Light – Reflection and Refraction — Problems & Practice

Chapter 9 | NCERT Class 10 Science

Original practice inspired by NCERT / board style. Full solutions included. Not verbatim textbook exercises.

Easy

Problem 1

The radius of curvature of a concave mirror is $24$ cm. Find its focal length. State the sign of $f$ under New Cartesian Sign Convention if light is incident from the left.

Solution: $f = \frac{R}{2} = \frac{24}{2} = 12 cm$ For a concave mirror, $f = - 12$ cm.

Problem 2

A ray of light goes from air into water. Does it bend towards or away from the normal? Why?

Solution: Water is optically denser than air ( $n_{w} > n_{air}$ ), so light slows down and bends towards the normal.

Problem 3

Find the power of a convex lens of focal length $50$ cm.

Solution: $f = + 0.50 m, P = \frac{1}{f} = \frac{1}{0.50} = + 2.0 D$

Problem 4

State two reasons why convex mirrors are preferred as rear-view mirrors in vehicles.

Solution: (i) They always form a virtual, erect, diminished image.
(ii) They provide a wider field of view than a plane mirror of the same size.

Medium

Problem 5

An object is placed $30$ cm in front of a concave mirror of focal length $15$ cm. Find the image distance and magnification. State the nature of the image.

Solution: $u = - 30$ cm, $f = - 15$ cm. $\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = - \frac{1}{15} - (- \frac{1}{30}) = - \frac{1}{15} + \frac{1}{30} = - \frac{1}{30}$ $v = - 30 cm, m = - \frac{v}{u} = - \frac{- 30}{- 30} = - 1$ Image is real, inverted, same size, at the centre of curvature.

Problem 6

Light enters from air to glass of refractive index $1.50$ . Speed of light in vacuum is $3 \times 1 0^{8} m s^{- 1}$ . Find the speed of light in glass.

Solution: $v = \frac{c}{n} = \frac{3 \times 1 0 ^{8}}{1.50} = 2 \times 1 0^{8} m s^{- 1}$

Problem 7

A $2.0$ cm tall object is placed $20$ cm from a convex lens of focal length $10$ cm. Find the position and size of the image.

Solution: $u = - 20$ cm, $f = + 10$ cm, $h = + 2.0$ cm. $\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{10} + \frac{1}{- 20} = \frac{1}{20} \Rightarrow v = + 20 cm$ $m = \frac{v}{u} = \frac{20}{- 20} = - 1 \Rightarrow h^{'} = mh = - 2.0 cm$ Real, inverted image of same size, $20$ cm on the other side of the lens (at $2 F$ ).

Problem 8

A concave lens forms an image $12$ cm from the lens on the same side as the object. If $f = - 18$ cm, find the object distance and magnification.

Solution: $v = - 12$ cm, $f = - 18$ cm. $\frac{1}{u} = \frac{1}{v} - \frac{1}{f} = - \frac{1}{12} - (- \frac{1}{18}) = - \frac{1}{12} + \frac{1}{18} = - \frac{1}{36}$ $u = - 36 cm, m = \frac{v}{u} = \frac{- 12}{- 36} = + \frac{1}{3}$ Virtual, erect, diminished to one-third.

Hard

Problem 9

A convex mirror of radius of curvature $2.0$ m is used as a rear-view mirror. A truck is $6.0$ m from the mirror. Find the position and magnification of the image.

Solution: $R = + 2.0$ m ⇒ $f = + 1.0$ m; $u = - 6.0$ m. $\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = 1 - (- \frac{1}{6}) = 1 + \frac{1}{6} = \frac{7}{6}$ $v = + \frac{6}{7} m \approx + 0.86 m$ $m = - \frac{v}{u} = - \frac{6/7}{- 6} = + \frac{1}{7} \approx + 0.14$ Virtual, erect, diminished image $0.86$ m behind the mirror.

Problem 10

Two thin lenses of powers $+ 2.5$ D and $- 1.0$ D are placed in contact. Find the net power and the focal length of the combination.

Solution: $P = 2.5 + (- 1.0) = + 1.5 D$ $f = \frac{1}{P} = \frac{1}{1.5} = \frac{2}{3} m \approx 0.67 m (convex equivalent)$

Problem 11

An object $3.0$ cm high is placed $15$ cm from a concave mirror of focal length $10$ cm. Where should a screen be placed to get a sharp image? Find the height of the image.

Solution: $u = - 15$ cm, $f = - 10$ cm, $h = + 3.0$ cm. $\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = - \frac{1}{10} + \frac{1}{15} = - \frac{1}{30} \Rightarrow v = - 30 cm$ Screen $30$ cm in front of the mirror. $m = - \frac{v}{u} = - \frac{- 30}{- 15} = - 2 \Rightarrow h^{'} = - 6.0 cm$ Real, inverted, $6.0$ cm tall.

Self-check

If you missed…	Revisit
Sign of $f$ / $v$	New Cartesian Sign Convention
Wrong formula (+ vs −)	Mirror vs lens formula
Magnification sign	$m = - v / u$ (mirror); $m = v / u$ (lens)
Power units	Convert $f$ to metres
Nature of image	Match signs to image tables