Mirror Formula and Magnification | Kumar Physics Classes
Ray Optics • Mirror Formula • Magnification

Mirror Formula and Magnification

Mirror Formula • Derivation • Sign Convention • Magnification • Linear Magnification • Relation Between m, u and v • Image Height Formula • Graphical Interpretation • Numericals • PYQs

FormulaMagnificationNumericalsPYQsCase Studies

1. Introduction to Mirror Formula

The mirror formula is one of the most important results in ray optics because it connects object distance, image distance and focal length in a single equation. It is used repeatedly in CBSE, NEET, JEE Main, JEE Advanced, IB Physics, ICSE, ISC, IGCSE and A-Level Physics for image formation, sign convention and numerical problem solving.

Why it matters
It predicts image position without drawing a full ray diagram every time.
Exam use
It appears in direct numericals, conceptual sign questions and mixed magnification problems.
Core idea
Once u, v and f are linked, image height follows from magnification.

2. New Cartesian Sign Convention

All distances are measured from the pole P of the mirror. The principal axis is taken as the x-axis. The direction of incident light is positive; the opposite direction is negative.

Pole P
The geometrical centre of the reflecting surface.
Principal Axis
The straight line passing through P and C.
Object distance u
Distance of object from P. For a real object in front of the mirror, u is negative.
Image distance v
Distance of image from P. Real image in front gives negative v; virtual image behind gives positive v.
Focal length f
For concave mirror f is negative; for convex mirror f is positive.
Heights
Height above the principal axis is positive and below it is negative.
+x direction of incident light-x Silvered side PFC Object u is measured from P to object: u < 0 f < 0 for concave mirror R < 0 for concave mirror

New Cartesian sign convention for mirror formula.

3. Mirror Formula

The mirror formula relates focal length, image distance and object distance:

1/f = 1/v + 1/u

Here f is focal length, v is image distance and u is object distance. Correct signs must be substituted according to the New Cartesian Sign Convention.

Concave mirrorf < 0
Convex mirrorf > 0
Real objectu < 0
Virtual image behind mirrorv > 0

4. Derivation of Mirror Formula

For paraxial rays, the object, image and mirror geometry form similar triangles. Using these similar triangles with the sign convention gives the mirror formula.

PFC Object h Image h' Similar triangles give h'/h = -v/u and lead to 1/f = 1/v + 1/u.

Similar triangles behind mirror formula and magnification.

Step 1: Similar triangles

From object-image triangles:

h'/h = -v/u

The negative sign appears because a real image is inverted.

Step 2: Focus triangle relation

Using the ray parallel to principal axis and reflecting through focus:

h'/h = -(v-f)/f

Step 3: Equate ratios

-v/u = -(v-f)/f

Cancel negative signs.

Step 4: Rearrangement

vf = u(v-f)uv = uf + vf1/f = 1/v + 1/u

Real Image Case

For a concave mirror forming a real image, u, v and f are usually negative. The formula remains the same after signs are inserted.

Virtual Image Case

For a concave mirror with object between F and P, v is positive because the image is behind the mirror. The same formula still works.

5–8. Magnification and Image Height Formula

Magnification

Magnification tells how many times the image is compared with the object.

m = h'/h

Linear Magnification

Using similar triangles:

m = -v/u

Relation Between m, u and v

m = -v/u

If |m| > 1, image is magnified. If |m| < 1, image is diminished.

Image Height Formula

h'/h = -v/uh' = -vh/u

This is useful when image height is asked directly.

Positive Magnification

Positive m means image is erect. Virtual images usually have positive magnification.

Negative Magnification

Negative m means image is inverted. Real images usually have negative magnification.

9. Graphical Interpretation

Magnification can be understood from object and image heights. Similar triangles decide the ratio of h' to h.

Principal axis Object h Real inverted image: m < 0 Virtual erect image: m > 0 Sign of m tells orientation; magnitude of m tells enlarged or diminished image.

Graphical meaning of positive and negative magnification.

10. Common Mistakes

Sign convention
Putting u positive for a real object in front of the mirror.
Image distance
Forgetting that virtual image behind the mirror has positive v.
Focal length
Using positive f for concave mirror.
Magnification sign
Ignoring the negative sign in m = -v/u.
Height sign
Calling an inverted image positive height.
Unit mismatch
Mixing cm and m in the same calculation.

11. Quick Revision Notes

Mirror Formula
  • 1/f = 1/v + 1/u
  • Use signs carefully.
  • All distances from pole P.
Magnification
  • m = h'/h
  • m = -v/u
  • m positive means erect.
Image Size
  • |m| > 1 magnified
  • |m| < 1 diminished
  • h' = -vh/u

12. Formula Sheet

Mirror Formula1/f = 1/v + 1/u
Magnificationm = h'/h
Linear Magnificationm = -v/u
Image Heighth' = -vh/u
Focal Lengthf = uv/(u+v)
Image Distancev = fu/(u-f)
Object Distanceu = fv/(v-f)
Radius RelationR = 2f

Numerical Section

Each numerical includes Given, Formula, Substitution, Calculation and Final Answer.

NEET-level Numericals

NEET-level Numerical 1. A spherical mirror has u = -30 cm and f = -15 cm. Find v, m and image height if h = 4 cm.

Given: u = -30 cm, f = -15 cm, h = 4 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-15)(-30) / (-30 - -15).
  2. Calculate: v = -30.00 cm.
  3. Magnification: m = -(-30.00) / (-30) = -1.00.
  4. Image height: h' = (-1.00)(4) = -4.00 cm.
  5. Since v is negative, the image is real and inverted.
Final Answer: v = -30.00 cm, m = -1.00, h' = -4.00 cm
NEET-level Numerical 2. A spherical mirror has u = -45 cm and f = -15 cm. Find v, m and image height if h = 5 cm.

Given: u = -45 cm, f = -15 cm, h = 5 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-15)(-45) / (-45 - -15).
  2. Calculate: v = -22.50 cm.
  3. Magnification: m = -(-22.50) / (-45) = -0.50.
  4. Image height: h' = (-0.50)(5) = -2.50 cm.
  5. Since v is negative, the image is real and inverted.
Final Answer: v = -22.50 cm, m = -0.50, h' = -2.50 cm
NEET-level Numerical 3. A spherical mirror has u = -20 cm and f = -15 cm. Find v, m and image height if h = 6 cm.

Given: u = -20 cm, f = -15 cm, h = 6 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-15)(-20) / (-20 - -15).
  2. Calculate: v = -60.00 cm.
  3. Magnification: m = -(-60.00) / (-20) = -3.00.
  4. Image height: h' = (-3.00)(6) = -18.00 cm.
  5. Since v is negative, the image is real and inverted.
Final Answer: v = -60.00 cm, m = -3.00, h' = -18.00 cm
NEET-level Numerical 4. A spherical mirror has u = -12 cm and f = -15 cm. Find v, m and image height if h = 3 cm.

Given: u = -12 cm, f = -15 cm, h = 3 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-15)(-12) / (-12 - -15).
  2. Calculate: v = 60.00 cm.
  3. Magnification: m = -(60.00) / (-12) = 5.00.
  4. Image height: h' = (5.00)(3) = 15.00 cm.
  5. Since v is positive, the image is virtual and erect.
Final Answer: v = 60.00 cm, m = 5.00, h' = 15.00 cm
NEET-level Numerical 5. A spherical mirror has u = -60 cm and f = -20 cm. Find v, m and image height if h = 4 cm.

Given: u = -60 cm, f = -20 cm, h = 4 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-20)(-60) / (-60 - -20).
  2. Calculate: v = -30.00 cm.
  3. Magnification: m = -(-30.00) / (-60) = -0.50.
  4. Image height: h' = (-0.50)(4) = -2.00 cm.
  5. Since v is negative, the image is real and inverted.
Final Answer: v = -30.00 cm, m = -0.50, h' = -2.00 cm
NEET-level Numerical 6. A spherical mirror has u = -25 cm and f = -10 cm. Find v, m and image height if h = 5 cm.

Given: u = -25 cm, f = -10 cm, h = 5 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-10)(-25) / (-25 - -10).
  2. Calculate: v = -16.67 cm.
  3. Magnification: m = -(-16.67) / (-25) = -0.67.
  4. Image height: h' = (-0.67)(5) = -3.33 cm.
  5. Since v is negative, the image is real and inverted.
Final Answer: v = -16.67 cm, m = -0.67, h' = -3.33 cm
NEET-level Numerical 7. A spherical mirror has u = -18 cm and f = 20 cm. Find v, m and image height if h = 6 cm.

Given: u = -18 cm, f = 20 cm, h = 6 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (20)(-18) / (-18 - 20).
  2. Calculate: v = 9.47 cm.
  3. Magnification: m = -(9.47) / (-18) = 0.53.
  4. Image height: h' = (0.53)(6) = 3.16 cm.
  5. Since v is positive, the image is virtual and erect.
Final Answer: v = 9.47 cm, m = 0.53, h' = 3.16 cm
NEET-level Numerical 8. A spherical mirror has u = -35 cm and f = 25 cm. Find v, m and image height if h = 3 cm.

Given: u = -35 cm, f = 25 cm, h = 3 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (25)(-35) / (-35 - 25).
  2. Calculate: v = 14.58 cm.
  3. Magnification: m = -(14.58) / (-35) = 0.42.
  4. Image height: h' = (0.42)(3) = 1.25 cm.
  5. Since v is positive, the image is virtual and erect.
Final Answer: v = 14.58 cm, m = 0.42, h' = 1.25 cm
NEET-level Numerical 9. A spherical mirror has u = -16 cm and f = -12 cm. Find v, m and image height if h = 4 cm.

Given: u = -16 cm, f = -12 cm, h = 4 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-12)(-16) / (-16 - -12).
  2. Calculate: v = -48.00 cm.
  3. Magnification: m = -(-48.00) / (-16) = -3.00.
  4. Image height: h' = (-3.00)(4) = -12.00 cm.
  5. Since v is negative, the image is real and inverted.
Final Answer: v = -48.00 cm, m = -3.00, h' = -12.00 cm
NEET-level Numerical 10. A spherical mirror has u = -80 cm and f = -20 cm. Find v, m and image height if h = 5 cm.

Given: u = -80 cm, f = -20 cm, h = 5 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-20)(-80) / (-80 - -20).
  2. Calculate: v = -26.67 cm.
  3. Magnification: m = -(-26.67) / (-80) = -0.33.
  4. Image height: h' = (-0.33)(5) = -1.67 cm.
  5. Since v is negative, the image is real and inverted.
Final Answer: v = -26.67 cm, m = -0.33, h' = -1.67 cm
NEET-level Numerical 11. A spherical mirror has u = -32 cm and f = -16 cm. Find v, m and image height if h = 6 cm.

Given: u = -32 cm, f = -16 cm, h = 6 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-16)(-32) / (-32 - -16).
  2. Calculate: v = -32.00 cm.
  3. Magnification: m = -(-32.00) / (-32) = -1.00.
  4. Image height: h' = (-1.00)(6) = -6.00 cm.
  5. Since v is negative, the image is real and inverted.
Final Answer: v = -32.00 cm, m = -1.00, h' = -6.00 cm
NEET-level Numerical 12. A spherical mirror has u = -47 cm and f = -16 cm. Find v, m and image height if h = 3 cm.

Given: u = -47 cm, f = -16 cm, h = 3 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-16)(-47) / (-47 - -16).
  2. Calculate: v = -24.26 cm.
  3. Magnification: m = -(-24.26) / (-47) = -0.52.
  4. Image height: h' = (-0.52)(3) = -1.55 cm.
  5. Since v is negative, the image is real and inverted.
Final Answer: v = -24.26 cm, m = -0.52, h' = -1.55 cm
NEET-level Numerical 13. A spherical mirror has u = -22 cm and f = -16 cm. Find v, m and image height if h = 4 cm.

Given: u = -22 cm, f = -16 cm, h = 4 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-16)(-22) / (-22 - -16).
  2. Calculate: v = -58.67 cm.
  3. Magnification: m = -(-58.67) / (-22) = -2.67.
  4. Image height: h' = (-2.67)(4) = -10.67 cm.
  5. Since v is negative, the image is real and inverted.
Final Answer: v = -58.67 cm, m = -2.67, h' = -10.67 cm
NEET-level Numerical 14. A spherical mirror has u = -14 cm and f = -16 cm. Find v, m and image height if h = 5 cm.

Given: u = -14 cm, f = -16 cm, h = 5 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-16)(-14) / (-14 - -16).
  2. Calculate: v = 112.00 cm.
  3. Magnification: m = -(112.00) / (-14) = 8.00.
  4. Image height: h' = (8.00)(5) = 40.00 cm.
  5. Since v is positive, the image is virtual and erect.
Final Answer: v = 112.00 cm, m = 8.00, h' = 40.00 cm
NEET-level Numerical 15. A spherical mirror has u = -62 cm and f = -21 cm. Find v, m and image height if h = 6 cm.

Given: u = -62 cm, f = -21 cm, h = 6 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-21)(-62) / (-62 - -21).
  2. Calculate: v = -31.76 cm.
  3. Magnification: m = -(-31.76) / (-62) = -0.51.
  4. Image height: h' = (-0.51)(6) = -3.07 cm.
  5. Since v is negative, the image is real and inverted.
Final Answer: v = -31.76 cm, m = -0.51, h' = -3.07 cm
NEET-level Numerical 16. A spherical mirror has u = -27 cm and f = -11 cm. Find v, m and image height if h = 3 cm.

Given: u = -27 cm, f = -11 cm, h = 3 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-11)(-27) / (-27 - -11).
  2. Calculate: v = -18.56 cm.
  3. Magnification: m = -(-18.56) / (-27) = -0.69.
  4. Image height: h' = (-0.69)(3) = -2.06 cm.
  5. Since v is negative, the image is real and inverted.
Final Answer: v = -18.56 cm, m = -0.69, h' = -2.06 cm
NEET-level Numerical 17. A spherical mirror has u = -20 cm and f = 21 cm. Find v, m and image height if h = 4 cm.

Given: u = -20 cm, f = 21 cm, h = 4 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (21)(-20) / (-20 - 21).
  2. Calculate: v = 10.24 cm.
  3. Magnification: m = -(10.24) / (-20) = 0.51.
  4. Image height: h' = (0.51)(4) = 2.05 cm.
  5. Since v is positive, the image is virtual and erect.
Final Answer: v = 10.24 cm, m = 0.51, h' = 2.05 cm
NEET-level Numerical 18. A spherical mirror has u = -37 cm and f = 26 cm. Find v, m and image height if h = 5 cm.

Given: u = -37 cm, f = 26 cm, h = 5 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (26)(-37) / (-37 - 26).
  2. Calculate: v = 15.27 cm.
  3. Magnification: m = -(15.27) / (-37) = 0.41.
  4. Image height: h' = (0.41)(5) = 2.06 cm.
  5. Since v is positive, the image is virtual and erect.
Final Answer: v = 15.27 cm, m = 0.41, h' = 2.06 cm
NEET-level Numerical 19. A spherical mirror has u = -18 cm and f = -13 cm. Find v, m and image height if h = 6 cm.

Given: u = -18 cm, f = -13 cm, h = 6 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-13)(-18) / (-18 - -13).
  2. Calculate: v = -46.80 cm.
  3. Magnification: m = -(-46.80) / (-18) = -2.60.
  4. Image height: h' = (-2.60)(6) = -15.60 cm.
  5. Since v is negative, the image is real and inverted.
Final Answer: v = -46.80 cm, m = -2.60, h' = -15.60 cm
NEET-level Numerical 20. A spherical mirror has u = -82 cm and f = -21 cm. Find v, m and image height if h = 3 cm.

Given: u = -82 cm, f = -21 cm, h = 3 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-21)(-82) / (-82 - -21).
  2. Calculate: v = -28.23 cm.
  3. Magnification: m = -(-28.23) / (-82) = -0.34.
  4. Image height: h' = (-0.34)(3) = -1.03 cm.
  5. Since v is negative, the image is real and inverted.
Final Answer: v = -28.23 cm, m = -0.34, h' = -1.03 cm
NEET-level Numerical 21. A spherical mirror has u = -34 cm and f = -17 cm. Find v, m and image height if h = 4 cm.

Given: u = -34 cm, f = -17 cm, h = 4 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-17)(-34) / (-34 - -17).
  2. Calculate: v = -34.00 cm.
  3. Magnification: m = -(-34.00) / (-34) = -1.00.
  4. Image height: h' = (-1.00)(4) = -4.00 cm.
  5. Since v is negative, the image is real and inverted.
Final Answer: v = -34.00 cm, m = -1.00, h' = -4.00 cm
NEET-level Numerical 22. A spherical mirror has u = -49 cm and f = -17 cm. Find v, m and image height if h = 5 cm.

Given: u = -49 cm, f = -17 cm, h = 5 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-17)(-49) / (-49 - -17).
  2. Calculate: v = -26.03 cm.
  3. Magnification: m = -(-26.03) / (-49) = -0.53.
  4. Image height: h' = (-0.53)(5) = -2.66 cm.
  5. Since v is negative, the image is real and inverted.
Final Answer: v = -26.03 cm, m = -0.53, h' = -2.66 cm
NEET-level Numerical 23. A spherical mirror has u = -24 cm and f = -17 cm. Find v, m and image height if h = 6 cm.

Given: u = -24 cm, f = -17 cm, h = 6 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-17)(-24) / (-24 - -17).
  2. Calculate: v = -58.29 cm.
  3. Magnification: m = -(-58.29) / (-24) = -2.43.
  4. Image height: h' = (-2.43)(6) = -14.57 cm.
  5. Since v is negative, the image is real and inverted.
Final Answer: v = -58.29 cm, m = -2.43, h' = -14.57 cm
NEET-level Numerical 24. A spherical mirror has u = -16 cm and f = -17 cm. Find v, m and image height if h = 3 cm.

Given: u = -16 cm, f = -17 cm, h = 3 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-17)(-16) / (-16 - -17).
  2. Calculate: v = 272.00 cm.
  3. Magnification: m = -(272.00) / (-16) = 17.00.
  4. Image height: h' = (17.00)(3) = 51.00 cm.
  5. Since v is positive, the image is virtual and erect.
Final Answer: v = 272.00 cm, m = 17.00, h' = 51.00 cm
NEET-level Numerical 25. A spherical mirror has u = -64 cm and f = -22 cm. Find v, m and image height if h = 4 cm.

Given: u = -64 cm, f = -22 cm, h = 4 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-22)(-64) / (-64 - -22).
  2. Calculate: v = -33.52 cm.
  3. Magnification: m = -(-33.52) / (-64) = -0.52.
  4. Image height: h' = (-0.52)(4) = -2.10 cm.
  5. Since v is negative, the image is real and inverted.
Final Answer: v = -33.52 cm, m = -0.52, h' = -2.10 cm
NEET-level Numerical 26. A spherical mirror has u = -29 cm and f = -12 cm. Find v, m and image height if h = 5 cm.

Given: u = -29 cm, f = -12 cm, h = 5 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-12)(-29) / (-29 - -12).
  2. Calculate: v = -20.47 cm.
  3. Magnification: m = -(-20.47) / (-29) = -0.71.
  4. Image height: h' = (-0.71)(5) = -3.53 cm.
  5. Since v is negative, the image is real and inverted.
Final Answer: v = -20.47 cm, m = -0.71, h' = -3.53 cm
NEET-level Numerical 27. A spherical mirror has u = -22 cm and f = 22 cm. Find v, m and image height if h = 6 cm.

Given: u = -22 cm, f = 22 cm, h = 6 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (22)(-22) / (-22 - 22).
  2. Calculate: v = 11.00 cm.
  3. Magnification: m = -(11.00) / (-22) = 0.50.
  4. Image height: h' = (0.50)(6) = 3.00 cm.
  5. Since v is positive, the image is virtual and erect.
Final Answer: v = 11.00 cm, m = 0.50, h' = 3.00 cm
NEET-level Numerical 28. A spherical mirror has u = -39 cm and f = 27 cm. Find v, m and image height if h = 3 cm.

Given: u = -39 cm, f = 27 cm, h = 3 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (27)(-39) / (-39 - 27).
  2. Calculate: v = 15.95 cm.
  3. Magnification: m = -(15.95) / (-39) = 0.41.
  4. Image height: h' = (0.41)(3) = 1.23 cm.
  5. Since v is positive, the image is virtual and erect.
Final Answer: v = 15.95 cm, m = 0.41, h' = 1.23 cm
NEET-level Numerical 29. A spherical mirror has u = -20 cm and f = -14 cm. Find v, m and image height if h = 4 cm.

Given: u = -20 cm, f = -14 cm, h = 4 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-14)(-20) / (-20 - -14).
  2. Calculate: v = -46.67 cm.
  3. Magnification: m = -(-46.67) / (-20) = -2.33.
  4. Image height: h' = (-2.33)(4) = -9.33 cm.
  5. Since v is negative, the image is real and inverted.
Final Answer: v = -46.67 cm, m = -2.33, h' = -9.33 cm
NEET-level Numerical 30. A spherical mirror has u = -84 cm and f = -22 cm. Find v, m and image height if h = 5 cm.

Given: u = -84 cm, f = -22 cm, h = 5 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-22)(-84) / (-84 - -22).
  2. Calculate: v = -29.81 cm.
  3. Magnification: m = -(-29.81) / (-84) = -0.35.
  4. Image height: h' = (-0.35)(5) = -1.77 cm.
  5. Since v is negative, the image is real and inverted.
Final Answer: v = -29.81 cm, m = -0.35, h' = -1.77 cm

JEE Main Numericals

JEE Main Numerical 1. A spherical mirror has u = -30 cm and f = -15 cm. Find v, m and image height if h = 4 cm.

Given: u = -30 cm, f = -15 cm, h = 4 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-15)(-30) / (-30 - -15).
  2. Calculate: v = -30.00 cm.
  3. Magnification: m = -(-30.00) / (-30) = -1.00.
  4. Image height: h' = (-1.00)(4) = -4.00 cm.
  5. Since v is negative, the image is real and inverted.
Final Answer: v = -30.00 cm, m = -1.00, h' = -4.00 cm
JEE Main Numerical 2. A spherical mirror has u = -45 cm and f = -15 cm. Find v, m and image height if h = 5 cm.

Given: u = -45 cm, f = -15 cm, h = 5 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-15)(-45) / (-45 - -15).
  2. Calculate: v = -22.50 cm.
  3. Magnification: m = -(-22.50) / (-45) = -0.50.
  4. Image height: h' = (-0.50)(5) = -2.50 cm.
  5. Since v is negative, the image is real and inverted.
Final Answer: v = -22.50 cm, m = -0.50, h' = -2.50 cm
JEE Main Numerical 3. A spherical mirror has u = -20 cm and f = -15 cm. Find v, m and image height if h = 6 cm.

Given: u = -20 cm, f = -15 cm, h = 6 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-15)(-20) / (-20 - -15).
  2. Calculate: v = -60.00 cm.
  3. Magnification: m = -(-60.00) / (-20) = -3.00.
  4. Image height: h' = (-3.00)(6) = -18.00 cm.
  5. Since v is negative, the image is real and inverted.
Final Answer: v = -60.00 cm, m = -3.00, h' = -18.00 cm
JEE Main Numerical 4. A spherical mirror has u = -12 cm and f = -15 cm. Find v, m and image height if h = 3 cm.

Given: u = -12 cm, f = -15 cm, h = 3 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-15)(-12) / (-12 - -15).
  2. Calculate: v = 60.00 cm.
  3. Magnification: m = -(60.00) / (-12) = 5.00.
  4. Image height: h' = (5.00)(3) = 15.00 cm.
  5. Since v is positive, the image is virtual and erect.
Final Answer: v = 60.00 cm, m = 5.00, h' = 15.00 cm
JEE Main Numerical 5. A spherical mirror has u = -60 cm and f = -20 cm. Find v, m and image height if h = 4 cm.

Given: u = -60 cm, f = -20 cm, h = 4 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-20)(-60) / (-60 - -20).
  2. Calculate: v = -30.00 cm.
  3. Magnification: m = -(-30.00) / (-60) = -0.50.
  4. Image height: h' = (-0.50)(4) = -2.00 cm.
  5. Since v is negative, the image is real and inverted.
Final Answer: v = -30.00 cm, m = -0.50, h' = -2.00 cm
JEE Main Numerical 6. A spherical mirror has u = -25 cm and f = -10 cm. Find v, m and image height if h = 5 cm.

Given: u = -25 cm, f = -10 cm, h = 5 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-10)(-25) / (-25 - -10).
  2. Calculate: v = -16.67 cm.
  3. Magnification: m = -(-16.67) / (-25) = -0.67.
  4. Image height: h' = (-0.67)(5) = -3.33 cm.
  5. Since v is negative, the image is real and inverted.
Final Answer: v = -16.67 cm, m = -0.67, h' = -3.33 cm
JEE Main Numerical 7. A spherical mirror has u = -18 cm and f = 20 cm. Find v, m and image height if h = 6 cm.

Given: u = -18 cm, f = 20 cm, h = 6 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (20)(-18) / (-18 - 20).
  2. Calculate: v = 9.47 cm.
  3. Magnification: m = -(9.47) / (-18) = 0.53.
  4. Image height: h' = (0.53)(6) = 3.16 cm.
  5. Since v is positive, the image is virtual and erect.
Final Answer: v = 9.47 cm, m = 0.53, h' = 3.16 cm
JEE Main Numerical 8. A spherical mirror has u = -35 cm and f = 25 cm. Find v, m and image height if h = 3 cm.

Given: u = -35 cm, f = 25 cm, h = 3 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (25)(-35) / (-35 - 25).
  2. Calculate: v = 14.58 cm.
  3. Magnification: m = -(14.58) / (-35) = 0.42.
  4. Image height: h' = (0.42)(3) = 1.25 cm.
  5. Since v is positive, the image is virtual and erect.
Final Answer: v = 14.58 cm, m = 0.42, h' = 1.25 cm
JEE Main Numerical 9. A spherical mirror has u = -16 cm and f = -12 cm. Find v, m and image height if h = 4 cm.

Given: u = -16 cm, f = -12 cm, h = 4 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-12)(-16) / (-16 - -12).
  2. Calculate: v = -48.00 cm.
  3. Magnification: m = -(-48.00) / (-16) = -3.00.
  4. Image height: h' = (-3.00)(4) = -12.00 cm.
  5. Since v is negative, the image is real and inverted.
Final Answer: v = -48.00 cm, m = -3.00, h' = -12.00 cm
JEE Main Numerical 10. A spherical mirror has u = -80 cm and f = -20 cm. Find v, m and image height if h = 5 cm.

Given: u = -80 cm, f = -20 cm, h = 5 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-20)(-80) / (-80 - -20).
  2. Calculate: v = -26.67 cm.
  3. Magnification: m = -(-26.67) / (-80) = -0.33.
  4. Image height: h' = (-0.33)(5) = -1.67 cm.
  5. Since v is negative, the image is real and inverted.
Final Answer: v = -26.67 cm, m = -0.33, h' = -1.67 cm
JEE Main Numerical 11. A spherical mirror has u = -32 cm and f = -16 cm. Find v, m and image height if h = 6 cm.

Given: u = -32 cm, f = -16 cm, h = 6 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-16)(-32) / (-32 - -16).
  2. Calculate: v = -32.00 cm.
  3. Magnification: m = -(-32.00) / (-32) = -1.00.
  4. Image height: h' = (-1.00)(6) = -6.00 cm.
  5. Since v is negative, the image is real and inverted.
Final Answer: v = -32.00 cm, m = -1.00, h' = -6.00 cm
JEE Main Numerical 12. A spherical mirror has u = -47 cm and f = -16 cm. Find v, m and image height if h = 3 cm.

Given: u = -47 cm, f = -16 cm, h = 3 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-16)(-47) / (-47 - -16).
  2. Calculate: v = -24.26 cm.
  3. Magnification: m = -(-24.26) / (-47) = -0.52.
  4. Image height: h' = (-0.52)(3) = -1.55 cm.
  5. Since v is negative, the image is real and inverted.
Final Answer: v = -24.26 cm, m = -0.52, h' = -1.55 cm
JEE Main Numerical 13. A spherical mirror has u = -22 cm and f = -16 cm. Find v, m and image height if h = 4 cm.

Given: u = -22 cm, f = -16 cm, h = 4 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-16)(-22) / (-22 - -16).
  2. Calculate: v = -58.67 cm.
  3. Magnification: m = -(-58.67) / (-22) = -2.67.
  4. Image height: h' = (-2.67)(4) = -10.67 cm.
  5. Since v is negative, the image is real and inverted.
Final Answer: v = -58.67 cm, m = -2.67, h' = -10.67 cm
JEE Main Numerical 14. A spherical mirror has u = -14 cm and f = -16 cm. Find v, m and image height if h = 5 cm.

Given: u = -14 cm, f = -16 cm, h = 5 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-16)(-14) / (-14 - -16).
  2. Calculate: v = 112.00 cm.
  3. Magnification: m = -(112.00) / (-14) = 8.00.
  4. Image height: h' = (8.00)(5) = 40.00 cm.
  5. Since v is positive, the image is virtual and erect.
Final Answer: v = 112.00 cm, m = 8.00, h' = 40.00 cm
JEE Main Numerical 15. A spherical mirror has u = -62 cm and f = -21 cm. Find v, m and image height if h = 6 cm.

Given: u = -62 cm, f = -21 cm, h = 6 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-21)(-62) / (-62 - -21).
  2. Calculate: v = -31.76 cm.
  3. Magnification: m = -(-31.76) / (-62) = -0.51.
  4. Image height: h' = (-0.51)(6) = -3.07 cm.
  5. Since v is negative, the image is real and inverted.
Final Answer: v = -31.76 cm, m = -0.51, h' = -3.07 cm
JEE Main Numerical 16. A spherical mirror has u = -27 cm and f = -11 cm. Find v, m and image height if h = 3 cm.

Given: u = -27 cm, f = -11 cm, h = 3 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-11)(-27) / (-27 - -11).
  2. Calculate: v = -18.56 cm.
  3. Magnification: m = -(-18.56) / (-27) = -0.69.
  4. Image height: h' = (-0.69)(3) = -2.06 cm.
  5. Since v is negative, the image is real and inverted.
Final Answer: v = -18.56 cm, m = -0.69, h' = -2.06 cm
JEE Main Numerical 17. A spherical mirror has u = -20 cm and f = 21 cm. Find v, m and image height if h = 4 cm.

Given: u = -20 cm, f = 21 cm, h = 4 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (21)(-20) / (-20 - 21).
  2. Calculate: v = 10.24 cm.
  3. Magnification: m = -(10.24) / (-20) = 0.51.
  4. Image height: h' = (0.51)(4) = 2.05 cm.
  5. Since v is positive, the image is virtual and erect.
Final Answer: v = 10.24 cm, m = 0.51, h' = 2.05 cm
JEE Main Numerical 18. A spherical mirror has u = -37 cm and f = 26 cm. Find v, m and image height if h = 5 cm.

Given: u = -37 cm, f = 26 cm, h = 5 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (26)(-37) / (-37 - 26).
  2. Calculate: v = 15.27 cm.
  3. Magnification: m = -(15.27) / (-37) = 0.41.
  4. Image height: h' = (0.41)(5) = 2.06 cm.
  5. Since v is positive, the image is virtual and erect.
Final Answer: v = 15.27 cm, m = 0.41, h' = 2.06 cm
JEE Main Numerical 19. A spherical mirror has u = -18 cm and f = -13 cm. Find v, m and image height if h = 6 cm.

Given: u = -18 cm, f = -13 cm, h = 6 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-13)(-18) / (-18 - -13).
  2. Calculate: v = -46.80 cm.
  3. Magnification: m = -(-46.80) / (-18) = -2.60.
  4. Image height: h' = (-2.60)(6) = -15.60 cm.
  5. Since v is negative, the image is real and inverted.
Final Answer: v = -46.80 cm, m = -2.60, h' = -15.60 cm
JEE Main Numerical 20. A spherical mirror has u = -82 cm and f = -21 cm. Find v, m and image height if h = 3 cm.

Given: u = -82 cm, f = -21 cm, h = 3 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-21)(-82) / (-82 - -21).
  2. Calculate: v = -28.23 cm.
  3. Magnification: m = -(-28.23) / (-82) = -0.34.
  4. Image height: h' = (-0.34)(3) = -1.03 cm.
  5. Since v is negative, the image is real and inverted.
Final Answer: v = -28.23 cm, m = -0.34, h' = -1.03 cm
JEE Main Numerical 21. A spherical mirror has u = -34 cm and f = -17 cm. Find v, m and image height if h = 4 cm.

Given: u = -34 cm, f = -17 cm, h = 4 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-17)(-34) / (-34 - -17).
  2. Calculate: v = -34.00 cm.
  3. Magnification: m = -(-34.00) / (-34) = -1.00.
  4. Image height: h' = (-1.00)(4) = -4.00 cm.
  5. Since v is negative, the image is real and inverted.
Final Answer: v = -34.00 cm, m = -1.00, h' = -4.00 cm
JEE Main Numerical 22. A spherical mirror has u = -49 cm and f = -17 cm. Find v, m and image height if h = 5 cm.

Given: u = -49 cm, f = -17 cm, h = 5 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-17)(-49) / (-49 - -17).
  2. Calculate: v = -26.03 cm.
  3. Magnification: m = -(-26.03) / (-49) = -0.53.
  4. Image height: h' = (-0.53)(5) = -2.66 cm.
  5. Since v is negative, the image is real and inverted.
Final Answer: v = -26.03 cm, m = -0.53, h' = -2.66 cm
JEE Main Numerical 23. A spherical mirror has u = -24 cm and f = -17 cm. Find v, m and image height if h = 6 cm.

Given: u = -24 cm, f = -17 cm, h = 6 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-17)(-24) / (-24 - -17).
  2. Calculate: v = -58.29 cm.
  3. Magnification: m = -(-58.29) / (-24) = -2.43.
  4. Image height: h' = (-2.43)(6) = -14.57 cm.
  5. Since v is negative, the image is real and inverted.
Final Answer: v = -58.29 cm, m = -2.43, h' = -14.57 cm
JEE Main Numerical 24. A spherical mirror has u = -16 cm and f = -17 cm. Find v, m and image height if h = 3 cm.

Given: u = -16 cm, f = -17 cm, h = 3 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-17)(-16) / (-16 - -17).
  2. Calculate: v = 272.00 cm.
  3. Magnification: m = -(272.00) / (-16) = 17.00.
  4. Image height: h' = (17.00)(3) = 51.00 cm.
  5. Since v is positive, the image is virtual and erect.
Final Answer: v = 272.00 cm, m = 17.00, h' = 51.00 cm
JEE Main Numerical 25. A spherical mirror has u = -64 cm and f = -22 cm. Find v, m and image height if h = 4 cm.

Given: u = -64 cm, f = -22 cm, h = 4 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-22)(-64) / (-64 - -22).
  2. Calculate: v = -33.52 cm.
  3. Magnification: m = -(-33.52) / (-64) = -0.52.
  4. Image height: h' = (-0.52)(4) = -2.10 cm.
  5. Since v is negative, the image is real and inverted.
Final Answer: v = -33.52 cm, m = -0.52, h' = -2.10 cm

JEE Advanced Numericals

JEE Advanced Numerical 1. A spherical mirror has u = -30 cm and f = -15 cm. Find v, m and image height if h = 4 cm.

Given: u = -30 cm, f = -15 cm, h = 4 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-15)(-30) / (-30 - -15).
  2. Calculate: v = -30.00 cm.
  3. Magnification: m = -(-30.00) / (-30) = -1.00.
  4. Image height: h' = (-1.00)(4) = -4.00 cm.
  5. Since v is negative, the image is real and inverted.
Final Answer: v = -30.00 cm, m = -1.00, h' = -4.00 cm
JEE Advanced Numerical 2. A spherical mirror has u = -45 cm and f = -15 cm. Find v, m and image height if h = 5 cm.

Given: u = -45 cm, f = -15 cm, h = 5 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-15)(-45) / (-45 - -15).
  2. Calculate: v = -22.50 cm.
  3. Magnification: m = -(-22.50) / (-45) = -0.50.
  4. Image height: h' = (-0.50)(5) = -2.50 cm.
  5. Since v is negative, the image is real and inverted.
Final Answer: v = -22.50 cm, m = -0.50, h' = -2.50 cm
JEE Advanced Numerical 3. A spherical mirror has u = -20 cm and f = -15 cm. Find v, m and image height if h = 6 cm.

Given: u = -20 cm, f = -15 cm, h = 6 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-15)(-20) / (-20 - -15).
  2. Calculate: v = -60.00 cm.
  3. Magnification: m = -(-60.00) / (-20) = -3.00.
  4. Image height: h' = (-3.00)(6) = -18.00 cm.
  5. Since v is negative, the image is real and inverted.
Final Answer: v = -60.00 cm, m = -3.00, h' = -18.00 cm
JEE Advanced Numerical 4. A spherical mirror has u = -12 cm and f = -15 cm. Find v, m and image height if h = 3 cm.

Given: u = -12 cm, f = -15 cm, h = 3 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-15)(-12) / (-12 - -15).
  2. Calculate: v = 60.00 cm.
  3. Magnification: m = -(60.00) / (-12) = 5.00.
  4. Image height: h' = (5.00)(3) = 15.00 cm.
  5. Since v is positive, the image is virtual and erect.
Final Answer: v = 60.00 cm, m = 5.00, h' = 15.00 cm
JEE Advanced Numerical 5. A spherical mirror has u = -60 cm and f = -20 cm. Find v, m and image height if h = 4 cm.

Given: u = -60 cm, f = -20 cm, h = 4 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-20)(-60) / (-60 - -20).
  2. Calculate: v = -30.00 cm.
  3. Magnification: m = -(-30.00) / (-60) = -0.50.
  4. Image height: h' = (-0.50)(4) = -2.00 cm.
  5. Since v is negative, the image is real and inverted.
Final Answer: v = -30.00 cm, m = -0.50, h' = -2.00 cm
JEE Advanced Numerical 6. A spherical mirror has u = -25 cm and f = -10 cm. Find v, m and image height if h = 5 cm.

Given: u = -25 cm, f = -10 cm, h = 5 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-10)(-25) / (-25 - -10).
  2. Calculate: v = -16.67 cm.
  3. Magnification: m = -(-16.67) / (-25) = -0.67.
  4. Image height: h' = (-0.67)(5) = -3.33 cm.
  5. Since v is negative, the image is real and inverted.
Final Answer: v = -16.67 cm, m = -0.67, h' = -3.33 cm
JEE Advanced Numerical 7. A spherical mirror has u = -18 cm and f = 20 cm. Find v, m and image height if h = 6 cm.

Given: u = -18 cm, f = 20 cm, h = 6 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (20)(-18) / (-18 - 20).
  2. Calculate: v = 9.47 cm.
  3. Magnification: m = -(9.47) / (-18) = 0.53.
  4. Image height: h' = (0.53)(6) = 3.16 cm.
  5. Since v is positive, the image is virtual and erect.
Final Answer: v = 9.47 cm, m = 0.53, h' = 3.16 cm
JEE Advanced Numerical 8. A spherical mirror has u = -35 cm and f = 25 cm. Find v, m and image height if h = 3 cm.

Given: u = -35 cm, f = 25 cm, h = 3 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (25)(-35) / (-35 - 25).
  2. Calculate: v = 14.58 cm.
  3. Magnification: m = -(14.58) / (-35) = 0.42.
  4. Image height: h' = (0.42)(3) = 1.25 cm.
  5. Since v is positive, the image is virtual and erect.
Final Answer: v = 14.58 cm, m = 0.42, h' = 1.25 cm
JEE Advanced Numerical 9. A spherical mirror has u = -16 cm and f = -12 cm. Find v, m and image height if h = 4 cm.

Given: u = -16 cm, f = -12 cm, h = 4 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-12)(-16) / (-16 - -12).
  2. Calculate: v = -48.00 cm.
  3. Magnification: m = -(-48.00) / (-16) = -3.00.
  4. Image height: h' = (-3.00)(4) = -12.00 cm.
  5. Since v is negative, the image is real and inverted.
Final Answer: v = -48.00 cm, m = -3.00, h' = -12.00 cm
JEE Advanced Numerical 10. A spherical mirror has u = -80 cm and f = -20 cm. Find v, m and image height if h = 5 cm.

Given: u = -80 cm, f = -20 cm, h = 5 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-20)(-80) / (-80 - -20).
  2. Calculate: v = -26.67 cm.
  3. Magnification: m = -(-26.67) / (-80) = -0.33.
  4. Image height: h' = (-0.33)(5) = -1.67 cm.
  5. Since v is negative, the image is real and inverted.
Final Answer: v = -26.67 cm, m = -0.33, h' = -1.67 cm
JEE Advanced Numerical 11. A spherical mirror has u = -32 cm and f = -16 cm. Find v, m and image height if h = 6 cm.

Given: u = -32 cm, f = -16 cm, h = 6 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-16)(-32) / (-32 - -16).
  2. Calculate: v = -32.00 cm.
  3. Magnification: m = -(-32.00) / (-32) = -1.00.
  4. Image height: h' = (-1.00)(6) = -6.00 cm.
  5. Since v is negative, the image is real and inverted.
Final Answer: v = -32.00 cm, m = -1.00, h' = -6.00 cm
JEE Advanced Numerical 12. A spherical mirror has u = -47 cm and f = -16 cm. Find v, m and image height if h = 3 cm.

Given: u = -47 cm, f = -16 cm, h = 3 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-16)(-47) / (-47 - -16).
  2. Calculate: v = -24.26 cm.
  3. Magnification: m = -(-24.26) / (-47) = -0.52.
  4. Image height: h' = (-0.52)(3) = -1.55 cm.
  5. Since v is negative, the image is real and inverted.
Final Answer: v = -24.26 cm, m = -0.52, h' = -1.55 cm
JEE Advanced Numerical 13. A spherical mirror has u = -22 cm and f = -16 cm. Find v, m and image height if h = 4 cm.

Given: u = -22 cm, f = -16 cm, h = 4 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-16)(-22) / (-22 - -16).
  2. Calculate: v = -58.67 cm.
  3. Magnification: m = -(-58.67) / (-22) = -2.67.
  4. Image height: h' = (-2.67)(4) = -10.67 cm.
  5. Since v is negative, the image is real and inverted.
Final Answer: v = -58.67 cm, m = -2.67, h' = -10.67 cm
JEE Advanced Numerical 14. A spherical mirror has u = -14 cm and f = -16 cm. Find v, m and image height if h = 5 cm.

Given: u = -14 cm, f = -16 cm, h = 5 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-16)(-14) / (-14 - -16).
  2. Calculate: v = 112.00 cm.
  3. Magnification: m = -(112.00) / (-14) = 8.00.
  4. Image height: h' = (8.00)(5) = 40.00 cm.
  5. Since v is positive, the image is virtual and erect.
Final Answer: v = 112.00 cm, m = 8.00, h' = 40.00 cm
JEE Advanced Numerical 15. A spherical mirror has u = -62 cm and f = -21 cm. Find v, m and image height if h = 6 cm.

Given: u = -62 cm, f = -21 cm, h = 6 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-21)(-62) / (-62 - -21).
  2. Calculate: v = -31.76 cm.
  3. Magnification: m = -(-31.76) / (-62) = -0.51.
  4. Image height: h' = (-0.51)(6) = -3.07 cm.
  5. Since v is negative, the image is real and inverted.
Final Answer: v = -31.76 cm, m = -0.51, h' = -3.07 cm
JEE Advanced Numerical 16. A spherical mirror has u = -27 cm and f = -11 cm. Find v, m and image height if h = 3 cm.

Given: u = -27 cm, f = -11 cm, h = 3 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-11)(-27) / (-27 - -11).
  2. Calculate: v = -18.56 cm.
  3. Magnification: m = -(-18.56) / (-27) = -0.69.
  4. Image height: h' = (-0.69)(3) = -2.06 cm.
  5. Since v is negative, the image is real and inverted.
Final Answer: v = -18.56 cm, m = -0.69, h' = -2.06 cm
JEE Advanced Numerical 17. A spherical mirror has u = -20 cm and f = 21 cm. Find v, m and image height if h = 4 cm.

Given: u = -20 cm, f = 21 cm, h = 4 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (21)(-20) / (-20 - 21).
  2. Calculate: v = 10.24 cm.
  3. Magnification: m = -(10.24) / (-20) = 0.51.
  4. Image height: h' = (0.51)(4) = 2.05 cm.
  5. Since v is positive, the image is virtual and erect.
Final Answer: v = 10.24 cm, m = 0.51, h' = 2.05 cm
JEE Advanced Numerical 18. A spherical mirror has u = -37 cm and f = 26 cm. Find v, m and image height if h = 5 cm.

Given: u = -37 cm, f = 26 cm, h = 5 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (26)(-37) / (-37 - 26).
  2. Calculate: v = 15.27 cm.
  3. Magnification: m = -(15.27) / (-37) = 0.41.
  4. Image height: h' = (0.41)(5) = 2.06 cm.
  5. Since v is positive, the image is virtual and erect.
Final Answer: v = 15.27 cm, m = 0.41, h' = 2.06 cm
JEE Advanced Numerical 19. A spherical mirror has u = -18 cm and f = -13 cm. Find v, m and image height if h = 6 cm.

Given: u = -18 cm, f = -13 cm, h = 6 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-13)(-18) / (-18 - -13).
  2. Calculate: v = -46.80 cm.
  3. Magnification: m = -(-46.80) / (-18) = -2.60.
  4. Image height: h' = (-2.60)(6) = -15.60 cm.
  5. Since v is negative, the image is real and inverted.
Final Answer: v = -46.80 cm, m = -2.60, h' = -15.60 cm
JEE Advanced Numerical 20. A spherical mirror has u = -82 cm and f = -21 cm. Find v, m and image height if h = 3 cm.

Given: u = -82 cm, f = -21 cm, h = 3 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-21)(-82) / (-82 - -21).
  2. Calculate: v = -28.23 cm.
  3. Magnification: m = -(-28.23) / (-82) = -0.34.
  4. Image height: h' = (-0.34)(3) = -1.03 cm.
  5. Since v is negative, the image is real and inverted.
Final Answer: v = -28.23 cm, m = -0.34, h' = -1.03 cm

CBSE Board Numericals

CBSE Board Numerical 1. A spherical mirror has u = -30 cm and f = -15 cm. Find v, m and image height if h = 4 cm.

Given: u = -30 cm, f = -15 cm, h = 4 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-15)(-30) / (-30 - -15).
  2. Calculate: v = -30.00 cm.
  3. Magnification: m = -(-30.00) / (-30) = -1.00.
  4. Image height: h' = (-1.00)(4) = -4.00 cm.
  5. Since v is negative, the image is real and inverted.
Final Answer: v = -30.00 cm, m = -1.00, h' = -4.00 cm
CBSE Board Numerical 2. A spherical mirror has u = -45 cm and f = -15 cm. Find v, m and image height if h = 5 cm.

Given: u = -45 cm, f = -15 cm, h = 5 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-15)(-45) / (-45 - -15).
  2. Calculate: v = -22.50 cm.
  3. Magnification: m = -(-22.50) / (-45) = -0.50.
  4. Image height: h' = (-0.50)(5) = -2.50 cm.
  5. Since v is negative, the image is real and inverted.
Final Answer: v = -22.50 cm, m = -0.50, h' = -2.50 cm
CBSE Board Numerical 3. A spherical mirror has u = -20 cm and f = -15 cm. Find v, m and image height if h = 6 cm.

Given: u = -20 cm, f = -15 cm, h = 6 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-15)(-20) / (-20 - -15).
  2. Calculate: v = -60.00 cm.
  3. Magnification: m = -(-60.00) / (-20) = -3.00.
  4. Image height: h' = (-3.00)(6) = -18.00 cm.
  5. Since v is negative, the image is real and inverted.
Final Answer: v = -60.00 cm, m = -3.00, h' = -18.00 cm
CBSE Board Numerical 4. A spherical mirror has u = -12 cm and f = -15 cm. Find v, m and image height if h = 3 cm.

Given: u = -12 cm, f = -15 cm, h = 3 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-15)(-12) / (-12 - -15).
  2. Calculate: v = 60.00 cm.
  3. Magnification: m = -(60.00) / (-12) = 5.00.
  4. Image height: h' = (5.00)(3) = 15.00 cm.
  5. Since v is positive, the image is virtual and erect.
Final Answer: v = 60.00 cm, m = 5.00, h' = 15.00 cm
CBSE Board Numerical 5. A spherical mirror has u = -60 cm and f = -20 cm. Find v, m and image height if h = 4 cm.

Given: u = -60 cm, f = -20 cm, h = 4 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-20)(-60) / (-60 - -20).
  2. Calculate: v = -30.00 cm.
  3. Magnification: m = -(-30.00) / (-60) = -0.50.
  4. Image height: h' = (-0.50)(4) = -2.00 cm.
  5. Since v is negative, the image is real and inverted.
Final Answer: v = -30.00 cm, m = -0.50, h' = -2.00 cm
CBSE Board Numerical 6. A spherical mirror has u = -25 cm and f = -10 cm. Find v, m and image height if h = 5 cm.

Given: u = -25 cm, f = -10 cm, h = 5 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-10)(-25) / (-25 - -10).
  2. Calculate: v = -16.67 cm.
  3. Magnification: m = -(-16.67) / (-25) = -0.67.
  4. Image height: h' = (-0.67)(5) = -3.33 cm.
  5. Since v is negative, the image is real and inverted.
Final Answer: v = -16.67 cm, m = -0.67, h' = -3.33 cm
CBSE Board Numerical 7. A spherical mirror has u = -18 cm and f = 20 cm. Find v, m and image height if h = 6 cm.

Given: u = -18 cm, f = 20 cm, h = 6 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (20)(-18) / (-18 - 20).
  2. Calculate: v = 9.47 cm.
  3. Magnification: m = -(9.47) / (-18) = 0.53.
  4. Image height: h' = (0.53)(6) = 3.16 cm.
  5. Since v is positive, the image is virtual and erect.
Final Answer: v = 9.47 cm, m = 0.53, h' = 3.16 cm
CBSE Board Numerical 8. A spherical mirror has u = -35 cm and f = 25 cm. Find v, m and image height if h = 3 cm.

Given: u = -35 cm, f = 25 cm, h = 3 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (25)(-35) / (-35 - 25).
  2. Calculate: v = 14.58 cm.
  3. Magnification: m = -(14.58) / (-35) = 0.42.
  4. Image height: h' = (0.42)(3) = 1.25 cm.
  5. Since v is positive, the image is virtual and erect.
Final Answer: v = 14.58 cm, m = 0.42, h' = 1.25 cm
CBSE Board Numerical 9. A spherical mirror has u = -16 cm and f = -12 cm. Find v, m and image height if h = 4 cm.

Given: u = -16 cm, f = -12 cm, h = 4 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-12)(-16) / (-16 - -12).
  2. Calculate: v = -48.00 cm.
  3. Magnification: m = -(-48.00) / (-16) = -3.00.
  4. Image height: h' = (-3.00)(4) = -12.00 cm.
  5. Since v is negative, the image is real and inverted.
Final Answer: v = -48.00 cm, m = -3.00, h' = -12.00 cm
CBSE Board Numerical 10. A spherical mirror has u = -80 cm and f = -20 cm. Find v, m and image height if h = 5 cm.

Given: u = -80 cm, f = -20 cm, h = 5 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-20)(-80) / (-80 - -20).
  2. Calculate: v = -26.67 cm.
  3. Magnification: m = -(-26.67) / (-80) = -0.33.
  4. Image height: h' = (-0.33)(5) = -1.67 cm.
  5. Since v is negative, the image is real and inverted.
Final Answer: v = -26.67 cm, m = -0.33, h' = -1.67 cm
CBSE Board Numerical 11. A spherical mirror has u = -32 cm and f = -16 cm. Find v, m and image height if h = 6 cm.

Given: u = -32 cm, f = -16 cm, h = 6 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-16)(-32) / (-32 - -16).
  2. Calculate: v = -32.00 cm.
  3. Magnification: m = -(-32.00) / (-32) = -1.00.
  4. Image height: h' = (-1.00)(6) = -6.00 cm.
  5. Since v is negative, the image is real and inverted.
Final Answer: v = -32.00 cm, m = -1.00, h' = -6.00 cm
CBSE Board Numerical 12. A spherical mirror has u = -47 cm and f = -16 cm. Find v, m and image height if h = 3 cm.

Given: u = -47 cm, f = -16 cm, h = 3 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-16)(-47) / (-47 - -16).
  2. Calculate: v = -24.26 cm.
  3. Magnification: m = -(-24.26) / (-47) = -0.52.
  4. Image height: h' = (-0.52)(3) = -1.55 cm.
  5. Since v is negative, the image is real and inverted.
Final Answer: v = -24.26 cm, m = -0.52, h' = -1.55 cm
CBSE Board Numerical 13. A spherical mirror has u = -22 cm and f = -16 cm. Find v, m and image height if h = 4 cm.

Given: u = -22 cm, f = -16 cm, h = 4 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-16)(-22) / (-22 - -16).
  2. Calculate: v = -58.67 cm.
  3. Magnification: m = -(-58.67) / (-22) = -2.67.
  4. Image height: h' = (-2.67)(4) = -10.67 cm.
  5. Since v is negative, the image is real and inverted.
Final Answer: v = -58.67 cm, m = -2.67, h' = -10.67 cm
CBSE Board Numerical 14. A spherical mirror has u = -14 cm and f = -16 cm. Find v, m and image height if h = 5 cm.

Given: u = -14 cm, f = -16 cm, h = 5 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-16)(-14) / (-14 - -16).
  2. Calculate: v = 112.00 cm.
  3. Magnification: m = -(112.00) / (-14) = 8.00.
  4. Image height: h' = (8.00)(5) = 40.00 cm.
  5. Since v is positive, the image is virtual and erect.
Final Answer: v = 112.00 cm, m = 8.00, h' = 40.00 cm
CBSE Board Numerical 15. A spherical mirror has u = -62 cm and f = -21 cm. Find v, m and image height if h = 6 cm.

Given: u = -62 cm, f = -21 cm, h = 6 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-21)(-62) / (-62 - -21).
  2. Calculate: v = -31.76 cm.
  3. Magnification: m = -(-31.76) / (-62) = -0.51.
  4. Image height: h' = (-0.51)(6) = -3.07 cm.
  5. Since v is negative, the image is real and inverted.
Final Answer: v = -31.76 cm, m = -0.51, h' = -3.07 cm

IB Physics Numericals

IB Physics Numerical 1. A spherical mirror has u = -30 cm and f = -15 cm. Find v, m and image height if h = 4 cm.

Given: u = -30 cm, f = -15 cm, h = 4 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-15)(-30) / (-30 - -15).
  2. Calculate: v = -30.00 cm.
  3. Magnification: m = -(-30.00) / (-30) = -1.00.
  4. Image height: h' = (-1.00)(4) = -4.00 cm.
  5. Since v is negative, the image is real and inverted.
Final Answer: v = -30.00 cm, m = -1.00, h' = -4.00 cm
IB Physics Numerical 2. A spherical mirror has u = -45 cm and f = -15 cm. Find v, m and image height if h = 5 cm.

Given: u = -45 cm, f = -15 cm, h = 5 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-15)(-45) / (-45 - -15).
  2. Calculate: v = -22.50 cm.
  3. Magnification: m = -(-22.50) / (-45) = -0.50.
  4. Image height: h' = (-0.50)(5) = -2.50 cm.
  5. Since v is negative, the image is real and inverted.
Final Answer: v = -22.50 cm, m = -0.50, h' = -2.50 cm
IB Physics Numerical 3. A spherical mirror has u = -20 cm and f = -15 cm. Find v, m and image height if h = 6 cm.

Given: u = -20 cm, f = -15 cm, h = 6 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-15)(-20) / (-20 - -15).
  2. Calculate: v = -60.00 cm.
  3. Magnification: m = -(-60.00) / (-20) = -3.00.
  4. Image height: h' = (-3.00)(6) = -18.00 cm.
  5. Since v is negative, the image is real and inverted.
Final Answer: v = -60.00 cm, m = -3.00, h' = -18.00 cm
IB Physics Numerical 4. A spherical mirror has u = -12 cm and f = -15 cm. Find v, m and image height if h = 3 cm.

Given: u = -12 cm, f = -15 cm, h = 3 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-15)(-12) / (-12 - -15).
  2. Calculate: v = 60.00 cm.
  3. Magnification: m = -(60.00) / (-12) = 5.00.
  4. Image height: h' = (5.00)(3) = 15.00 cm.
  5. Since v is positive, the image is virtual and erect.
Final Answer: v = 60.00 cm, m = 5.00, h' = 15.00 cm
IB Physics Numerical 5. A spherical mirror has u = -60 cm and f = -20 cm. Find v, m and image height if h = 4 cm.

Given: u = -60 cm, f = -20 cm, h = 4 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-20)(-60) / (-60 - -20).
  2. Calculate: v = -30.00 cm.
  3. Magnification: m = -(-30.00) / (-60) = -0.50.
  4. Image height: h' = (-0.50)(4) = -2.00 cm.
  5. Since v is negative, the image is real and inverted.
Final Answer: v = -30.00 cm, m = -0.50, h' = -2.00 cm
IB Physics Numerical 6. A spherical mirror has u = -25 cm and f = -10 cm. Find v, m and image height if h = 5 cm.

Given: u = -25 cm, f = -10 cm, h = 5 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-10)(-25) / (-25 - -10).
  2. Calculate: v = -16.67 cm.
  3. Magnification: m = -(-16.67) / (-25) = -0.67.
  4. Image height: h' = (-0.67)(5) = -3.33 cm.
  5. Since v is negative, the image is real and inverted.
Final Answer: v = -16.67 cm, m = -0.67, h' = -3.33 cm
IB Physics Numerical 7. A spherical mirror has u = -18 cm and f = 20 cm. Find v, m and image height if h = 6 cm.

Given: u = -18 cm, f = 20 cm, h = 6 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (20)(-18) / (-18 - 20).
  2. Calculate: v = 9.47 cm.
  3. Magnification: m = -(9.47) / (-18) = 0.53.
  4. Image height: h' = (0.53)(6) = 3.16 cm.
  5. Since v is positive, the image is virtual and erect.
Final Answer: v = 9.47 cm, m = 0.53, h' = 3.16 cm
IB Physics Numerical 8. A spherical mirror has u = -35 cm and f = 25 cm. Find v, m and image height if h = 3 cm.

Given: u = -35 cm, f = 25 cm, h = 3 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (25)(-35) / (-35 - 25).
  2. Calculate: v = 14.58 cm.
  3. Magnification: m = -(14.58) / (-35) = 0.42.
  4. Image height: h' = (0.42)(3) = 1.25 cm.
  5. Since v is positive, the image is virtual and erect.
Final Answer: v = 14.58 cm, m = 0.42, h' = 1.25 cm
IB Physics Numerical 9. A spherical mirror has u = -16 cm and f = -12 cm. Find v, m and image height if h = 4 cm.

Given: u = -16 cm, f = -12 cm, h = 4 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-12)(-16) / (-16 - -12).
  2. Calculate: v = -48.00 cm.
  3. Magnification: m = -(-48.00) / (-16) = -3.00.
  4. Image height: h' = (-3.00)(4) = -12.00 cm.
  5. Since v is negative, the image is real and inverted.
Final Answer: v = -48.00 cm, m = -3.00, h' = -12.00 cm
IB Physics Numerical 10. A spherical mirror has u = -80 cm and f = -20 cm. Find v, m and image height if h = 5 cm.

Given: u = -80 cm, f = -20 cm, h = 5 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-20)(-80) / (-80 - -20).
  2. Calculate: v = -26.67 cm.
  3. Magnification: m = -(-26.67) / (-80) = -0.33.
  4. Image height: h' = (-0.33)(5) = -1.67 cm.
  5. Since v is negative, the image is real and inverted.
Final Answer: v = -26.67 cm, m = -0.33, h' = -1.67 cm
IB Physics Numerical 11. A spherical mirror has u = -32 cm and f = -16 cm. Find v, m and image height if h = 6 cm.

Given: u = -32 cm, f = -16 cm, h = 6 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-16)(-32) / (-32 - -16).
  2. Calculate: v = -32.00 cm.
  3. Magnification: m = -(-32.00) / (-32) = -1.00.
  4. Image height: h' = (-1.00)(6) = -6.00 cm.
  5. Since v is negative, the image is real and inverted.
Final Answer: v = -32.00 cm, m = -1.00, h' = -6.00 cm
IB Physics Numerical 12. A spherical mirror has u = -47 cm and f = -16 cm. Find v, m and image height if h = 3 cm.

Given: u = -47 cm, f = -16 cm, h = 3 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-16)(-47) / (-47 - -16).
  2. Calculate: v = -24.26 cm.
  3. Magnification: m = -(-24.26) / (-47) = -0.52.
  4. Image height: h' = (-0.52)(3) = -1.55 cm.
  5. Since v is negative, the image is real and inverted.
Final Answer: v = -24.26 cm, m = -0.52, h' = -1.55 cm
IB Physics Numerical 13. A spherical mirror has u = -22 cm and f = -16 cm. Find v, m and image height if h = 4 cm.

Given: u = -22 cm, f = -16 cm, h = 4 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-16)(-22) / (-22 - -16).
  2. Calculate: v = -58.67 cm.
  3. Magnification: m = -(-58.67) / (-22) = -2.67.
  4. Image height: h' = (-2.67)(4) = -10.67 cm.
  5. Since v is negative, the image is real and inverted.
Final Answer: v = -58.67 cm, m = -2.67, h' = -10.67 cm
IB Physics Numerical 14. A spherical mirror has u = -14 cm and f = -16 cm. Find v, m and image height if h = 5 cm.

Given: u = -14 cm, f = -16 cm, h = 5 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-16)(-14) / (-14 - -16).
  2. Calculate: v = 112.00 cm.
  3. Magnification: m = -(112.00) / (-14) = 8.00.
  4. Image height: h' = (8.00)(5) = 40.00 cm.
  5. Since v is positive, the image is virtual and erect.
Final Answer: v = 112.00 cm, m = 8.00, h' = 40.00 cm
IB Physics Numerical 15. A spherical mirror has u = -62 cm and f = -21 cm. Find v, m and image height if h = 6 cm.

Given: u = -62 cm, f = -21 cm, h = 6 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-21)(-62) / (-62 - -21).
  2. Calculate: v = -31.76 cm.
  3. Magnification: m = -(-31.76) / (-62) = -0.51.
  4. Image height: h' = (-0.51)(6) = -3.07 cm.
  5. Since v is negative, the image is real and inverted.
Final Answer: v = -31.76 cm, m = -0.51, h' = -3.07 cm

IGCSE Numericals

IGCSE Numerical 1. A spherical mirror has u = -30 cm and f = -15 cm. Find v, m and image height if h = 4 cm.

Given: u = -30 cm, f = -15 cm, h = 4 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-15)(-30) / (-30 - -15).
  2. Calculate: v = -30.00 cm.
  3. Magnification: m = -(-30.00) / (-30) = -1.00.
  4. Image height: h' = (-1.00)(4) = -4.00 cm.
  5. Since v is negative, the image is real and inverted.
Final Answer: v = -30.00 cm, m = -1.00, h' = -4.00 cm
IGCSE Numerical 2. A spherical mirror has u = -45 cm and f = -15 cm. Find v, m and image height if h = 5 cm.

Given: u = -45 cm, f = -15 cm, h = 5 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-15)(-45) / (-45 - -15).
  2. Calculate: v = -22.50 cm.
  3. Magnification: m = -(-22.50) / (-45) = -0.50.
  4. Image height: h' = (-0.50)(5) = -2.50 cm.
  5. Since v is negative, the image is real and inverted.
Final Answer: v = -22.50 cm, m = -0.50, h' = -2.50 cm
IGCSE Numerical 3. A spherical mirror has u = -20 cm and f = -15 cm. Find v, m and image height if h = 6 cm.

Given: u = -20 cm, f = -15 cm, h = 6 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-15)(-20) / (-20 - -15).
  2. Calculate: v = -60.00 cm.
  3. Magnification: m = -(-60.00) / (-20) = -3.00.
  4. Image height: h' = (-3.00)(6) = -18.00 cm.
  5. Since v is negative, the image is real and inverted.
Final Answer: v = -60.00 cm, m = -3.00, h' = -18.00 cm
IGCSE Numerical 4. A spherical mirror has u = -12 cm and f = -15 cm. Find v, m and image height if h = 3 cm.

Given: u = -12 cm, f = -15 cm, h = 3 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-15)(-12) / (-12 - -15).
  2. Calculate: v = 60.00 cm.
  3. Magnification: m = -(60.00) / (-12) = 5.00.
  4. Image height: h' = (5.00)(3) = 15.00 cm.
  5. Since v is positive, the image is virtual and erect.
Final Answer: v = 60.00 cm, m = 5.00, h' = 15.00 cm
IGCSE Numerical 5. A spherical mirror has u = -60 cm and f = -20 cm. Find v, m and image height if h = 4 cm.

Given: u = -60 cm, f = -20 cm, h = 4 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-20)(-60) / (-60 - -20).
  2. Calculate: v = -30.00 cm.
  3. Magnification: m = -(-30.00) / (-60) = -0.50.
  4. Image height: h' = (-0.50)(4) = -2.00 cm.
  5. Since v is negative, the image is real and inverted.
Final Answer: v = -30.00 cm, m = -0.50, h' = -2.00 cm
IGCSE Numerical 6. A spherical mirror has u = -25 cm and f = -10 cm. Find v, m and image height if h = 5 cm.

Given: u = -25 cm, f = -10 cm, h = 5 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-10)(-25) / (-25 - -10).
  2. Calculate: v = -16.67 cm.
  3. Magnification: m = -(-16.67) / (-25) = -0.67.
  4. Image height: h' = (-0.67)(5) = -3.33 cm.
  5. Since v is negative, the image is real and inverted.
Final Answer: v = -16.67 cm, m = -0.67, h' = -3.33 cm
IGCSE Numerical 7. A spherical mirror has u = -18 cm and f = 20 cm. Find v, m and image height if h = 6 cm.

Given: u = -18 cm, f = 20 cm, h = 6 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (20)(-18) / (-18 - 20).
  2. Calculate: v = 9.47 cm.
  3. Magnification: m = -(9.47) / (-18) = 0.53.
  4. Image height: h' = (0.53)(6) = 3.16 cm.
  5. Since v is positive, the image is virtual and erect.
Final Answer: v = 9.47 cm, m = 0.53, h' = 3.16 cm
IGCSE Numerical 8. A spherical mirror has u = -35 cm and f = 25 cm. Find v, m and image height if h = 3 cm.

Given: u = -35 cm, f = 25 cm, h = 3 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (25)(-35) / (-35 - 25).
  2. Calculate: v = 14.58 cm.
  3. Magnification: m = -(14.58) / (-35) = 0.42.
  4. Image height: h' = (0.42)(3) = 1.25 cm.
  5. Since v is positive, the image is virtual and erect.
Final Answer: v = 14.58 cm, m = 0.42, h' = 1.25 cm
IGCSE Numerical 9. A spherical mirror has u = -16 cm and f = -12 cm. Find v, m and image height if h = 4 cm.

Given: u = -16 cm, f = -12 cm, h = 4 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-12)(-16) / (-16 - -12).
  2. Calculate: v = -48.00 cm.
  3. Magnification: m = -(-48.00) / (-16) = -3.00.
  4. Image height: h' = (-3.00)(4) = -12.00 cm.
  5. Since v is negative, the image is real and inverted.
Final Answer: v = -48.00 cm, m = -3.00, h' = -12.00 cm
IGCSE Numerical 10. A spherical mirror has u = -80 cm and f = -20 cm. Find v, m and image height if h = 5 cm.

Given: u = -80 cm, f = -20 cm, h = 5 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-20)(-80) / (-80 - -20).
  2. Calculate: v = -26.67 cm.
  3. Magnification: m = -(-26.67) / (-80) = -0.33.
  4. Image height: h' = (-0.33)(5) = -1.67 cm.
  5. Since v is negative, the image is real and inverted.
Final Answer: v = -26.67 cm, m = -0.33, h' = -1.67 cm
IGCSE Numerical 11. A spherical mirror has u = -32 cm and f = -16 cm. Find v, m and image height if h = 6 cm.

Given: u = -32 cm, f = -16 cm, h = 6 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-16)(-32) / (-32 - -16).
  2. Calculate: v = -32.00 cm.
  3. Magnification: m = -(-32.00) / (-32) = -1.00.
  4. Image height: h' = (-1.00)(6) = -6.00 cm.
  5. Since v is negative, the image is real and inverted.
Final Answer: v = -32.00 cm, m = -1.00, h' = -6.00 cm
IGCSE Numerical 12. A spherical mirror has u = -47 cm and f = -16 cm. Find v, m and image height if h = 3 cm.

Given: u = -47 cm, f = -16 cm, h = 3 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-16)(-47) / (-47 - -16).
  2. Calculate: v = -24.26 cm.
  3. Magnification: m = -(-24.26) / (-47) = -0.52.
  4. Image height: h' = (-0.52)(3) = -1.55 cm.
  5. Since v is negative, the image is real and inverted.
Final Answer: v = -24.26 cm, m = -0.52, h' = -1.55 cm
IGCSE Numerical 13. A spherical mirror has u = -22 cm and f = -16 cm. Find v, m and image height if h = 4 cm.

Given: u = -22 cm, f = -16 cm, h = 4 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-16)(-22) / (-22 - -16).
  2. Calculate: v = -58.67 cm.
  3. Magnification: m = -(-58.67) / (-22) = -2.67.
  4. Image height: h' = (-2.67)(4) = -10.67 cm.
  5. Since v is negative, the image is real and inverted.
Final Answer: v = -58.67 cm, m = -2.67, h' = -10.67 cm
IGCSE Numerical 14. A spherical mirror has u = -14 cm and f = -16 cm. Find v, m and image height if h = 5 cm.

Given: u = -14 cm, f = -16 cm, h = 5 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-16)(-14) / (-14 - -16).
  2. Calculate: v = 112.00 cm.
  3. Magnification: m = -(112.00) / (-14) = 8.00.
  4. Image height: h' = (8.00)(5) = 40.00 cm.
  5. Since v is positive, the image is virtual and erect.
Final Answer: v = 112.00 cm, m = 8.00, h' = 40.00 cm
IGCSE Numerical 15. A spherical mirror has u = -62 cm and f = -21 cm. Find v, m and image height if h = 6 cm.

Given: u = -62 cm, f = -21 cm, h = 6 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-21)(-62) / (-62 - -21).
  2. Calculate: v = -31.76 cm.
  3. Magnification: m = -(-31.76) / (-62) = -0.51.
  4. Image height: h' = (-0.51)(6) = -3.07 cm.
  5. Since v is negative, the image is real and inverted.
Final Answer: v = -31.76 cm, m = -0.51, h' = -3.07 cm

A-Level Numericals

A-Level Numerical 1. A spherical mirror has u = -30 cm and f = -15 cm. Find v, m and image height if h = 4 cm.

Given: u = -30 cm, f = -15 cm, h = 4 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-15)(-30) / (-30 - -15).
  2. Calculate: v = -30.00 cm.
  3. Magnification: m = -(-30.00) / (-30) = -1.00.
  4. Image height: h' = (-1.00)(4) = -4.00 cm.
  5. Since v is negative, the image is real and inverted.
Final Answer: v = -30.00 cm, m = -1.00, h' = -4.00 cm
A-Level Numerical 2. A spherical mirror has u = -45 cm and f = -15 cm. Find v, m and image height if h = 5 cm.

Given: u = -45 cm, f = -15 cm, h = 5 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-15)(-45) / (-45 - -15).
  2. Calculate: v = -22.50 cm.
  3. Magnification: m = -(-22.50) / (-45) = -0.50.
  4. Image height: h' = (-0.50)(5) = -2.50 cm.
  5. Since v is negative, the image is real and inverted.
Final Answer: v = -22.50 cm, m = -0.50, h' = -2.50 cm
A-Level Numerical 3. A spherical mirror has u = -20 cm and f = -15 cm. Find v, m and image height if h = 6 cm.

Given: u = -20 cm, f = -15 cm, h = 6 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-15)(-20) / (-20 - -15).
  2. Calculate: v = -60.00 cm.
  3. Magnification: m = -(-60.00) / (-20) = -3.00.
  4. Image height: h' = (-3.00)(6) = -18.00 cm.
  5. Since v is negative, the image is real and inverted.
Final Answer: v = -60.00 cm, m = -3.00, h' = -18.00 cm
A-Level Numerical 4. A spherical mirror has u = -12 cm and f = -15 cm. Find v, m and image height if h = 3 cm.

Given: u = -12 cm, f = -15 cm, h = 3 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-15)(-12) / (-12 - -15).
  2. Calculate: v = 60.00 cm.
  3. Magnification: m = -(60.00) / (-12) = 5.00.
  4. Image height: h' = (5.00)(3) = 15.00 cm.
  5. Since v is positive, the image is virtual and erect.
Final Answer: v = 60.00 cm, m = 5.00, h' = 15.00 cm
A-Level Numerical 5. A spherical mirror has u = -60 cm and f = -20 cm. Find v, m and image height if h = 4 cm.

Given: u = -60 cm, f = -20 cm, h = 4 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-20)(-60) / (-60 - -20).
  2. Calculate: v = -30.00 cm.
  3. Magnification: m = -(-30.00) / (-60) = -0.50.
  4. Image height: h' = (-0.50)(4) = -2.00 cm.
  5. Since v is negative, the image is real and inverted.
Final Answer: v = -30.00 cm, m = -0.50, h' = -2.00 cm
A-Level Numerical 6. A spherical mirror has u = -25 cm and f = -10 cm. Find v, m and image height if h = 5 cm.

Given: u = -25 cm, f = -10 cm, h = 5 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-10)(-25) / (-25 - -10).
  2. Calculate: v = -16.67 cm.
  3. Magnification: m = -(-16.67) / (-25) = -0.67.
  4. Image height: h' = (-0.67)(5) = -3.33 cm.
  5. Since v is negative, the image is real and inverted.
Final Answer: v = -16.67 cm, m = -0.67, h' = -3.33 cm
A-Level Numerical 7. A spherical mirror has u = -18 cm and f = 20 cm. Find v, m and image height if h = 6 cm.

Given: u = -18 cm, f = 20 cm, h = 6 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (20)(-18) / (-18 - 20).
  2. Calculate: v = 9.47 cm.
  3. Magnification: m = -(9.47) / (-18) = 0.53.
  4. Image height: h' = (0.53)(6) = 3.16 cm.
  5. Since v is positive, the image is virtual and erect.
Final Answer: v = 9.47 cm, m = 0.53, h' = 3.16 cm
A-Level Numerical 8. A spherical mirror has u = -35 cm and f = 25 cm. Find v, m and image height if h = 3 cm.

Given: u = -35 cm, f = 25 cm, h = 3 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (25)(-35) / (-35 - 25).
  2. Calculate: v = 14.58 cm.
  3. Magnification: m = -(14.58) / (-35) = 0.42.
  4. Image height: h' = (0.42)(3) = 1.25 cm.
  5. Since v is positive, the image is virtual and erect.
Final Answer: v = 14.58 cm, m = 0.42, h' = 1.25 cm
A-Level Numerical 9. A spherical mirror has u = -16 cm and f = -12 cm. Find v, m and image height if h = 4 cm.

Given: u = -16 cm, f = -12 cm, h = 4 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-12)(-16) / (-16 - -12).
  2. Calculate: v = -48.00 cm.
  3. Magnification: m = -(-48.00) / (-16) = -3.00.
  4. Image height: h' = (-3.00)(4) = -12.00 cm.
  5. Since v is negative, the image is real and inverted.
Final Answer: v = -48.00 cm, m = -3.00, h' = -12.00 cm
A-Level Numerical 10. A spherical mirror has u = -80 cm and f = -20 cm. Find v, m and image height if h = 5 cm.

Given: u = -80 cm, f = -20 cm, h = 5 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-20)(-80) / (-80 - -20).
  2. Calculate: v = -26.67 cm.
  3. Magnification: m = -(-26.67) / (-80) = -0.33.
  4. Image height: h' = (-0.33)(5) = -1.67 cm.
  5. Since v is negative, the image is real and inverted.
Final Answer: v = -26.67 cm, m = -0.33, h' = -1.67 cm
A-Level Numerical 11. A spherical mirror has u = -32 cm and f = -16 cm. Find v, m and image height if h = 6 cm.

Given: u = -32 cm, f = -16 cm, h = 6 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-16)(-32) / (-32 - -16).
  2. Calculate: v = -32.00 cm.
  3. Magnification: m = -(-32.00) / (-32) = -1.00.
  4. Image height: h' = (-1.00)(6) = -6.00 cm.
  5. Since v is negative, the image is real and inverted.
Final Answer: v = -32.00 cm, m = -1.00, h' = -6.00 cm
A-Level Numerical 12. A spherical mirror has u = -47 cm and f = -16 cm. Find v, m and image height if h = 3 cm.

Given: u = -47 cm, f = -16 cm, h = 3 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-16)(-47) / (-47 - -16).
  2. Calculate: v = -24.26 cm.
  3. Magnification: m = -(-24.26) / (-47) = -0.52.
  4. Image height: h' = (-0.52)(3) = -1.55 cm.
  5. Since v is negative, the image is real and inverted.
Final Answer: v = -24.26 cm, m = -0.52, h' = -1.55 cm
A-Level Numerical 13. A spherical mirror has u = -22 cm and f = -16 cm. Find v, m and image height if h = 4 cm.

Given: u = -22 cm, f = -16 cm, h = 4 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-16)(-22) / (-22 - -16).
  2. Calculate: v = -58.67 cm.
  3. Magnification: m = -(-58.67) / (-22) = -2.67.
  4. Image height: h' = (-2.67)(4) = -10.67 cm.
  5. Since v is negative, the image is real and inverted.
Final Answer: v = -58.67 cm, m = -2.67, h' = -10.67 cm
A-Level Numerical 14. A spherical mirror has u = -14 cm and f = -16 cm. Find v, m and image height if h = 5 cm.

Given: u = -14 cm, f = -16 cm, h = 5 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-16)(-14) / (-14 - -16).
  2. Calculate: v = 112.00 cm.
  3. Magnification: m = -(112.00) / (-14) = 8.00.
  4. Image height: h' = (8.00)(5) = 40.00 cm.
  5. Since v is positive, the image is virtual and erect.
Final Answer: v = 112.00 cm, m = 8.00, h' = 40.00 cm
A-Level Numerical 15. A spherical mirror has u = -62 cm and f = -21 cm. Find v, m and image height if h = 6 cm.

Given: u = -62 cm, f = -21 cm, h = 6 cm

Formula:

1/f = 1/v + 1/uv = fu/(u - f)m = -v/uh' = mh
  1. Substitute in image-distance formula: v = (-21)(-62) / (-62 - -21).
  2. Calculate: v = -31.76 cm.
  3. Magnification: m = -(-31.76) / (-62) = -0.51.
  4. Image height: h' = (-0.51)(6) = -3.07 cm.
  5. Since v is negative, the image is real and inverted.
Final Answer: v = -31.76 cm, m = -0.51, h' = -3.07 cm

PYQ Section

CBSE PYQs

CBSE PYQs 1. A concave mirror forms an erect enlarged image. Where is the object?

The object must be between pole P and focus F.

Answer: The object must be between pole P and focus F.
CBSE PYQs 2. What is the sign of focal length for a convex mirror?

The focal length of a convex mirror is positive.

Answer: The focal length of a convex mirror is positive.
CBSE PYQs 3. If magnification is negative, what does it mean?

The image is inverted.

Answer: The image is inverted.
CBSE PYQs 4. Why is u negative for a real object?

Because the real object is placed opposite to the positive direction of incident light.

Answer: Because the real object is placed opposite to the positive direction of incident light.
CBSE PYQs 5. Which formula gives image height?

h' = -vh/u or h' = mh.

Answer: h' = -vh/u or h' = mh.
CBSE PYQs 6. A concave mirror forms an erect enlarged image. Where is the object?

The object must be between pole P and focus F.

Answer: The object must be between pole P and focus F.
CBSE PYQs 7. What is the sign of focal length for a convex mirror?

The focal length of a convex mirror is positive.

Answer: The focal length of a convex mirror is positive.
CBSE PYQs 8. If magnification is negative, what does it mean?

The image is inverted.

Answer: The image is inverted.
CBSE PYQs 9. Why is u negative for a real object?

Because the real object is placed opposite to the positive direction of incident light.

Answer: Because the real object is placed opposite to the positive direction of incident light.
CBSE PYQs 10. Which formula gives image height?

h' = -vh/u or h' = mh.

Answer: h' = -vh/u or h' = mh.

NEET PYQs

NEET PYQs 1. A concave mirror forms an erect enlarged image. Where is the object?

The object must be between pole P and focus F.

Answer: The object must be between pole P and focus F.
NEET PYQs 2. What is the sign of focal length for a convex mirror?

The focal length of a convex mirror is positive.

Answer: The focal length of a convex mirror is positive.
NEET PYQs 3. If magnification is negative, what does it mean?

The image is inverted.

Answer: The image is inverted.
NEET PYQs 4. Why is u negative for a real object?

Because the real object is placed opposite to the positive direction of incident light.

Answer: Because the real object is placed opposite to the positive direction of incident light.
NEET PYQs 5. Which formula gives image height?

h' = -vh/u or h' = mh.

Answer: h' = -vh/u or h' = mh.
NEET PYQs 6. A concave mirror forms an erect enlarged image. Where is the object?

The object must be between pole P and focus F.

Answer: The object must be between pole P and focus F.
NEET PYQs 7. What is the sign of focal length for a convex mirror?

The focal length of a convex mirror is positive.

Answer: The focal length of a convex mirror is positive.
NEET PYQs 8. If magnification is negative, what does it mean?

The image is inverted.

Answer: The image is inverted.
NEET PYQs 9. Why is u negative for a real object?

Because the real object is placed opposite to the positive direction of incident light.

Answer: Because the real object is placed opposite to the positive direction of incident light.
NEET PYQs 10. Which formula gives image height?

h' = -vh/u or h' = mh.

Answer: h' = -vh/u or h' = mh.

JEE Main PYQs

JEE Main PYQs 1. A concave mirror forms an erect enlarged image. Where is the object?

The object must be between pole P and focus F.

Answer: The object must be between pole P and focus F.
JEE Main PYQs 2. What is the sign of focal length for a convex mirror?

The focal length of a convex mirror is positive.

Answer: The focal length of a convex mirror is positive.
JEE Main PYQs 3. If magnification is negative, what does it mean?

The image is inverted.

Answer: The image is inverted.
JEE Main PYQs 4. Why is u negative for a real object?

Because the real object is placed opposite to the positive direction of incident light.

Answer: Because the real object is placed opposite to the positive direction of incident light.
JEE Main PYQs 5. Which formula gives image height?

h' = -vh/u or h' = mh.

Answer: h' = -vh/u or h' = mh.
JEE Main PYQs 6. A concave mirror forms an erect enlarged image. Where is the object?

The object must be between pole P and focus F.

Answer: The object must be between pole P and focus F.
JEE Main PYQs 7. What is the sign of focal length for a convex mirror?

The focal length of a convex mirror is positive.

Answer: The focal length of a convex mirror is positive.
JEE Main PYQs 8. If magnification is negative, what does it mean?

The image is inverted.

Answer: The image is inverted.
JEE Main PYQs 9. Why is u negative for a real object?

Because the real object is placed opposite to the positive direction of incident light.

Answer: Because the real object is placed opposite to the positive direction of incident light.
JEE Main PYQs 10. Which formula gives image height?

h' = -vh/u or h' = mh.

Answer: h' = -vh/u or h' = mh.

JEE Advanced PYQs

JEE Advanced PYQs 1. A concave mirror forms an erect enlarged image. Where is the object?

The object must be between pole P and focus F.

Answer: The object must be between pole P and focus F.
JEE Advanced PYQs 2. What is the sign of focal length for a convex mirror?

The focal length of a convex mirror is positive.

Answer: The focal length of a convex mirror is positive.
JEE Advanced PYQs 3. If magnification is negative, what does it mean?

The image is inverted.

Answer: The image is inverted.
JEE Advanced PYQs 4. Why is u negative for a real object?

Because the real object is placed opposite to the positive direction of incident light.

Answer: Because the real object is placed opposite to the positive direction of incident light.
JEE Advanced PYQs 5. Which formula gives image height?

h' = -vh/u or h' = mh.

Answer: h' = -vh/u or h' = mh.
JEE Advanced PYQs 6. A concave mirror forms an erect enlarged image. Where is the object?

The object must be between pole P and focus F.

Answer: The object must be between pole P and focus F.
JEE Advanced PYQs 7. What is the sign of focal length for a convex mirror?

The focal length of a convex mirror is positive.

Answer: The focal length of a convex mirror is positive.
JEE Advanced PYQs 8. If magnification is negative, what does it mean?

The image is inverted.

Answer: The image is inverted.
JEE Advanced PYQs 9. Why is u negative for a real object?

Because the real object is placed opposite to the positive direction of incident light.

Answer: Because the real object is placed opposite to the positive direction of incident light.
JEE Advanced PYQs 10. Which formula gives image height?

h' = -vh/u or h' = mh.

Answer: h' = -vh/u or h' = mh.

IB Physics Questions

IB Physics Questions 1. A concave mirror forms an erect enlarged image. Where is the object?

The object must be between pole P and focus F.

Answer: The object must be between pole P and focus F.
IB Physics Questions 2. What is the sign of focal length for a convex mirror?

The focal length of a convex mirror is positive.

Answer: The focal length of a convex mirror is positive.
IB Physics Questions 3. If magnification is negative, what does it mean?

The image is inverted.

Answer: The image is inverted.
IB Physics Questions 4. Why is u negative for a real object?

Because the real object is placed opposite to the positive direction of incident light.

Answer: Because the real object is placed opposite to the positive direction of incident light.
IB Physics Questions 5. Which formula gives image height?

h' = -vh/u or h' = mh.

Answer: h' = -vh/u or h' = mh.
IB Physics Questions 6. A concave mirror forms an erect enlarged image. Where is the object?

The object must be between pole P and focus F.

Answer: The object must be between pole P and focus F.
IB Physics Questions 7. What is the sign of focal length for a convex mirror?

The focal length of a convex mirror is positive.

Answer: The focal length of a convex mirror is positive.
IB Physics Questions 8. If magnification is negative, what does it mean?

The image is inverted.

Answer: The image is inverted.
IB Physics Questions 9. Why is u negative for a real object?

Because the real object is placed opposite to the positive direction of incident light.

Answer: Because the real object is placed opposite to the positive direction of incident light.
IB Physics Questions 10. Which formula gives image height?

h' = -vh/u or h' = mh.

Answer: h' = -vh/u or h' = mh.

ICSE / ISC Questions

ICSE / ISC Questions 1. A concave mirror forms an erect enlarged image. Where is the object?

The object must be between pole P and focus F.

Answer: The object must be between pole P and focus F.
ICSE / ISC Questions 2. What is the sign of focal length for a convex mirror?

The focal length of a convex mirror is positive.

Answer: The focal length of a convex mirror is positive.
ICSE / ISC Questions 3. If magnification is negative, what does it mean?

The image is inverted.

Answer: The image is inverted.
ICSE / ISC Questions 4. Why is u negative for a real object?

Because the real object is placed opposite to the positive direction of incident light.

Answer: Because the real object is placed opposite to the positive direction of incident light.
ICSE / ISC Questions 5. Which formula gives image height?

h' = -vh/u or h' = mh.

Answer: h' = -vh/u or h' = mh.
ICSE / ISC Questions 6. A concave mirror forms an erect enlarged image. Where is the object?

The object must be between pole P and focus F.

Answer: The object must be between pole P and focus F.
ICSE / ISC Questions 7. What is the sign of focal length for a convex mirror?

The focal length of a convex mirror is positive.

Answer: The focal length of a convex mirror is positive.
ICSE / ISC Questions 8. If magnification is negative, what does it mean?

The image is inverted.

Answer: The image is inverted.
ICSE / ISC Questions 9. Why is u negative for a real object?

Because the real object is placed opposite to the positive direction of incident light.

Answer: Because the real object is placed opposite to the positive direction of incident light.
ICSE / ISC Questions 10. Which formula gives image height?

h' = -vh/u or h' = mh.

Answer: h' = -vh/u or h' = mh.

IGCSE Questions

IGCSE Questions 1. A concave mirror forms an erect enlarged image. Where is the object?

The object must be between pole P and focus F.

Answer: The object must be between pole P and focus F.
IGCSE Questions 2. What is the sign of focal length for a convex mirror?

The focal length of a convex mirror is positive.

Answer: The focal length of a convex mirror is positive.
IGCSE Questions 3. If magnification is negative, what does it mean?

The image is inverted.

Answer: The image is inverted.
IGCSE Questions 4. Why is u negative for a real object?

Because the real object is placed opposite to the positive direction of incident light.

Answer: Because the real object is placed opposite to the positive direction of incident light.
IGCSE Questions 5. Which formula gives image height?

h' = -vh/u or h' = mh.

Answer: h' = -vh/u or h' = mh.
IGCSE Questions 6. A concave mirror forms an erect enlarged image. Where is the object?

The object must be between pole P and focus F.

Answer: The object must be between pole P and focus F.
IGCSE Questions 7. What is the sign of focal length for a convex mirror?

The focal length of a convex mirror is positive.

Answer: The focal length of a convex mirror is positive.
IGCSE Questions 8. If magnification is negative, what does it mean?

The image is inverted.

Answer: The image is inverted.
IGCSE Questions 9. Why is u negative for a real object?

Because the real object is placed opposite to the positive direction of incident light.

Answer: Because the real object is placed opposite to the positive direction of incident light.
IGCSE Questions 10. Which formula gives image height?

h' = -vh/u or h' = mh.

Answer: h' = -vh/u or h' = mh.

A-Level Questions

A-Level Questions 1. A concave mirror forms an erect enlarged image. Where is the object?

The object must be between pole P and focus F.

Answer: The object must be between pole P and focus F.
A-Level Questions 2. What is the sign of focal length for a convex mirror?

The focal length of a convex mirror is positive.

Answer: The focal length of a convex mirror is positive.
A-Level Questions 3. If magnification is negative, what does it mean?

The image is inverted.

Answer: The image is inverted.
A-Level Questions 4. Why is u negative for a real object?

Because the real object is placed opposite to the positive direction of incident light.

Answer: Because the real object is placed opposite to the positive direction of incident light.
A-Level Questions 5. Which formula gives image height?

h' = -vh/u or h' = mh.

Answer: h' = -vh/u or h' = mh.
A-Level Questions 6. A concave mirror forms an erect enlarged image. Where is the object?

The object must be between pole P and focus F.

Answer: The object must be between pole P and focus F.
A-Level Questions 7. What is the sign of focal length for a convex mirror?

The focal length of a convex mirror is positive.

Answer: The focal length of a convex mirror is positive.
A-Level Questions 8. If magnification is negative, what does it mean?

The image is inverted.

Answer: The image is inverted.
A-Level Questions 9. Why is u negative for a real object?

Because the real object is placed opposite to the positive direction of incident light.

Answer: Because the real object is placed opposite to the positive direction of incident light.
A-Level Questions 10. Which formula gives image height?

h' = -vh/u or h' = mh.

Answer: h' = -vh/u or h' = mh.

Case Study Section

Case Study 1: Rear-view mirror

Passage: Rear-view mirror uses a convex mirror because it gives an erect, diminished virtual image and a wider field of view.

principal axiswide field of view Silvered side PFC Object extension through F small virtual image Reflected rays diverge; dotted extensions meet behind the convex mirror.

Correct convex mirror ray diagram for rear-view mirror with wide field of view.

Question 1. Which mirror is used here?

Convex mirror; it gives wide field of view with a virtual, erect and diminished image.

Answer: Convex mirror; it gives wide field of view with a virtual, erect and diminished image.
Question 2. Where is the image formed?

The image is formed exactly where the dotted backward extensions meet behind the mirror.

Answer: Behind the mirror, virtual and erect.
Question 3. Which formula connects image size and distance?
m = h'/h = -v/uAnswer: Magnification formula.

Case Study 2: Shaving mirror

Passage: Shaving mirror uses a concave mirror with the object between P and F to produce a virtual, erect and magnified image.

principal axis Silvered side PFC Face / objectbetween P and F enlarged virtual image Concave mirror: object between P and F gives virtual, erect and enlarged image behind mirror.

Correct concave mirror ray diagram for shaving mirror.

Question 1. Which mirror is used here?

Concave mirror; object is placed between P and F, producing a virtual, erect and enlarged image.

Answer: Concave mirror; object is placed between P and F, producing a virtual, erect and enlarged image.
Question 2. Where is the image formed?

The image is formed exactly where the dotted backward extensions meet behind the mirror.

Answer: Behind the mirror, virtual and erect.
Question 3. Which formula connects image size and distance?
m = h'/h = -v/uAnswer: Magnification formula.

Case Study 3: Dentist mirror

Passage: Dentist mirror uses a concave mirror with the object between P and F to produce a virtual, erect and magnified image.

principal axis Silvered side PFC Tooth / objectbetween P and F enlarged virtual image Concave mirror: object between P and F gives virtual, erect and enlarged image behind mirror.

Correct concave mirror ray diagram for dentist mirror.

Question 1. Which mirror is used here?

Concave mirror; object is placed between P and F, producing a virtual, erect and enlarged image.

Answer: Concave mirror; object is placed between P and F, producing a virtual, erect and enlarged image.
Question 2. Where is the image formed?

The image is formed exactly where the dotted backward extensions meet behind the mirror.

Answer: Behind the mirror, virtual and erect.
Question 3. Which formula connects image size and distance?
m = h'/h = -v/uAnswer: Magnification formula.

Case Study 4: Vehicle mirrors

Passage: Vehicle mirrors use a convex mirror because they give erect, diminished virtual images and a wider field of view.

principal axisconvex mirror gives wide-angle vision Silvered side PFC Object extension through F small virtual image Reflected rays diverge; dotted extensions meet behind the convex mirror.

Correct convex mirror ray diagram for vehicle mirrors.

Question 1. Which mirror is used here?

Convex mirror; it gives wide field of view with a virtual, erect and diminished image.

Answer: Convex mirror; it gives wide field of view with a virtual, erect and diminished image.
Question 2. Where is the image formed?

The image is formed exactly where the dotted backward extensions meet behind the mirror.

Answer: Behind the mirror, virtual and erect.
Question 3. Which formula connects image size and distance?
m = h'/h = -v/uAnswer: Magnification formula.

Case Study 5: Security mirrors

Passage: Security mirrors use a convex mirror because they give erect, diminished virtual images and wide-angle coverage.

principal axiswide field of view Silvered side PFC Object extension through F small virtual image Reflected rays diverge; dotted extensions meet behind the convex mirror.

Correct convex mirror ray diagram for security mirrors.

Question 1. Which mirror is used here?

Convex mirror; it gives wide-angle coverage with a virtual, erect and diminished image.

Answer: Convex mirror; it gives wide-angle coverage with a virtual, erect and diminished image.
Question 2. Where is the image formed?

The image is formed exactly where the dotted backward extensions meet behind the mirror.

Answer: Behind the mirror, virtual and erect.
Question 3. Which formula connects image size and distance?
m = h'/h = -v/uAnswer: Magnification formula.

Conceptual Questions

Difficult Conceptual Questions

Difficult Conceptual Questions 1. A concave mirror forms an erect enlarged image. Where is the object?

The object must be between pole P and focus F.

Answer: The object must be between pole P and focus F.
Difficult Conceptual Questions 2. What is the sign of focal length for a convex mirror?

The focal length of a convex mirror is positive.

Answer: The focal length of a convex mirror is positive.
Difficult Conceptual Questions 3. If magnification is negative, what does it mean?

The image is inverted.

Answer: The image is inverted.
Difficult Conceptual Questions 4. Why is u negative for a real object?

Because the real object is placed opposite to the positive direction of incident light.

Answer: Because the real object is placed opposite to the positive direction of incident light.
Difficult Conceptual Questions 5. Which formula gives image height?

h' = -vh/u or h' = mh.

Answer: h' = -vh/u or h' = mh.
Difficult Conceptual Questions 6. A concave mirror forms an erect enlarged image. Where is the object?

The object must be between pole P and focus F.

Answer: The object must be between pole P and focus F.
Difficult Conceptual Questions 7. What is the sign of focal length for a convex mirror?

The focal length of a convex mirror is positive.

Answer: The focal length of a convex mirror is positive.

NEET Conceptual Questions

NEET Conceptual Questions 1. A concave mirror forms an erect enlarged image. Where is the object?

The object must be between pole P and focus F.

Answer: The object must be between pole P and focus F.
NEET Conceptual Questions 2. What is the sign of focal length for a convex mirror?

The focal length of a convex mirror is positive.

Answer: The focal length of a convex mirror is positive.
NEET Conceptual Questions 3. If magnification is negative, what does it mean?

The image is inverted.

Answer: The image is inverted.
NEET Conceptual Questions 4. Why is u negative for a real object?

Because the real object is placed opposite to the positive direction of incident light.

Answer: Because the real object is placed opposite to the positive direction of incident light.
NEET Conceptual Questions 5. Which formula gives image height?

h' = -vh/u or h' = mh.

Answer: h' = -vh/u or h' = mh.
NEET Conceptual Questions 6. A concave mirror forms an erect enlarged image. Where is the object?

The object must be between pole P and focus F.

Answer: The object must be between pole P and focus F.
NEET Conceptual Questions 7. What is the sign of focal length for a convex mirror?

The focal length of a convex mirror is positive.

Answer: The focal length of a convex mirror is positive.

JEE Advanced Conceptual Questions

JEE Advanced Conceptual Questions 1. A concave mirror forms an erect enlarged image. Where is the object?

The object must be between pole P and focus F.

Answer: The object must be between pole P and focus F.
JEE Advanced Conceptual Questions 2. What is the sign of focal length for a convex mirror?

The focal length of a convex mirror is positive.

Answer: The focal length of a convex mirror is positive.
JEE Advanced Conceptual Questions 3. If magnification is negative, what does it mean?

The image is inverted.

Answer: The image is inverted.
JEE Advanced Conceptual Questions 4. Why is u negative for a real object?

Because the real object is placed opposite to the positive direction of incident light.

Answer: Because the real object is placed opposite to the positive direction of incident light.
JEE Advanced Conceptual Questions 5. Which formula gives image height?

h' = -vh/u or h' = mh.

Answer: h' = -vh/u or h' = mh.
JEE Advanced Conceptual Questions 6. A concave mirror forms an erect enlarged image. Where is the object?

The object must be between pole P and focus F.

Answer: The object must be between pole P and focus F.
JEE Advanced Conceptual Questions 7. What is the sign of focal length for a convex mirror?

The focal length of a convex mirror is positive.

Answer: The focal length of a convex mirror is positive.

IB Conceptual Questions

IB Conceptual Questions 1. A concave mirror forms an erect enlarged image. Where is the object?

The object must be between pole P and focus F.

Answer: The object must be between pole P and focus F.
IB Conceptual Questions 2. What is the sign of focal length for a convex mirror?

The focal length of a convex mirror is positive.

Answer: The focal length of a convex mirror is positive.
IB Conceptual Questions 3. If magnification is negative, what does it mean?

The image is inverted.

Answer: The image is inverted.
IB Conceptual Questions 4. Why is u negative for a real object?

Because the real object is placed opposite to the positive direction of incident light.

Answer: Because the real object is placed opposite to the positive direction of incident light.
IB Conceptual Questions 5. Which formula gives image height?

h' = -vh/u or h' = mh.

Answer: h' = -vh/u or h' = mh.
IB Conceptual Questions 6. A concave mirror forms an erect enlarged image. Where is the object?

The object must be between pole P and focus F.

Answer: The object must be between pole P and focus F.
IB Conceptual Questions 7. What is the sign of focal length for a convex mirror?

The focal length of a convex mirror is positive.

Answer: The focal length of a convex mirror is positive.

IGCSE Conceptual Questions

IGCSE Conceptual Questions 1. A concave mirror forms an erect enlarged image. Where is the object?

The object must be between pole P and focus F.

Answer: The object must be between pole P and focus F.
IGCSE Conceptual Questions 2. What is the sign of focal length for a convex mirror?

The focal length of a convex mirror is positive.

Answer: The focal length of a convex mirror is positive.
IGCSE Conceptual Questions 3. If magnification is negative, what does it mean?

The image is inverted.

Answer: The image is inverted.
IGCSE Conceptual Questions 4. Why is u negative for a real object?

Because the real object is placed opposite to the positive direction of incident light.

Answer: Because the real object is placed opposite to the positive direction of incident light.
IGCSE Conceptual Questions 5. Which formula gives image height?

h' = -vh/u or h' = mh.

Answer: h' = -vh/u or h' = mh.
IGCSE Conceptual Questions 6. A concave mirror forms an erect enlarged image. Where is the object?

The object must be between pole P and focus F.

Answer: The object must be between pole P and focus F.
IGCSE Conceptual Questions 7. What is the sign of focal length for a convex mirror?

The focal length of a convex mirror is positive.

Answer: The focal length of a convex mirror is positive.

Need Help in Mirror Formula and Magnification?

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