Light Class 10 Numerical Problems: Solved Examples

Mastering Light Class 10 numerical problems is essential for scoring high in Physics. In this guide, we provide step-by-step solutions for mirror and lens formulas using the correct sign conventions.

Solved Light Class 10 Numerical Problems (Mirror & Lens)

The Sign Convention (Very Important)

Solved Light Class 10 Numerical Problems (Mirror & Lens)

Before solving any numerical, it is essential to understand the New Cartesian Sign Convention:

  • All distances are measured from the pole (mirror) or optical center (lens)
  • Distances measured in the direction of incident light → Positive (+)
  • Distances measured opposite to the direction of light → Negative (−)
  • Height above principal axis → Positive (+)
  • Height below principal axis → Negative (−)

👉 Tip: Add a diagram here to improve understanding and increase engagement.

Important Formulas

Use these formulas in almost every numerical:

Mirror Formula:

1f=1v+1u\frac{1}{f} = \frac{1}{v} + \frac{1}{u}

Lens Formula:

1f=1v−1u\frac{1}{f} = \frac{1}{v} – \frac{1}{u}

Magnification:

  • For mirrors:
    m=−vum = -\frac{v}{u}
  • For lenses:
    m=vum = \frac{v}{u}

Mirror Numerical (Solved Example)

Question:

An object is placed at a distance of 10 cm from a convex mirror. The focal length of the mirror is 15 cm. Find the position of the image.

Solution:

Given:
Object distance, u=−10 cmu = -10 \, cm
Focal length, f=+15 cmf = +15 \, cm (positive for convex mirror)

Using mirror formula:

1f=1v+1u\frac{1}{f} = \frac{1}{v} + \frac{1}{u}

Substitute values:

115=1v+1−10\frac{1}{15} = \frac{1}{v} + \frac{1}{-10}

115=1v−110\frac{1}{15} = \frac{1}{v} – \frac{1}{10}

110\frac{1}{10}

1v=115+110\frac{1}{v} = \frac{1}{15} + \frac{1}{10}

Take LCM:

1v=2+330=530=16\frac{1}{v} = \frac{2 + 3}{30} = \frac{5}{30} = \frac{1}{6}

So,

v=+6 cmv = +6 \, cm

Final Answer:

The image is formed 6 cm behind the mirror, and it is virtual and erect.

Quick Calculation Trick (Pro Tip)

You can directly use this shortcut formula to save time:

v=ufu+fv = \frac{uf}{u + f}

👉 This trick works for quick calculations in exams, but only if you apply the correct sign convention.

1. Concave Mirror (Image Position)

Question:
An object is placed 20 cm in front of a concave mirror of focal length 10 cm. Find the image distance.

Solution:
u=−20 cm,  f=−10 cmu = -20 \, cm,\; f = -10 \, cm

Using mirror formula:1f=1v+1u\frac{1}{f} = \frac{1}{v} + \frac{1}{u}1−10=1v+1−20\frac{1}{-10} = \frac{1}{v} + \frac{1}{-20}1−10=1v−120\frac{1}{-10} = \frac{1}{v} – \frac{1}{20}1v=−110+120=−120\frac{1}{v} = -\frac{1}{10} + \frac{1}{20} = -\frac{1}{20}v=−20 cmv = -20 \, cm

Answer: Image at 20 cm in front of mirror (real, inverted)

2. Convex Mirror (Magnification)

Question:
An object is placed 15 cm from a convex mirror (f = 10 cm). Find magnification.

Solution:
u=−15,  f=+10u = -15,\; f = +10110=1v+1−15\frac{1}{10} = \frac{1}{v} + \frac{1}{-15}1v=110+115=16\frac{1}{v} = \frac{1}{10} + \frac{1}{15} = \frac{1}{6}v=6 cmv = 6 \, cmm=−vu=−6−15=0.4m = -\frac{v}{u} = -\frac{6}{-15} = 0.4

Answer: Magnification = +0.4 (virtual, erect, diminished)

3. Concave Mirror (Focal Length)

Question:
An image is formed at 30 cm for an object at 15 cm. Find focal length.

Solution:
u=−15,  v=−30u = -15,\; v = -301f=1v+1u\frac{1}{f} = \frac{1}{v} + \frac{1}{u}1f=−130−115=−110\frac{1}{f} = -\frac{1}{30} – \frac{1}{15} = -\frac{1}{10}f=−10 cmf = -10 \, cm

Answer: f=−10 cmf = -10 \, cm

4. Convex Lens (Image Distance)

Question:
Object at 30 cm from convex lens (f = 15 cm). Find image distance.

Solution:
u=−30,  f=+15u = -30,\; f = +15115=1v−1−30\frac{1}{15} = \frac{1}{v} – \frac{1}{-30}115=1v+130\frac{1}{15} = \frac{1}{v} + \frac{1}{30}1v=115−130=130\frac{1}{v} = \frac{1}{15} – \frac{1}{30} = \frac{1}{30}v=30 cmv = 30 \, cm

Answer: Image at 30 cm (real, inverted)

5. Convex Lens (Magnification)

Question:
Object at 20 cm, image at 40 cm. Find magnification.

Solution:m=vu=40−20=−2m = \frac{v}{u} = \frac{40}{-20} = -2

Answer: Magnification = −2 (real, inverted, enlarged)

6. Concave Lens (Image Distance)

Question:
Object at 25 cm from concave lens (f = −10 cm). Find image distance.

Solution:
u=−25,  f=−10u = -25,\; f = -101−10=1v−1−25\frac{1}{-10} = \frac{1}{v} – \frac{1}{-25}1−10=1v+125\frac{1}{-10} = \frac{1}{v} + \frac{1}{25}1v=−110−125=−750\frac{1}{v} = -\frac{1}{10} – \frac{1}{25} = -\frac{7}{50}v=−7.14 cmv = -7.14 \, cm

Answer: Image at 7.14 cm (virtual, erect)

7. Convex Lens (Focal Length)

Question:
Object at 10 cm, image at 20 cm. Find focal length.

Solution:
u=−10,  v=+20u = -10,\; v = +201f=120−1−10\frac{1}{f} = \frac{1}{20} – \frac{1}{-10}1f=120+110=320\frac{1}{f} = \frac{1}{20} + \frac{1}{10} = \frac{3}{20}f=6.67 cmf = 6.67 \, cm

Answer: f=6.67 cmf = 6.67 \, cm

8. Concave Mirror (Magnification)

Question:
Object at 15 cm, image at 30 cm. Find magnification.

Solution:m=−vu=−−30−15=−2m = -\frac{v}{u} = -\frac{-30}{-15} = -2

Answer: Magnification = −2 (real, inverted)

9. Convex Mirror (Image Distance)

Question:
Object at 12 cm, focal length 8 cm. Find image position.

Solution:
u=−12,  f=+8u = -12,\; f = +818=1v+1−12\frac{1}{8} = \frac{1}{v} + \frac{1}{-12}1v=18+112=524\frac{1}{v} = \frac{1}{8} + \frac{1}{12} = \frac{5}{24}v=4.8 cmv = 4.8 \, cm

Answer: Image at 4.8 cm (virtual)

10. Concave Lens (Magnification)

Question:
Object at 30 cm, image at 10 cm. Find magnification.

Solution:m=vu=−10−30=13m = \frac{v}{u} = \frac{-10}{-30} = \frac{1}{3}

Answer: Magnification = +0.33 (virtual, erect, diminished)

Pro Tip for Students

Always remember:

  • Check sign convention first
  • Then apply formula
  • Then calculate carefully

This simple habit will help you avoid 90% of mistakes in exams.

NCERT TEXTBOOK EXAMPLES | Light Class 10 Numerical Problems

Example 9.1 (Convex Mirror)

Question:
A convex mirror used for rear-view on an automobile has a radius of curvature of 3.00 m. If a bus is located at 5.00 m from this mirror, find the position, nature and size of the image.

Solution:

Radius of curvature, R=3.0 mR = 3.0 \, m
So, f=R2=+1.5 mf = \frac{R}{2} = +1.5 \, m

Object distance, u=−5.0 mu = -5.0 \, m

Using mirror formula:1f=1v+1u\frac{1}{f} = \frac{1}{v} + \frac{1}{u}11.5=1v+1−5\frac{1}{1.5} = \frac{1}{v} + \frac{1}{-5}11.5=1v−15\frac{1}{1.5} = \frac{1}{v} – \frac{1}{5}1v=11.5+15=23+15=1315\frac{1}{v} = \frac{1}{1.5} + \frac{1}{5} = \frac{2}{3} + \frac{1}{5} = \frac{13}{15}v=1.15 mv = 1.15 \, m

Magnification:m=−vu=−1.15−5=+0.23m = -\frac{v}{u} = -\frac{1.15}{-5} = +0.23

Answer:

  • Image position: 1.15 m behind the mirror
  • Nature: Virtual, erect
  • Size: Diminished (m = 0.23)

Example 9.2 (Concave Mirror)

Question:
An object, 4.0 cm in size, is placed at 25.0 cm in front of a concave mirror of focal length 15.0 cm. Find image position, nature and size.

Solution:

u=−25 cm,  f=−15 cm,  h=4.0 cmu = -25 \, cm,\; f = -15 \, cm,\; h = 4.0 \, cm1−15=1v+1−25\frac{1}{-15} = \frac{1}{v} + \frac{1}{-25}1−15=1v−125\frac{1}{-15} = \frac{1}{v} – \frac{1}{25}1v=−115+125=−275\frac{1}{v} = -\frac{1}{15} + \frac{1}{25} = -\frac{2}{75}v=−37.5 cmv = -37.5 \, cm

Magnification:m=−vu=−−37.5−25=−1.5m = -\frac{v}{u} = -\frac{-37.5}{-25} = -1.5

Image height:h′=m×h=−1.5×4=−6 cmh’ = m \times h = -1.5 \times 4 = -6 \, cm

Answer:

  • Image position: 37.5 cm in front of mirror
  • Nature: Real, inverted
  • Size: 6 cm (enlarged)
  • Screen should be placed at 37.5 cm

Example 9.3 (Concave Lens)

Question:
A concave lens has focal length of 15 cm. At what distance should the object be placed so that image is formed at 10 cm? Also find magnification.

Solution:

f=−15 cm,  v=−10 cmf = -15 \, cm,\; v = -10 \, cm

Using lens formula:1f=1v−1u\frac{1}{f} = \frac{1}{v} – \frac{1}{u}1−15=1−10−1u\frac{1}{-15} = \frac{1}{-10} – \frac{1}{u}−115=−110−1u-\frac{1}{15} = -\frac{1}{10} – \frac{1}{u}−115+110=−1u-\frac{1}{15} + \frac{1}{10} = -\frac{1}{u}130=−1u\frac{1}{30} = -\frac{1}{u}u=−30 cmu = -30 \, cm

Magnification:m=vu=−10−30=13m = \frac{v}{u} = \frac{-10}{-30} = \frac{1}{3}

Answer:

  • Object distance: 30 cm in front of lens
  • Magnification: +0.33
  • Nature: Virtual, erect, diminished

Example 9.4 (Convex Lens)

Question:
A 2.0 cm tall object is placed 15 cm from a convex lens (f = 10 cm). Find position, nature, size and magnification.

Solution:

h=2 cm,  u=−15 cm,  f=+10 cmh = 2 \, cm,\; u = -15 \, cm,\; f = +10 \, cm110=1v−1−15\frac{1}{10} = \frac{1}{v} – \frac{1}{-15}110=1v+115\frac{1}{10} = \frac{1}{v} + \frac{1}{15}1v=110−115=130\frac{1}{v} = \frac{1}{10} – \frac{1}{15} = \frac{1}{30}v=30 cmv = 30 \, cm

Magnification:m=vu=30−15=−2m = \frac{v}{u} = \frac{30}{-15} = -2

Image height:h′=m×h=−2×2=−4 cmh’ = m \times h = -2 \times 2 = -4 \, cm

Answer:

  • Image position: 30 cm on opposite side of lens
  • Nature: Real, inverted
  • Size: 4 cm (enlarged)
  • Magnification: −2
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