Solved Examples

About This Section

This is a collection of 30+ solved examples covering every topic of Chapter 9 — spherical mirrors, the mirror formula and magnification, refraction and refractive index, lenses, the lens formula, and the power of a lens. All NCERT in-text and exercise problems are worked here with full sign conventions.

Tips for Board Exam Preparation:

1. Always write the given data with correct signs first.
2. Memorise: mirror formula has a plus; lens formula has a minus.
3. State the nature (real/virtual, erect/inverted, enlarged/diminished) at the end of every numerical.
4. Practise every NCERT exercise.

Let's begin!

Example 1: NCERT — Radius to Focal Length

The radius of curvature of a spherical mirror is 20 cm. What is its focal length?

Solution: $f = \dfrac{R}{2} = \dfrac{20}{2} = 10$ cm.

Example 2: NCERT — Focal Length of a Convex Mirror

Find the focal length of a convex mirror of radius of curvature 32 cm.

Solution: $f = \dfrac{R}{2} = \dfrac{32}{2} = 16$ cm (positive for a convex mirror).

Example 3: NCERT — 3x Magnified Real Image

A concave mirror produces a 3× magnified real image of an object at 10 cm. Where is the image?

Solution: Real image ⇒ $m = -3 = -\dfrac{v}{u}$, with $u = -10$ cm. So $v = -30$ cm — the image is 30 cm in front of the mirror.

Example 4: NCERT — Convex Rear-View Mirror

A convex mirror of R = 3.00 m has a bus at 5.00 m. Find the image.

Solution: $f = +1.50$ m, $u = -5.00$ m. $\dfrac{1}{v} = \dfrac{1}{1.50} + \dfrac{1}{5.00} = \dfrac{6.5}{7.5}$, so $v = +1.15$ m. $m = -\dfrac{v}{u} = +0.23$. Image is virtual, erect, diminished, 1.15 m behind the mirror.

Example 5: NCERT — Concave Mirror, Screen Position

Object 4.0 cm at 25.0 cm before a concave mirror of f = 15.0 cm. Find v, nature and size.

Solution: $u=-25$, $f=-15$. $\dfrac{1}{v} = -\dfrac{1}{15}+\dfrac{1}{25} = \dfrac{-2}{75}$, so $v=-37.5$ cm (real). $h' = -\dfrac{v}{u}h = -6.0$ cm. Real, inverted, enlarged, 6.0 cm tall, screen at 37.5 cm.

Example 6: NCERT Exercise — Convex Mirror, Object at 10 cm

Object 10 cm from a convex mirror of f = 15 cm. Find position and nature.

Solution: $u=-10$, $f=+15$. $\dfrac{1}{v} = \dfrac{1}{15}+\dfrac{1}{10} = \dfrac{5}{30}$, so $v = +6$ cm. Image virtual, erect, diminished, 6 cm behind the mirror.

Example 7: NCERT Exercise — Convex Mirror, Object 5 cm tall

Object 5.0 cm at 20 cm before a convex mirror of R = 30 cm. Find image position, nature and size.

Solution: $f = R/2 = +15$ cm, $u=-20$. $\dfrac{1}{v} = \dfrac{1}{15}+\dfrac{1}{20} = \dfrac{4+3}{60}=\dfrac{7}{60}$, so $v = +8.57$ cm. $m = -\dfrac{v}{u} = +0.43$, $h' = 0.43\times5 = +2.15$ cm. Virtual, erect, diminished, ~2.2 cm, 8.6 cm behind mirror.

Example 8: NCERT Exercise — Concave Mirror, 7 cm object

Object 7.0 cm at 27 cm before a concave mirror of f = 18 cm. Find screen distance, size and nature.

Solution: $u=-27$, $f=-18$. $\dfrac{1}{v} = -\dfrac{1}{18}+\dfrac{1}{27}=\dfrac{-1}{54}$, so $v=-54$ cm (real). $h' = -\dfrac{v}{u}h = -14$ cm. Real, inverted, enlarged, 14 cm tall, screen at 54 cm.

Example 9: NCERT — Define Principal Focus (concave mirror)

Define the principal focus of a concave mirror.

Solution: It is the point on the principal axis where rays parallel to the principal axis converge after reflection from the concave mirror.

Example 10: NCERT — Mirror for Erect, Enlarged Image

Name a mirror giving an erect, enlarged image.

Solution: A concave mirror, with the object placed between the pole and the focus (used as a shaving/dentist's mirror).

Example 11: NCERT — Convex Mirror as Rear-View

Why is a convex mirror preferred as a rear-view mirror?

Solution: It always gives an erect, diminished image and has a wider field of view, so the driver sees a larger area of traffic behind.

Example 12: NCERT Exercise — Concave Mirror Image (virtual, erect, larger)

A concave mirror gives a virtual, erect, larger image. Where is the object? (a) between F and C (b) at C (c) beyond C (d) between P and F.

Solution: (d) between the pole P and the focus F — the only position giving a virtual, erect, enlarged image.

Example 13: NCERT Exercise — Focal Length -15 cm Mirror and Lens

A spherical mirror and a thin lens each have focal length -15 cm. They are likely to be — ?

Solution: Both concave. A negative focal length means a concave mirror and a concave lens (by sign convention).

Example 14: NCERT Exercise — Always-Erect Image Mirror

No matter how far you stand, your image is erect. The mirror is —?

Solution: Either plane or convex — both always give erect images (a concave mirror gives erect images only for objects within F).

Example 15: NCERT Exercise — Magnification of a Plane Mirror

The magnification of a plane mirror is +1. What does this mean?

Solution: The image is the same size as the object ($|m|=1$) and is virtual and erect (positive sign). A plane mirror neither enlarges nor inverts.

Example 16: NCERT — Air to Water Bending

A ray goes from air into water obliquely. Which way does it bend? Why?

Solution: It bends towards the normal, because water is optically denser than air, so light slows down (rarer → denser).

Example 17: NCERT — Speed of Light in Glass

Glass has refractive index 1.50. Find the speed of light in it ($c = 3\times10^8$ m/s).

Solution: $v = \dfrac{c}{n} = \dfrac{3\times10^8}{1.50} = 2\times10^8$ m/s.

Example 18: NCERT — Meaning of n = 2.42 (diamond)

What does 'refractive index of diamond = 2.42' mean?

Solution: Light travels 2.42 times faster in vacuum/air than in diamond (diamond is optically the densest of common media).

Example 19: NCERT — Fastest Medium

In kerosene (1.44), turpentine (1.47) or water (1.33), where is light fastest?

Solution: In water, since it has the lowest refractive index ($v = c/n$; lowest n ⇒ highest speed).

Example 20: NCERT — Highest and Lowest Optical Density

From the table, which medium has the highest and lowest optical density?

Solution: Diamond (n = 2.42) has the highest optical density; air (n = 1.0003) the lowest.

Example 21: Refractive Index from Speeds

Speed of light is $2.4\times10^8$ m/s in medium 1 and $2.0\times10^8$ m/s in medium 2. Find $n_{21}$.

Solution: $n_{21} = \dfrac{v_1}{v_2} = \dfrac{2.4\times10^8}{2.0\times10^8} = 1.2$.

Example 22: Emergent Ray through a Glass Slab

Why is the emergent ray from a glass slab parallel to the incident ray?

Solution: Bending towards the normal at the first face and away at the second (parallel) face are equal and opposite, so the emergent ray is parallel to the incident ray, only laterally shifted.

Example 23: NCERT — Concave Lens Numerical

Concave lens f = 15 cm forms an image at 10 cm. Find object distance and magnification.

Solution: $v=-10$, $f=-15$. $\dfrac{1}{u} = \dfrac{1}{v}-\dfrac{1}{f} = -\dfrac{1}{10}+\dfrac{1}{15} = -\dfrac{1}{30}$, so $u=-30$ cm. $m = \dfrac{v}{u} = +0.33$ (virtual, erect, diminished).

Example 24: NCERT — Convex Lens Numerical

Object 2.0 cm at 15 cm before a convex lens of f = 10 cm. Find v, m, size, nature.

Solution: $u=-15$, $f=+10$. $\dfrac{1}{v} = \dfrac{1}{10}-\dfrac{1}{15}=\dfrac{1}{30}$, so $v=+30$ cm. $m = \dfrac{v}{u} = -2$, $h' = -4.0$ cm. Real, inverted, enlarged (2×), 4 cm, 30 cm on the other side.

Example 25: NCERT Exercise — Converging Lens, Object at 25 cm

Object 5 cm at 25 cm before a converging lens of f = 10 cm. Find image.

Solution: $u=-25$, $f=+10$. $\dfrac{1}{v} = \dfrac{1}{10}-\dfrac{1}{25} = \dfrac{3}{50}$, so $v=+16.7$ cm. $m = \dfrac{v}{u} = -0.67$, $h' = -3.3$ cm. Real, inverted, diminished, ~3.3 cm.

Example 26: NCERT Exercise — Concave Lens, Image at 10 cm

Concave lens f = 15 cm forms an image 10 cm from it. Find the object distance.

Solution: $v=-10$, $f=-15$. $\dfrac{1}{u} = \dfrac{1}{v}-\dfrac{1}{f} = -\dfrac{1}{10}+\dfrac{1}{15} = -\dfrac{1}{30}$, so $u=-30$ cm.

Example 27: NCERT — Define 1 Dioptre

Define 1 dioptre.

Solution: One dioptre is the power of a lens of focal length 1 metre; $1\ \text{D} = 1\ \text{m}^{-1}$.

Example 28: NCERT — Power of Concave Lens (f = 2 m)

Find the power of a concave lens of focal length 2 m.

Solution: $f = -2$ m, so $P = \dfrac{1}{f} = \dfrac{1}{-2} = -0.5$ D.

Example 29: NCERT Exercise — Focal Length from -2.0 D

Find the focal length of a lens of power -2.0 D and its type.

Solution: $f = \dfrac{1}{P} = \dfrac{1}{-2.0} = -0.5$ m $= -50$ cm; it is a concave lens.

Example 30: NCERT Exercise — Corrective Lens +1.5 D

A lens of power +1.5 D. Find focal length and type.

Solution: $f = \dfrac{1}{1.5} = +0.67$ m $\approx +67$ cm; it is a convex (converging) lens.

Example 31: NCERT — Real, Same-Size Image (convex lens)

Where should an object be placed before a convex lens to get a real image of the same size?

Solution: At 2F₁ (twice the focal length). The image forms at 2F₂, same size, real and inverted.

Example 32: Combination of Lenses

Two lenses of +2.0 D and +0.25 D are in contact. Find net power and focal length.

Solution: $P = 2.0 + 0.25 = +2.25$ D; $f = \dfrac{1}{2.25} \approx +0.44$ m (a convex combination).

Example 33: Convex Lens Object at 2F

A convex lens of f = 10 cm has an object at 20 cm. Find v and magnification.

Solution: $u=-20$, $f=+10$. $\dfrac{1}{v} = \dfrac{1}{10}-\dfrac{1}{20} = \dfrac{1}{20}$, so $v=+20$ cm. $m = \dfrac{v}{u} = -1$ (real, inverted, same size — object at 2F).