salon.
Case Study
Plane mirror in salon can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Answer: Plane mirror in salon can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Case 10.5. Explain reflection point 5 for Plane mirror in salon.
Case Study
Plane mirror in salon can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Answer: Plane mirror in salon can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Case Study 11: Multiple mirror setup
Scenario: Multiple mirror setup uses reflection of light. The observed behaviour depends on whether reflection is regular or diffuse, how the mirror is placed and how rays obey the law i=r.
Representative reflection diagram for Multiple mirror setup.
Case 11.1. Explain reflection point 1 for Multiple mirror setup.
Case Study
Multiple mirror setup can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Answer: Multiple mirror setup can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Case 11.2. Explain reflection point 2 for Multiple mirror setup.
Case Study
Multiple mirror setup can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Answer: Multiple mirror setup can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Case 11.3. Explain reflection point 3 for Multiple mirror setup.
Case Study
Multiple mirror setup can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Answer: Multiple mirror setup can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Case 11.4. Explain reflection point 4 for Multiple mirror setup.
Case Study
Multiple mirror setup can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Answer: Multiple mirror setup can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Case 11.5. Explain reflection point 5 for Multiple mirror setup.
Case Study
Multiple mirror setup can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Answer: Multiple mirror setup can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Case Study 12: Reflecting telescope
Scenario: Reflecting telescope uses reflection of light. The observed behaviour depends on whether reflection is regular or diffuse, how the mirror is placed and how rays obey the law i=r.
Representative reflection diagram for Reflecting telescope.
Case 12.1. Explain reflection point 1 for Reflecting telescope.
Case Study
Reflecting telescope can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Answer: Reflecting telescope can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Case 12.2. Explain reflection point 2 for Reflecting telescope.
Case Study
Reflecting telescope can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Answer: Reflecting telescope can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Case 12.3. Explain reflection point 3 for Reflecting telescope.
Case Study
Reflecting telescope can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Answer: Reflecting telescope can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Case 12.4. Explain reflection point 4 for Reflecting telescope.
Case Study
Reflecting telescope can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Answer: Reflecting telescope can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Case 12.5. Explain reflection point 5 for Reflecting telescope.
Case Study
Reflecting telescope can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Answer: Reflecting telescope can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Case Study 13: Sign board visibility
Scenario: Sign board visibility uses reflection of light. The observed behaviour depends on whether reflection is regular or diffuse, how the mirror is placed and how rays obey the law i=r.
Representative reflection diagram for Sign board visibility.
Case 13.1. Explain reflection point 1 for Sign board visibility.
Case Study
Sign board visibility can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Answer: Sign board visibility can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Case 13.2. Explain reflection point 2 for Sign board visibility.
Case Study
Sign board visibility can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Answer: Sign board visibility can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Case 13.3. Explain reflection point 3 for Sign board visibility.
Case Study
Sign board visibility can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Answer: Sign board visibility can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Case 13.4. Explain reflection point 4 for Sign board visibility.
Case Study
Sign board visibility can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Answer: Sign board visibility can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Case 13.5. Explain reflection point 5 for Sign board visibility.
Case Study
Sign board visibility can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Answer: Sign board visibility can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Case Study 14: Reflective jacket
Scenario: Reflective jacket uses reflection of light. The observed behaviour depends on whether reflection is regular or diffuse, how the mirror is placed and how rays obey the law i=r.
Representative reflection diagram for Reflective jacket.
Case 14.1. Explain reflection point 1 for Reflective jacket.
Case Study
Reflective jacket can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Answer: Reflective jacket can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Case 14.2. Explain reflection point 2 for Reflective jacket.
Case Study
Reflective jacket can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Answer: Reflective jacket can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Case 14.3. Explain reflection point 3 for Reflective jacket.
Case Study
Reflective jacket can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Answer: Reflective jacket can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Case 14.4. Explain reflection point 4 for Reflective jacket.
Case Study
Reflective jacket can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Answer: Reflective jacket can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Case 14.5. Explain reflection point 5 for Reflective jacket.
Case Study
Reflective jacket can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Answer: Reflective jacket can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Case Study 15: Bathroom mirror
Scenario: Bathroom mirror uses reflection of light. The observed behaviour depends on whether reflection is regular or diffuse, how the mirror is placed and how rays obey the law i=r.
Representative reflection diagram for Bathroom mirror.
Case 15.1. Explain reflection point 1 for Bathroom mirror.
Case Study
Bathroom mirror can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Answer: Bathroom mirror can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Case 15.2. Explain reflection point 2 for Bathroom mirror.
Case Study
Bathroom mirror can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Answer: Bathroom mirror can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Case 15.3. Explain reflection point 3 for Bathroom mirror.
Case Study
Bathroom mirror can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Answer: Bathroom mirror can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Case 15.4. Explain reflection point 4 for Bathroom mirror.
Case Study
Bathroom mirror can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Answer: Bathroom mirror can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Case 15.5. Explain reflection point 5 for Bathroom mirror.
Case Study
Bathroom mirror can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Answer: Bathroom mirror can be analysed by drawing incident ray, reflected ray and normal, then applying the law of reflection and image formation principles.
Section 14: Quick Revision Notes
Important formulas
i = r
mirror rotation: θ → 2θ
N = 360/θ − 1
plane mirror magnification = +1
Important concepts
Angles are measured from normal.
Plane mirror image is virtual and erect.
Diffuse reflection obeys laws locally.
Object-image distance changes at 2v for perpendicular motion.
Exam tips
Always draw normal.
Mark i and r clearly.
Use symmetry for plane mirrors.
For two mirrors, check angle and object position.
Most repeated questions: laws of reflection, lateral inversion, mirror rotation, number of images and regular vs diffuse reflection.
Section 15: Kumar Sir Guidance Section
If Reflection of Light, Plane Mirrors, Multiple Images, Mirror Rotation or any Ray Optics concept is not clear, students may contact Kumar Sir for one-to-one personalised Physics guidance.
