Ray Optics • Reflection • Refraction • Formula Sheet • Exam Practice

Ray Optics – Introduction and Complete Chapter Roadmap

Premium coaching-style notes for CBSE, NEET, JEE Main, JEE Advanced, IB Physics, IGCSE Physics and A-Level Physics with concepts, formulas, diagrams, solved numericals, assertion-reason practice, case studies and revision cards.

BasicsReflection vs RefractionFormulasNumericalsPYQsCase Studies

Section 1: How to Use This Ray Optics Page

Use this page like a coaching class worksheet: concept first, formula second, diagram third, then numerical practice and exam-style questions.

Step 1
Read the concept and identify the medium, boundary, normal and ray direction.
Step 2
Match the concept with the formula sheet and draw a small ray diagram before calculating.
Step 3
Solve numericals, then attempt NEET, JEE, board and international-style questions.

Section 2: Why Ray Optics is Important for NEET, JEE, CBSE, IB, IGCSE and A-Level

Ray Optics connects daily-life observations with scoring exam topics. It explains mirrors, lenses, the human eye, optical instruments, prisms, fibres, mirages and apparent depth.

NEET and CBSE
Questions often test direct formulas, ray diagrams, total internal reflection and conceptual clarity.
JEE Main and Advanced
Problems combine sign convention, refraction geometry, critical angle, lens combinations and multi-step reasoning.
IB, IGCSE and A-Level
Structured questions focus on explanation, experiments, graphs, uncertainty and real-world interpretation.

Section 3: What is Ray Optics?

Ray Optics, also called Geometrical Optics, is the branch of Physics in which light is represented by straight lines called rays. A ray shows the direction in which light energy travels.

The ray model is valid when the wavelength of light is very small compared with the size of the obstacle, aperture, mirror or lens. In ordinary mirrors, lenses and optical instruments, visible light has a wavelength around 10-7 m, while the instrument size is much larger, so diffraction and interference are often negligible.

Light has wave nature and photon nature, but Ray Optics focuses on the geometrical path of light. Rectilinear propagation means that in a homogeneous medium, light travels in straight lines. This explains shadows, pinhole cameras, mirror images and many lens systems.

Light represented as straight raysValid when wavelength is very small compared with objects
Definition
Ray optics studies light using rays and geometrical construction.
Ray approximation
Valid when λ is much smaller than the size of obstacle or aperture.
Rectilinear propagation
In a homogeneous medium, light travels in straight lines.

Section 4: Reflection vs Refraction

Reflection

Reflection is the phenomenon in which light returns to the same medium after striking a surface. For a smooth surface, the reflected rays are regular and can form images.

i = r

Applications: plane mirrors, spherical mirrors, periscopes, reflecting telescopes, rear-view mirrors and mirror systems.

Refraction

Refraction is the bending of light when it passes from one transparent medium to another because the speed of light changes.

μ1 sin i = μ2 sin r

Applications: lenses, prisms, optical fibres, atmospheric refraction, human eye and optical instruments.

Incident rayReflected rayRefracted rayMedium 1Medium 2

At a boundary, part of light may reflect and part may refract into the second medium.

PointReflectionRefraction
DefinitionLight returns to the same medium.Light enters another medium and bends.
Medium changeNo medium change after reflection.Medium changes from one transparent medium to another.
Speed changeSpeed remains same because medium is same.Speed changes due to different optical density.
Direction changeDirection changes according to i = r.Direction changes according to Snell's law.
FrequencyFrequency remains same.Frequency remains same.
WavelengthWavelength remains same in same medium.Wavelength changes when speed changes.
ApplicationsMirrors, periscopes, reflectors.Lenses, fibres, prisms, eye, atmosphere.

Section 5: Refractive Index

The refractive index of a medium tells us how much the speed of light is reduced in that medium compared with vacuum. It is a measure of optical density.

μ = c/v
μ
Refractive index of the medium.
c
Speed of light in vacuum, 3 × 108 m/s.
v
Speed of light in the given medium.

If μ is large, the speed of light in that medium is small. Glass has greater refractive index than air, so light slows down in glass. Diamond has a very high refractive index, which contributes to its brilliance along with total internal reflection.

Section 6: Frequency and Wavelength During Refraction

When light enters another medium, frequency remains constant because frequency is determined by the source of light. The boundary cannot change how many wave crests are emitted per second by the source. However, the speed changes due to the optical properties of the medium. Since v = fλ, wavelength also changes.

Frequency remains constantf1 = f2

Frequency is fixed by source.

Speed in mediumv = c/μ

Speed decreases in optically denser medium.

Wavelength relationλ = v/f

Wavelength changes when speed changes.

Medium wavelengthλmedium = λair/μ

For air to medium approximately.

Air: larger wavelengthMedium: smaller wavelengthFrequency remains same, speed and wavelength change.

Numerical Example 1

Light of wavelength 600 nm enters glass of refractive index 1.5. Frequency remains same.

λglass = 600/1.5 = 400 nm

Numerical Example 2

Speed of light in water of refractive index 1.33 is:

v = 3 × 108/1.33 = 2.26 × 108 m/s

Section 7: Important Formula Sheet

Reflectioni = r

Angle of incidence equals angle of reflection.

Refractive Indexμ = c/v

Ratio of speed in vacuum to speed in medium.

Relative Refractive Indexμ21 = μ2/μ1

Index of medium 2 relative to medium 1.

Snell's Lawμ1 sin i = μ2 sin r

Basic law of refraction.

Critical Anglesin C = 1/μ

For denser medium to air.

Total Internal Reflectioni > C

Light must go from denser to rarer medium.

Mirror Formula1/f = 1/v + 1/u

Use Cartesian sign convention.

Lens PowerP = 1/f

f in metre gives power in dioptre.

Section 8: Geometrical Optics vs Wave Optics

PointGeometrical OpticsWave Optics
NatureUses rays and straight-line propagation.Uses wavefronts, phase and superposition.
InterferenceIgnored.Explained in detail.
DiffractionIgnored when aperture is large.Important when aperture is comparable with wavelength.
PolarisationGenerally not central.Explained as transverse wave phenomenon.
ApplicationsMirrors, lenses, instruments, eye.Young's experiment, diffraction, polarisation.
MathematicsGeometry, trigonometry and sign convention.Wave equations, phase and superposition.

Section 9: Chapter Roadmap

Reflection at Plane Mirror
Builds image formation, lateral inversion and basic ray diagram skill.
Spherical Mirrors
Introduces sign convention, mirror formula, magnification and real versus virtual images.
Refraction at Plane Surface
Covers Snell's law, apparent depth, real depth and ray bending.
Glass Slab and Prism
Adds lateral shift, deviation, dispersion and minimum deviation formula.
Total Internal Reflection
Explains critical angle, optical fibres, mirage and diamond brilliance.
Lenses and Instruments
Covers lens formula, power, combinations, microscope, telescope and human eye basics.

Section 10: Solved Numericals

These JEE Main-level numericals are unique practice questions covering speed, refractive index, Snell's law, critical angle, apparent depth, lateral shift, mirrors, lenses, power and prisms.

JEE Main Ray Optics Numericals

JEE Main 1. Speed of light in glass
JEE Main Numerical

Given data: Refractive index of glass = 1.5; c = 3 × 108 m/s.

Formula used: v = c/μ

Step-by-step solution:

  1. v = 3 × 108 / 1.5
  2. v = 2.0 × 108 m/s
Final answer: 2.0 × 108 m/s
JEE Main 2. Refraction angle in glass
JEE Main Numerical

Given data: Light goes from air to glass (μ = 1.5) at i = 30°.

Formula used: μ1 sin i = μ2 sin r

Step-by-step solution:

  1. 1 × sin 30° = 1.5 sin r
  2. sin r = 0.5/1.5 = 1/3
  3. r = sin-1(1/3) = 19.5°
Final answer: r = 19.5°
JEE Main 3. Critical angle of glass
JEE Main Numerical

Given data: Glass has refractive index 1.5 and is surrounded by air.

Formula used: sin C = 1/μ

Step-by-step solution:

  1. sin C = 1/1.5 = 0.667
  2. C = sin-1(0.667)
Final answer: C = 41.8°
JEE Main 4. Apparent depth in water
JEE Main Numerical

Given data: A point object is 12 cm below water surface; μwater = 4/3.

Formula used: apparent depth = real depth/μ

Step-by-step solution:

  1. apparent depth = 12/(4/3)
  2. apparent depth = 9 cm
Final answer: 9 cm
JEE Main 5. Lateral shift in a glass slab
JEE Main Numerical

Given data: Slab thickness t = 6 cm, i = 45°, r = 28.1°.

Formula used: d = t sin(i - r)/cos r

Step-by-step solution:

  1. d = 6 sin(16.9°)/cos(28.1°)
  2. d = 6(0.291)/0.882
  3. d = 1.98 cm
Final answer: about 2.0 cm
JEE Main 6. Concave mirror image distance
JEE Main Numerical

Given data: Concave mirror f = -20 cm; object distance u = -30 cm.

Formula used: 1/f = 1/v + 1/u

Step-by-step solution:

  1. -1/20 = 1/v - 1/30
  2. 1/v = -1/20 + 1/30 = -1/60
  3. v = -60 cm
Final answer: v = -60 cm
JEE Main 7. Convex mirror image distance
JEE Main Numerical

Given data: Convex mirror f = +20 cm; object distance u = -30 cm.

Formula used: 1/f = 1/v + 1/u

Step-by-step solution:

  1. 1/20 = 1/v - 1/30
  2. 1/v = 1/20 + 1/30 = 1/12
  3. v = +12 cm
Final answer: v = +12 cm
JEE Main 8. Convex lens image distance
JEE Main Numerical

Given data: Convex lens f = +15 cm; object distance u = -30 cm.

Formula used: 1/f = 1/v - 1/u

Step-by-step solution:

  1. 1/15 = 1/v + 1/30
  2. 1/v = 1/15 - 1/30 = 1/30
  3. v = 30 cm
Final answer: v = +30 cm
JEE Main 9. Concave lens image distance
JEE Main Numerical

Given data: Concave lens f = -20 cm; object distance u = -40 cm.

Formula used: 1/f = 1/v - 1/u

Step-by-step solution:

  1. -1/20 = 1/v + 1/40
  2. 1/v = -1/20 - 1/40 = -3/40
  3. v = -13.3 cm
Final answer: v = -13.3 cm
JEE Main 10. Power of a convex lens
JEE Main Numerical

Given data: Focal length f = 25 cm.

Formula used: P = 1/f in metre

Step-by-step solution:

  1. f = 0.25 m
  2. P = 1/0.25
Final answer: P = +4 D
JEE Main 11. Combination of two thin lenses
JEE Main Numerical

Given data: Two lenses of powers +5 D and -2 D are in contact.

Formula used: P = P1 + P2

Step-by-step solution:

  1. P = 5 - 2 = 3 D
  2. f = 1/P = 1/3 m = 0.333 m
Final answer: Equivalent power = +3 D, focal length = 33.3 cm
JEE Main 12. Minimum deviation of a prism
JEE Main Numerical

Given data: Equilateral prism A = 60°, refractive index μ = √2.

Formula used: μ = sin[(A + Dm)/2]/sin(A/2)

Step-by-step solution:

  1. √2 = sin[(60° + Dm)/2]/sin30°
  2. sin[(60° + Dm)/2] = 0.707
  3. (60° + Dm)/2 = 45°
Final answer: Dm = 30°
JEE Main 13. Wavelength inside glass
JEE Main Numerical

Given data: Light of wavelength 600 nm enters glass of μ = 1.5.

Formula used: λmedium = λair/μ

Step-by-step solution:

  1. λglass = 600/1.5
  2. λglass = 400 nm
Final answer: 400 nm
JEE Main 14. Relative refractive index
JEE Main Numerical

Given data: Absolute refractive indices are μwater = 1.33 and μglass = 1.50.

Formula used: μglass, water = μglass/μwater

Step-by-step solution:

  1. μglass, water = 1.50/1.33
  2. μglass, water = 1.13
Final answer: 1.13
JEE Main 15. Critical angle for water-air
JEE Main Numerical

Given data: Water has μ = 4/3.

Formula used: sin C = 1/μ

Step-by-step solution:

  1. sin C = 3/4
  2. C = sin-1(0.75)
Final answer: C = 48.6°
JEE Main 16. Plane mirror relative speed
JEE Main Numerical

Given data: A student walks towards a plane mirror with speed 2 m/s.

Formula used: distance between object and image changes at 2v

Step-by-step solution:

  1. Image approaches the mirror with equal speed on the other side
  2. Object-image separation therefore decreases at 2 × 2 m/s
Final answer: 4 m/s
JEE Main 17. Apparent depth through two layers
JEE Main Numerical

Given data: Water layer = 6 cm, μ = 4/3; glass layer = 3 cm, μ = 1.5.

Formula used: apparent depth = Σ(thickness/refractive index)

Step-by-step solution:

  1. Water contribution = 6/(4/3) = 4.5 cm
  2. Glass contribution = 3/1.5 = 2.0 cm
  3. Total apparent depth = 6.5 cm
Final answer: 6.5 cm
JEE Main 18. Magnification by a convex lens
JEE Main Numerical

Given data: Convex lens f = 10 cm; object distance u = -20 cm.

Formula used: 1/f = 1/v - 1/u and m = v/u

Step-by-step solution:

  1. 1/10 = 1/v + 1/20
  2. 1/v = 1/20, so v = 20 cm
  3. m = 20/(-20) = -1
Final answer: m = -1; real inverted image of same size
JEE Main 19. Object at centre of curvature
JEE Main Numerical

Given data: Concave mirror has radius R = 40 cm; object is at centre of curvature.

Formula used: f = R/2 and object at C forms image at C

Step-by-step solution:

  1. f = -20 cm and u = -40 cm
  2. Using mirror formula gives v = -40 cm
Final answer: Image at 40 cm in front of mirror
JEE Main 20. Small-angle prism deviation
JEE Main Numerical

Given data: A thin prism has A = 10° and μ = 1.5.

Formula used: δ = (μ - 1)A

Step-by-step solution:

  1. δ = (1.5 - 1) × 10°
  2. δ = 5°
Final answer: 5°

Section 11: Previous Year / Exam-style Questions

Use this section for original, non-repetitive exam-style practice focused on the concepts students are repeatedly tested on.

NEET Ray Optics Questions

NEET 1. Ray approximation is most reliable when
NEET MCQ
  1. The aperture is smaller than the wavelength
  2. The wavelength of light is much smaller than the size of the obstacle or aperture
  3. Light travels only in vacuum
  4. Diffraction is the dominant effect

Correct answer: B

Explanation: Ray optics ignores diffraction, so it works best when the wavelength is negligible compared with mirrors, lenses, slits or obstacles.

Answer: B
NEET 2. In reflection from a plane mirror, the angle of incidence is measured between
NEET MCQ
  1. Incident ray and mirror surface
  2. Reflected ray and mirror surface
  3. Incident ray and normal
  4. Incident ray and reflected ray

Correct answer: C

Explanation: All angles in reflection and refraction are measured from the normal drawn at the point of incidence.

Answer: C
NEET 3. When monochromatic light enters glass from air, which quantity remains unchanged?
NEET MCQ
  1. Speed
  2. Wavelength
  3. Frequency
  4. Direction

Correct answer: C

Explanation: Frequency is fixed by the source. Speed and wavelength change when the medium changes.

Answer: C
NEET 4. The absolute refractive index of a medium is
NEET MCQ
  1. v/c
  2. c/v
  3. sin r/sin i
  4. wavelength in medium divided by wavelength in vacuum

Correct answer: B

Explanation: Absolute refractive index is μ = c/v, where c is speed in vacuum and v is speed in the medium.

Answer: B
NEET 5. Snell's law for light going from medium 1 to medium 2 is
NEET MCQ
  1. μ1 sin r = μ2 sin i
  2. μ1 sin i = μ2 sin r
  3. i = r
  4. μ1 cos i = μ2 cos r

Correct answer: B

Explanation: Snell's law connects refractive indices with the sines of the angles measured from the normal.

Answer: B
NEET 6. A medium with larger refractive index has
NEET MCQ
  1. Greater speed of light
  2. Lower optical density
  3. Smaller speed of light
  4. No effect on light speed

Correct answer: C

Explanation: Since v = c/μ, a larger refractive index means smaller speed in that medium.

Answer: C
NEET 7. If light enters a denser medium obliquely from a rarer medium, it generally bends
NEET MCQ
  1. Away from the normal
  2. Towards the normal
  3. Along the boundary
  4. Back into the first medium always

Correct answer: B

Explanation: Light slows down in the denser medium, so the refracted ray bends towards the normal.

Answer: B
NEET 8. Critical angle is the angle of incidence in denser medium for which angle of refraction is
NEET MCQ
  1. 0°
  2. 30°
  3. 45°
  4. 90°

Correct answer: D

Explanation: At the critical angle, the refracted ray just grazes the boundary, so r = 90°.

Answer: D
NEET 9. Total internal reflection can occur when light travels
NEET MCQ
  1. From rarer to denser medium only
  2. From denser to rarer medium with i > C
  3. In any direction if the surface is polished
  4. Only in opaque media

Correct answer: B

Explanation: Both conditions are required: denser to rarer travel and angle of incidence greater than the critical angle.

Answer: B
NEET 10. A coin at the bottom of water appears raised because
NEET MCQ
  1. Reflection makes the water surface act like a mirror
  2. Refraction bends rays away from the normal as they enter air
  3. Frequency increases in air
  4. Water absorbs lower rays

Correct answer: B

Explanation: Rays from the coin bend away from the normal on passing from water to air; their backward extensions meet at a shallower apparent position.

Answer: B
NEET 11. Optical fibres guide light mainly by
NEET MCQ
  1. Dispersion
  2. Polarisation
  3. Total internal reflection
  4. Diffuse reflection

Correct answer: C

Explanation: Light repeatedly undergoes total internal reflection at the core-cladding boundary, staying inside the fibre.

Answer: C
NEET 12. Diamond sparkles strongly because it has
NEET MCQ
  1. Very small refractive index
  2. High refractive index and small critical angle
  3. No reflection at its surfaces
  4. Very low optical density

Correct answer: B

Explanation: High refractive index gives diamond a small critical angle, increasing internal reflections.

Answer: B
NEET 13. Mirage on a hot road is mainly due to
NEET MCQ
  1. Regular reflection from dry road
  2. Multiple refractions and possible total internal reflection in air layers
  3. Diffraction around dust particles
  4. Absorption by hot air

Correct answer: B

Explanation: Air near the road is hotter and optically rarer, so rays bend continuously and may undergo total internal reflection.

Answer: B
NEET 14. If yellow light enters glass from air, its wavelength in glass
NEET MCQ
  1. Increases
  2. Decreases
  3. Becomes zero
  4. Remains exactly same

Correct answer: B

Explanation: Frequency is constant while speed decreases, so λ = v/f decreases.

Answer: B
NEET 15. Image formed by a plane mirror is
NEET MCQ
  1. Real, inverted and diminished
  2. Virtual, erect and same size
  3. Real, erect and magnified
  4. Virtual, inverted and enlarged

Correct answer: B

Explanation: A plane mirror forms a virtual, erect, laterally inverted image at the same distance behind the mirror.

Answer: B
NEET 16. A convex mirror always forms an image that is
NEET MCQ
  1. Real and enlarged
  2. Virtual, erect and diminished
  3. Real and same size
  4. Inverted and behind the mirror

Correct answer: B

Explanation: For a real object, a convex mirror forms a virtual, erect and diminished image behind the mirror.

Answer: B
NEET 17. SI unit of lens power is
NEET MCQ
  1. metre
  2. dioptre
  3. candela
  4. radian

Correct answer: B

Explanation: Lens power P = 1/f when f is in metres. Its unit is dioptre.

Answer: B
NEET 18. Stars twinkle because of
NEET MCQ
  1. Reflection from clouds
  2. Atmospheric refraction through changing air layers
  3. Total absorption by atmosphere
  4. Dispersion only in vacuum

Correct answer: B

Explanation: The refractive index of air changes with height and temperature, causing small random changes in apparent star position and brightness.

Answer: B
NEET 19. A ray emerging from a parallel glass slab is
NEET MCQ
  1. Always perpendicular to the slab
  2. Parallel to the incident ray but laterally shifted
  3. Opposite to the incident ray
  4. Never refracted

Correct answer: B

Explanation: The two slab surfaces are parallel, so the emergent ray becomes parallel to the incident ray with lateral displacement.

Answer: B
NEET 20. The normal at the point of incidence is used because
NEET MCQ
  1. It is the actual path of light
  2. It is the reference line for measuring i and r
  3. It is always horizontal
  4. It replaces Snell's law

Correct answer: B

Explanation: Reflection and refraction laws define angles with respect to the normal, not with respect to the surface.

Answer: B

JEE Advanced Style Original Questions

JEE Advanced Original 1. Layered media apparent depth
JEE Advanced Original

Difficulty: Hard

A point object is below two horizontal layers: 6 cm of water (μ = 4/3) above 4 cm of oil (μ = 1.25). Viewed normally from air, find the apparent depth.

Complete solution:

  1. For near-normal viewing, each layer contributes t/μ.
  2. Water contribution = 6/(4/3) = 4.5 cm.
  3. Oil contribution = 4/1.25 = 3.2 cm.
  4. Total apparent depth = 7.7 cm.

Key concept tested: Apparent depth in stratified media

Final answer: 7.7 cm
JEE Advanced Original 2. Multiple-correct Snell graph
JEE Advanced Original

Difficulty: Moderate

A graph is drawn with sin i on the vertical axis and sin r on the horizontal axis for air to glass refraction. The slope is 1.50. Choose the correct statements: A. Refractive index of glass is 1.50. B. Speed in glass is 3.0 × 108 m/s. C. Frequency remains unchanged. D. Wavelength in glass is 1.50 times the wavelength in air.

Complete solution:

  1. Snell law gives sin i = μ sin r for air to glass, so slope = μ = 1.50.
  2. Speed is c/μ = 2.0 × 108 m/s, so B is false.
  3. Frequency remains unchanged, so C is true.
  4. Wavelength becomes λ/1.50, so D is false.

Key concept tested: Graph interpretation of Snell's law

Final answer: A and C
JEE Advanced Original 3. Prism plus TIR condition
JEE Advanced Original

Difficulty: Hard

A ray is incident normally on the first face of a prism of angle 60°. The prism has μ = √3 and is surrounded by air. Will the ray emerge from the second face?

Complete solution:

  1. Normal incidence at the first face means no bending there.
  2. Inside the prism, the incidence angle at the second face equals the prism angle, 60°.
  3. Critical angle C satisfies sin C = 1/√3, so C = 35.3°.
  4. Since 60° > C, total internal reflection occurs at the second face.

Key concept tested: Combining prism geometry with critical angle

Final answer: The ray does not emerge from the second face; it undergoes TIR.
JEE Advanced Original 4. Integer-type lens combination
JEE Advanced Original

Difficulty: Moderate

Two thin lenses of powers +7 D and -4 D are kept in contact. Give the equivalent power as an integer in dioptres.

Complete solution:

  1. For thin lenses in contact, powers add algebraically.
  2. P = +7 - 4 = +3 D.

Key concept tested: Power addition of thin lenses

Final answer: 3
JEE Advanced Original 5. Moving object-image relation
JEE Advanced Original

Difficulty: Hard

For a convex lens of focal length 20 cm, an object is at u = -40 cm and moves towards the lens at 2 cm/s. Find the instantaneous speed of the image.

Complete solution:

  1. Lens formula: 1/f = 1/v - 1/u. At u = -40 cm, v = +40 cm.
  2. Differentiate with f constant: -dv/v2 + du/u2 = 0, so dv/du = v2/u2 = 1.
  3. As the object moves towards the lens, u increases at 2 cm/s.
  4. Therefore v increases at 2 cm/s, so the image moves away from the lens at 2 cm/s.

Key concept tested: Differentiation of lens formula

Final answer: 2 cm/s away from the lens
JEE Advanced Original 6. Optical fibre numerical aperture
JEE Advanced Original

Difficulty: Hard

A fibre has core refractive index 1.50 and cladding refractive index 1.47. Find the numerical aperture in air and the acceptance angle.

Complete solution:

  1. Numerical aperture NA = √(n12 - n22).
  2. NA = √(2.25 - 2.1609) = √0.0891 = 0.299.
  3. In air, sin θmax = NA.
  4. θmax = sin-1(0.299) = 17.4°.

Key concept tested: Acceptance cone and total internal reflection

Final answer: NA = 0.299; acceptance angle = 17.4°
JEE Advanced Original 7. Multiple-correct TIR traps
JEE Advanced Original

Difficulty: Moderate

For total internal reflection at a glass-air boundary, choose correct statements: A. Light must be incident from glass side. B. It can occur at any angle in glass. C. At i > C, no refracted ray carries energy into air in ideal ray optics. D. It is possible from air to glass if glass is polished.

Complete solution:

  1. TIR requires denser to rarer travel, so A is true.
  2. The angle must exceed critical angle, so B is false.
  3. For i > C the refracted ray is absent in ideal ray optics, so C is true.
  4. Polishing does not change the denser-to-rarer condition, so D is false.

Key concept tested: Conditions for total internal reflection

Final answer: A and C
JEE Advanced Original 8. Integer-type critical angle
JEE Advanced Original

Difficulty: Moderate

For a transparent medium in air, the critical angle is 30°. Find its refractive index.

Complete solution:

  1. sin C = 1/μ.
  2. sin 30° = 1/2 = 1/μ.
  3. μ = 2.

Key concept tested: Critical angle relation

Final answer: 2
JEE Advanced Original 9. Paragraph ray-displacement problem
JEE Advanced Original

Difficulty: Hard

A ray passes through a parallel slab. The incident and emergent rays are parallel, but the emergent ray is shifted sideways. Explain why the deviation is zero but displacement is not zero.

Complete solution:

  1. At the first face the ray bends towards the normal; at the second parallel face it bends away from the normal by the same net angular amount.
  2. Because the two faces are parallel, angular deviation cancels.
  3. The path inside the slab is oblique, so the emergent ray starts from a laterally shifted point.

Key concept tested: Parallel slab geometry

Final answer: Net angular deviation is zero; lateral displacement is finite.
JEE Advanced Original 10. Mirror rotation integer
JEE Advanced Original

Difficulty: Moderate

A plane mirror is rotated by 7° while the incident ray direction is fixed. Through what angle does the reflected ray rotate?

Complete solution:

  1. When a plane mirror rotates by θ, its normal also rotates by θ.
  2. The angle of reflection changes symmetrically, so the reflected ray rotates by 2θ.
  3. Here 2θ = 14°.

Key concept tested: Reflection from a rotating plane mirror

Final answer: 14°

IB / IGCSE / A-Level Questions

IB/IGCSE/A-Level 1. Semicircular glass block method
IB / IGCSE / A-Level

Question: Describe how a semicircular glass block can be used to measure refractive index.

Answer: Direct the ray through the curved surface towards the centre so it enters normally. Measure i and r at the plane surface for several trials, then use μ = sin i/sin r or the gradient of a sin i versus sin r graph.

IB/IGCSE/A-Level 2. Practical graph interpretation
IB / IGCSE / A-Level

Question: A student plots sin i on the y-axis and sin r on the x-axis for air to glass and obtains gradient 1.52. What does the gradient represent?

Answer: The gradient represents the refractive index of glass relative to air, approximately 1.52.

IB/IGCSE/A-Level 3. Ray diagram reasoning
IB / IGCSE / A-Level

Question: A ray enters glass from air. State how the ray bends and why.

Answer: It bends towards the normal because light slows down in the optically denser glass.

IB/IGCSE/A-Level 4. Experimental uncertainty
IB / IGCSE / A-Level

Question: Name one common source of error while tracing rays through a glass block.

Answer: A common error is marking thick ray lines or pins inaccurately. This changes measured angles and affects the calculated refractive index.

IB/IGCSE/A-Level 5. Real-world optics
IB / IGCSE / A-Level

Question: Why does a straw look bent in a glass of water?

Answer: Light from the underwater part refracts at the water-air boundary. The eye traces the refracted rays backward, so the submerged part appears displaced.

IB/IGCSE/A-Level 6. Critical angle observation
IB / IGCSE / A-Level

Question: How would you identify the critical angle in a glass block experiment?

Answer: Increase the angle of incidence inside glass until the refracted ray just travels along the glass-air boundary. That incidence angle is the critical angle.

IB/IGCSE/A-Level 7. Optical fibre communication
IB / IGCSE / A-Level

Question: Give one reason optical fibres are useful in communication.

Answer: They carry signals over long distances by repeated total internal reflection, with low loss and little electromagnetic interference.

IB/IGCSE/A-Level 8. Regular and diffuse reflection
IB / IGCSE / A-Level

Question: Explain why a mirror forms a clear image but paper does not.

Answer: A mirror gives regular reflection from a smooth surface, so reflected rays keep an ordered pattern. Paper has a rough surface and gives diffuse reflection.

IB/IGCSE/A-Level 9. Apparent depth
IB / IGCSE / A-Level

Question: A pond looks shallower than it really is. Which boundary causes this effect?

Answer: The water-air boundary causes refraction. Rays bend away from the normal on emerging into air, making the bottom appear raised.

IB/IGCSE/A-Level 10. Atmospheric refraction
IB / IGCSE / A-Level

Question: Why can the Sun be seen slightly before actual sunrise?

Answer: Atmospheric refraction bends sunlight around the curved layers of air, raising the apparent position of the Sun.

IB/IGCSE/A-Level 11. Lens focal length experiment
IB / IGCSE / A-Level

Question: How can the approximate focal length of a convex lens be found using a distant object?

Answer: Focus the image of a distant object on a screen. The lens-screen distance is approximately the focal length because incoming rays are nearly parallel.

IB/IGCSE/A-Level 12. Convex lens image
IB / IGCSE / A-Level

Question: For an object beyond 2F of a convex lens, describe the image.

Answer: The image is real, inverted, diminished and formed between F and 2F on the other side.

IB/IGCSE/A-Level 13. Data question
IB / IGCSE / A-Level

Question: For a ray entering glass, i = 45° and r = 28°. Estimate the refractive index.

Answer: μ = sin45°/sin28° = 0.707/0.469 = 1.51.

IB/IGCSE/A-Level 14. Material choice
IB / IGCSE / A-Level

Question: Why are high-refractive-index materials useful in compact lenses?

Answer: A larger refractive index bends light more strongly, so the same focusing effect can be achieved with a thinner or shorter focal length lens.

IB/IGCSE/A-Level 15. Evaluation
IB / IGCSE / A-Level

Question: Why should angles be measured from the normal rather than from the surface?

Answer: The laws of reflection and refraction are defined with respect to the normal. Measuring from the surface gives complementary angles and leads to wrong substitution.

Section 12: Assertion-Reason Questions

Assertion-Reason 1. Assertion: Frequency of light remains constant during refraction. Reason: Frequency is fixed by the source of light.
Assertion-Reason

Answer: Both Assertion and Reason are true, and Reason correctly explains Assertion.

Explanation: The boundary can change speed and wavelength, but it cannot change the rate at which crests are emitted by the source.

Both Assertion and Reason are true, and Reason correctly explains Assertion.
Assertion-Reason 2. Assertion: Wavelength of light decreases when it enters glass from air. Reason: Speed decreases in glass while frequency remains unchanged.
Assertion-Reason

Answer: Both Assertion and Reason are true, and Reason correctly explains Assertion.

Explanation: Using v = fλ, lower speed at constant frequency means lower wavelength.

Both Assertion and Reason are true, and Reason correctly explains Assertion.
Assertion-Reason 3. Assertion: Total internal reflection is possible only from denser to rarer medium. Reason: Only then can the refracted ray bend away from the normal and reach r = 90°.
Assertion-Reason

Answer: Both Assertion and Reason are true, and Reason correctly explains Assertion.

Explanation: Critical angle is defined only for denser-to-rarer travel.

Both Assertion and Reason are true, and Reason correctly explains Assertion.
Assertion-Reason 4. Assertion: Critical angle decreases when refractive index of the denser medium increases. Reason: sin C = 1/μ for denser medium to air.
Assertion-Reason

Answer: Both Assertion and Reason are true, and Reason correctly explains Assertion.

Explanation: A larger μ gives a smaller value of sin C, hence a smaller critical angle.

Both Assertion and Reason are true, and Reason correctly explains Assertion.
Assertion-Reason 5. Assertion: Diamond shows strong brilliance. Reason: Its high refractive index gives a small critical angle and repeated internal reflection.
Assertion-Reason

Answer: Both Assertion and Reason are true, and Reason correctly explains Assertion.

Explanation: Diamond cuts are designed so many rays undergo total internal reflection before emerging.

Both Assertion and Reason are true, and Reason correctly explains Assertion.
Assertion-Reason 6. Assertion: Mirage is observed on hot roads. Reason: Air near the road can be optically rarer than the cooler air above.
Assertion-Reason

Answer: Both Assertion and Reason are true, and Reason correctly explains Assertion.

Explanation: The refractive index gradient bends rays upward and may support total internal reflection-like behavior.

Both Assertion and Reason are true, and Reason correctly explains Assertion.
Assertion-Reason 7. Assertion: A coin in water appears raised. Reason: Rays from the coin bend away from the normal while emerging into air.
Assertion-Reason

Answer: Both Assertion and Reason are true, and Reason correctly explains Assertion.

Explanation: The eye traces the refracted rays backward to a virtual point closer to the surface.

Both Assertion and Reason are true, and Reason correctly explains Assertion.
Assertion-Reason 8. Assertion: Optical fibres require the core to have higher refractive index than cladding. Reason: Light must meet a denser-to-rarer boundary for total internal reflection.
Assertion-Reason

Answer: Both Assertion and Reason are true, and Reason correctly explains Assertion.

Explanation: The core-cladding design keeps light trapped in the core.

Both Assertion and Reason are true, and Reason correctly explains Assertion.
Assertion-Reason 9. Assertion: Ray optics fails for very narrow apertures. Reason: Diffraction becomes significant when aperture size is comparable with wavelength.
Assertion-Reason

Answer: Both Assertion and Reason are true, and Reason correctly explains Assertion.

Explanation: Ray optics assumes diffraction is negligible.

Both Assertion and Reason are true, and Reason correctly explains Assertion.
Assertion-Reason 10. Assertion: The normal is drawn at the point of incidence. Reason: Reflection and refraction angles are measured with respect to that normal.
Assertion-Reason

Answer: Both Assertion and Reason are true, and Reason correctly explains Assertion.

Explanation: The normal provides the local reference line at the exact point where the ray meets the surface.

Both Assertion and Reason are true, and Reason correctly explains Assertion.

Section 13: Case Study Questions

Case Study 1: Fish appearing raised in water

Passage: A fish under water is seen by an observer standing in air. Light from the fish travels from water to air and bends away from the normal at the water surface. The observer traces the emergent rays backwards and sees a virtual image closer to the surface than the real fish.

Case 1: Fish appearing raised in water

Rays from the fish bend away from the normal as they leave water.

Case 1.1. Why does the fish appear closer to the surface?
Case Study

Answer: Because rays emerging from water into air bend away from the normal, and their backward extensions meet at a shallower virtual image.

Explanation: The observer assumes light has travelled straight in air, so the virtual image is located where the refracted rays seem to originate.

Case 1.2. If the real depth is 80 cm and μwater = 4/3, what is the apparent depth for near-normal viewing?
Case Study

Answer: Apparent depth = 80/(4/3) = 60 cm.

Explanation: For near-normal viewing from air, apparent depth equals real depth divided by the refractive index of water.

Case 1.3. Which quantity of light remains constant at the water-air boundary?
Case Study

Answer: Frequency remains constant because it is determined by the source.

Explanation: Speed increases and wavelength increases when light enters air, but the wave crests must cross the boundary at the same rate.

Case Study 2: Coin in water

Passage: A coin placed at the bottom of a bowl becomes visible when water is poured into the bowl. Refraction at the water-air surface raises the apparent position of the coin, allowing rays to reach the eye.

Case 2: Coin in water

The coin appears raised because its virtual image is formed above its actual position.

Case 2.1. What optical effect makes the coin visible after adding water?
Case Study

Answer: Refraction at the water-air surface bends rays from the coin so that they can enter the observer's eye.

Explanation: Before adding water, those rays may miss the eye; after adding water, bending at the surface changes their direction.

Case 2.2. Is the seen coin image real or virtual?
Case Study

Answer: It is virtual because the observer sees the backward extension of refracted rays.

Explanation: The rays do not actually meet at the apparent raised position; only their extensions meet there.

Case 2.3. What happens to the apparent depth if water is replaced by a liquid of larger refractive index?
Case Study

Answer: Apparent depth decreases further because apparent depth = real depth/refractive index.

Explanation: A larger refractive index produces stronger bending at the surface and a shallower apparent position for the same real depth.

Case Study 3: Mirage on a hot road

Passage: On a hot day, air close to the road is hotter and less optically dense than air above it. Light from the sky bends gradually while passing through these layers and can reach the eye as if it came from the road surface.

Case 3: Mirage on a hot road

A vertical refractive-index gradient bends rays upward near the hot road.

Case 3.1. Why does the road look wet?
Case Study

Answer: The eye receives bent rays from the sky and interprets them as coming from the road, creating a water-like image.

Explanation: The apparent image is often of the bright sky, so the brain mistakes it for reflection from water on the road.

Case 3.2. Which medium layer is optically rarer near the road?
Case Study

Answer: The hotter air close to the road is optically rarer.

Explanation: Higher temperature lowers air density and slightly lowers refractive index near the road surface.

Case 3.3. Is mirage mainly a reflection from water?
Case Study

Answer: No. It is caused by atmospheric refraction and sometimes total internal reflection-like bending in air layers.

Explanation: There is no water layer; the ray path curves because refractive index changes continuously with height.

Case Study 4: Optical fibre communication

Passage: An optical fibre has a core of slightly higher refractive index surrounded by cladding of lower refractive index. A light signal entering within the acceptance cone repeatedly reflects inside the core and can travel long distances.

Case 4: Optical fibre communication

Light is guided by repeated total internal reflection at the core-cladding boundary.

Case 4.1. Why must the core have higher refractive index than the cladding?
Case Study

Answer: The ray must travel from denser core to rarer cladding for total internal reflection to occur.

Explanation: At each core-cladding hit, the incidence angle must exceed the critical angle for the signal to remain trapped.

Case 4.2. What happens if the ray enters outside the acceptance cone?
Case Study

Answer: It may not satisfy the TIR condition and can leak into the cladding.

Explanation: The acceptance cone limits entry angles so the internal incidence angle stays large enough for total internal reflection.

Case 4.3. Name one advantage of optical fibre communication.
Case Study

Answer: It provides low-loss signal transmission and is less affected by electromagnetic interference.

Explanation: The information is carried as light pulses inside glass rather than as electrical currents in metal wires.

Case Study 5: Diamond brilliance

Passage: Diamond has a refractive index about 2.42, much larger than glass. Its small critical angle and cut faces make many rays undergo repeated total internal reflection before emerging from the top faces.

Case 5: Diamond brilliance

A small critical angle keeps many rays trapped until they emerge brightly.

Case 5.1. Why is the critical angle of diamond small?
Case Study

Answer: Since sin C = 1/μ, a large refractive index gives a small critical angle.

Explanation: With μ about 2.42, even moderately oblique internal rays can satisfy i > C.

Case 5.2. How does cutting improve brilliance?
Case Study

Answer: Proper cutting directs internal rays so that they undergo total internal reflection and emerge towards the observer.

Explanation: If the faces are poorly angled, rays escape through the bottom instead of returning through the top.

Case 5.3. Estimate the critical angle if μ = 2.42.
Case Study

Answer: sin C = 1/2.42 = 0.413, so C = 24.4° approximately.

Explanation: The small value explains why diamond traps many rays more effectively than ordinary glass.

Section 14: Conceptual Questions

Conceptual 1. Why is ray optics also called geometrical optics?
Conceptual

Answer: Because it uses straight-line rays, angles, triangles and geometrical construction to predict the path of light.

Conceptual 2. Why do sharp shadows support rectilinear propagation?
Conceptual

Answer: A sharp shadow forms when light travels approximately in straight lines and cannot bend strongly around a large object.

Conceptual 3. Why are angles measured from the normal?
Conceptual

Answer: The normal is perpendicular to the surface and gives a consistent reference for both reflection and refraction laws.

Conceptual 4. Does a plane mirror really reverse left and right?
Conceptual

Answer: A plane mirror reverses front-back direction. Left-right reversal is a perception caused by how we turn ourselves to compare with the image.

Conceptual 5. Why does refraction occur?
Conceptual

Answer: Refraction occurs because light changes speed when it enters a medium with different optical density.

Conceptual 6. Why does light bend towards the normal in glass?
Conceptual

Answer: Glass has a higher refractive index than air, so light slows down and bends towards the normal.

Conceptual 7. What changes when light enters glass from air?
Conceptual

Answer: Speed and wavelength decrease; direction may change if incidence is oblique; frequency remains constant.

Conceptual 8. Why does frequency remain unchanged?
Conceptual

Answer: Frequency is controlled by the source and must remain continuous across the boundary.

Conceptual 9. Why does the bottom of a pool look raised?
Conceptual

Answer: Rays from the bottom refract away from the normal at the water-air surface, so their backward extensions meet at a shallower point.

Conceptual 10. Why is critical angle not defined for rarer-to-denser travel?
Conceptual

Answer: In rarer-to-denser travel, the refracted ray bends towards the normal and cannot reach 90°.

Conceptual 11. What happens after the angle exceeds critical angle?
Conceptual

Answer: The ray is totally internally reflected and no ordinary refracted ray emerges into the rarer medium.

Conceptual 12. Why do optical fibres have cladding?
Conceptual

Answer: Cladding with lower refractive index creates the boundary needed for total internal reflection and protects the core.

Conceptual 13. Why is diamond more brilliant than ordinary glass?
Conceptual

Answer: Diamond has a much higher refractive index, smaller critical angle and carefully cut faces that promote repeated internal reflection.

Conceptual 14. Why is a mirage not a real water layer?
Conceptual

Answer: It is an optical illusion caused by bending of light through hot air layers, not reflection from actual water.

Conceptual 15. Why do stars twinkle but planets usually do not?
Conceptual

Answer: Stars appear point-like, so atmospheric refraction changes their brightness noticeably. Planets have a finite apparent disc, so fluctuations average out.

Conceptual 16. Why is the emergent ray parallel in a glass slab?
Conceptual

Answer: The two refracting surfaces are parallel, so angular deviation at the first surface is cancelled at the second.

Conceptual 17. Why are convex mirrors used in vehicles?
Conceptual

Answer: They provide a wide field of view and form erect diminished images of objects behind the vehicle.

Conceptual 18. Why is a concave mirror useful for shaving?
Conceptual

Answer: When the face is within focal length, the concave mirror forms a virtual, erect and magnified image.

Conceptual 19. What does negative power of a lens mean?
Conceptual

Answer: Negative power means the lens is diverging, usually a concave lens with negative focal length.

Conceptual 20. How is deviation by a prism different from dispersion?
Conceptual

Answer: Deviation is bending of a ray; dispersion is splitting of white light into colours because refractive index depends on wavelength.

Section 15: Quick Revision Notes

Important formulas
  • i = r
  • μ = c/v
  • μ1 sin i = μ2 sin r
  • sin C = 1/μ
  • P = 1/f
Important concepts
  • Frequency remains constant during refraction
  • Speed changes in a new medium
  • Wavelength changes when speed changes
  • TIR needs denser to rarer travel
  • Angles are measured from the normal
Common mistakes
  • Changing frequency during refraction
  • Using wrong medium order
  • Forgetting sign convention
  • Applying TIR from rarer to denser medium
  • Using surface angle instead of normal angle

NEET/JEE Tip: In every ray optics problem, first identify media, draw the normal, mark angles with the normal, then apply the formula.

Section 16: Kumar Sir Guidance Section

If any concept in Ray Optics is not clear, students can contact Kumar Sir for one-to-one personalised Physics guidance for CBSE, NEET, JEE Main, JEE Advanced, IB Physics, IGCSE Physics and A-Level Physics.

Phone: +91-9958461445

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