Light – Reflection & Refraction | Class 10 Science (NCERT Ch 9)

CBSE Class 10 › Natural Phenomena

Light – Reflection & Refraction

NCERT Chapter 9 • Mirrors, Lenses, and Optical Laws

NCERT 2025–26 Foundation Physics Ray Optics

Illustration showing reflection in mirrors and refraction through lenses.

Light travels in straight lines. This fundamental property leads to phenomena like shadows, reflection, and refraction. In this chapter, we explore how light behaves when it strikes mirrors or passes through transparent media like glass and water.

1. Reflection of Light

Reflection is the bouncing back of light when it strikes a polished surface. It follows two specific laws:

First Law: The angle of incidence equals the angle of reflection ( $\angle i = \angle r$ ).
Second Law: The incident ray, the normal, and the reflected ray all lie in the same plane.

Spherical Mirrors

Mirrors whose reflecting surfaces are spherical are called spherical mirrors.

Concave Mirror: Reflecting surface curved inwards. It converges light.
Convex Mirror: Reflecting surface curved outwards. It diverges light.

Diagram showing Pole, Focus, and Center of Curvature for Concave and Convex mirrors. — Light rays converging in a concave mirror vs diverging in a convex mirror.

Key Definition:

The distance between the Pole (P) and the Principal Focus (F) is the Focal Length (f). For small apertures, the radius of curvature $R$ is twice the focal length:
$R = 2f$ .

Image Formation by Concave Mirror

Concave mirrors can form real or virtual images depending on the object’s position.

Object Position	Image Position	Nature of Image
At Infinity	At Focus F	Real, Inverted, Point-sized
Beyond C	Between F and C	Real, Inverted, Diminished
At C	At C	Real, Inverted, Same Size
Between C and F	Beyond C	Real, Inverted, Enlarged
Between P and F	Behind Mirror	Virtual, Erect, Enlarged

Image Formation by Convex Mirror

Convex mirrors always form virtual, erect, and diminished images regardless of object position.

Object Position	Image Position	Nature of Image
At Infinity	At Focus F (behind mirror)	Virtual, Erect, Point-sized
Anywhere in front	Between P and F (behind mirror)	Virtual, Erect, Diminished

Uses of Mirrors:

Concave: Shaving mirrors, headlights, solar furnaces.
Convex: Rear-view mirrors in vehicles (wider field of view).

Q1. Define the principal focus of a concave mirror.

The principal focus of a concave mirror is a point on its principal axis where a beam of light rays, initially parallel to the axis, converges after reflection from the mirror.

Q2. The radius of curvature of a spherical mirror is 20 cm. What is its focal length?

We know that $R = 2f$ .
Given $R = 20 \text{ cm}$ .
So, $f = \dfrac{R}{2} = \dfrac{20}{2} = 10 \text{ cm}$ .

Q3. Name a mirror that can give an erect and enlarged image of an object.

A Concave Mirror. This happens specifically when the object is placed between the Pole (P) and the Principal Focus (F).

Q4. Why do we prefer a convex mirror as a rear-view mirror in vehicles?

(1) They always give an erect (though diminished) image.
(2) They have a wider field of view because they are curved outwards, allowing the driver to see more traffic behind them.

2. Mirror Formula and Magnification

To solve numerical problems, we use the New Cartesian Sign Convention. The pole (P) is the origin. Distances measured in the direction of incident light are positive; distances against it are negative.

Mirror Formula

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

Where:

$u$ = Object distance (always negative)
$v$ = Image distance
$f$ = Focal length (Negative for Concave, Positive for Convex)

Magnification (m)

Magnification determines how much larger or smaller the image is compared to the object.

Magnification

m = \dfrac{h'}{h} = -\dfrac{v}{u}

Note: If $m$ is negative, the image is Real and Inverted. If $m$ is positive, the image is Virtual and Erect.

Q1. Find the focal length of a convex mirror whose radius of curvature is 32 cm.

For any spherical mirror, $f = \dfrac{R}{2}$ .
Given $R = +32 \text{ cm}$ (convex mirror → positive R),
$f = \dfrac{32}{2} = +16 \text{ cm}$ .

Q2. A concave mirror produces three times magnified (enlarged) real image of an object placed at 10 cm in front of it. Where is the image located?

Given: $m = -3$ (real image → negative magnification), $u = -10 \text{ cm}$ .
Using $m = -\dfrac{v}{u}$ :
$-3 = -\dfrac{v}{-10} \Rightarrow v = -30 \text{ cm}$ .
The image is 30 cm in front of the mirror (real).

3. Refraction of Light

Refraction is the bending of light when it passes from one transparent medium to another due to a change in its speed.

Refraction through a Glass Slab

When light passes through a rectangular glass slab, it bends twice. The emergent ray is parallel to the incident ray but laterally displaced.

Laws of Refraction

The incident ray, refracted ray, and normal lie in the same plane.
Snell’s Law: The ratio of sine of angle of incidence to sine of angle of refraction is constant.

\dfrac{\sin i}{\sin r} = n_{21} = \dfrac{n_2}{n_1}

4. Refractive Index

The refractive index ( $n_{21}$ ) represents the speed of light in medium 1 divided by the speed of light in medium 2.

Absolute Refractive Index ( $n_m$ ): Speed of light in air (c) / Speed in medium (v).
$n_m = c/v$

Q1. A ray of light travelling in air enters obliquely into water. Does the light ray bend towards the normal or away from the normal? Why?

It bends towards the normal.
Reason: Water is optically denser than air. When light travels from a rarer medium (air) to a denser medium (water), its speed decreases, causing it to bend towards the normal.

Q2. Light enters from air to glass having refractive index 1.50. What is the speed of light in the glass? The speed of light in vacuum is $3 \times 10^8 \text{ ms}^{-1}$ .

Given: $n_g = 1.50$ , $c = 3 \times 10^8 \text{ m/s}$ .
Formula: $n_g = \dfrac{c}{v} \Rightarrow v = \dfrac{c}{n_g}$
$v = \dfrac{3 \times 10^8}{1.50} = 2 \times 10^8 \text{ m/s}$ .

Q3. Find out, from Table 9.3, the medium having highest optical density. Also find the medium with lowest optical density.

From the standard NCERT table:
Highest Optical Density: Diamond ( $n = 2.42$ ).
Lowest Optical Density: Air ( $n = 1.0003$ ).

Q4. You are given kerosene, turpentine and water. In which of these does the light travel fastest?

Refractive indices are: Water (1.33), Kerosene (1.44), Turpentine (1.47).
Since speed $v = \dfrac{c}{n}$ , speed is inversely proportional to refractive index.
Light travels fastest in Water because it has the lowest refractive index among the three.

Q5. The refractive index of diamond is 2.42. What is the meaning of this statement?

It means that the speed of light in diamond is $\dfrac{1}{2.42}$ times the speed of light in vacuum. (It effectively indicates that light travels very slowly in diamond compared to vacuum).

5. Spherical Lenses

A transparent material bound by at least one spherical surface is a lens.

Convex Lens (Converging): Thicker at the middle. Converges light to a focus.
Concave Lens (Diverging): Thicker at the edges. Diverges light.

Converging action of convex lens and diverging action of concave lens. — Convex lenses focus light (Real focus), while Concave lenses diverge light (Virtual focus).

6. Lens Formula, Magnification & Power

Similar to mirrors, lenses have a formula relating object distance, image distance, and focal length.

Lens Formula: $\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$

Magnification: $m = \dfrac{v}{u}$

Caution: Notice the negative sign in the lens formula and the positive sign in the magnification formula (opposite of mirrors).

Power of a Lens (P)

Power is the degree of convergence or divergence. It is the reciprocal of focal length.

Formula:

$P = \dfrac{1}{f}$
Unit: Dioptre (D). Focal length must be in meters.
Convex lens has Positive Power (+D). Concave lens has Negative Power (-D).

Combination of Lenses:

When lenses are in contact, total power is additive:
$P = P_1 + P_2 + P_3 + \cdots$

Q1. Define 1 dioptre of power of a lens.

1 dioptre is the power of a lens whose focal length is 1 metre. ( $1D = 1m^{-1}$ ).

Q2. A convex lens forms a real and inverted image of a needle at a distance of 50 cm from it. Where is the needle placed if the image is equal to the size of the object? Also, find the power of the lens.

Given: $v = +50 \text{ cm}$ . Image size = Object size means $m = -1$ (Real/Inverted).
Since $m = \dfrac{v}{u} \Rightarrow -1 = \dfrac{50}{u} \Rightarrow u = -50 \text{ cm}$ .
Using lens formula: $\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{50} - \dfrac{1}{-50} = \dfrac{2}{50} = \dfrac{1}{25}$ .
So $f = +25 \text{ cm} = +0.25 \text{ m}$ .
Power $P = \dfrac{1}{0.25} = +4 \text{ D}$ .

Q3. Find the power of a concave lens of focal length 2 m.

Focal length $f = -2 \text{ m}$ (concave lens → negative).
Power $P = \dfrac{1}{f} = \dfrac{1}{-2} = -0.5 \text{ D}$ .

7. Chapter Exercises

Practice these NCERT exercise questions to master the chapter:

Q1. Which one of the following materials cannot be used to make a lens?
(a) Water (b) Glass (c) Plastic (d) Clay

(d) Clay — because it is opaque and does not allow light to pass through.

Q2. The image formed by a concave mirror is observed to be virtual, erect and larger than the object. Where should be the position of the object?
(a) Between F and C
(b) At C
(c) Beyond C
(d) Between P and F

(d) Between P and F — only in this region is the image virtual, erect, and enlarged.

Q3. Where should an object be placed in front of a convex lens to get a real image of the size of the object?
(a) At F
(b) At 2F
(c) At infinity
(d) Between O and F

(b) At 2F — image is real, inverted, and same size at 2F on the other side.

Q4. A spherical mirror and a thin spherical lens have each a focal length of –15 cm. The mirror and the lens are likely to be
(a) both concave.
(b) both convex.
(c) mirror concave, lens convex.
(d) mirror convex, lens concave.

(a) both concave — focal length is negative for concave mirror and concave lens.

Q5. No matter how far you stand from a mirror, your image appears erect. The mirror is likely to be
(a) only plane.
(b) only concave.
(c) only convex.
(d) either plane or convex.

(d) either plane or convex — both always produce erect images.

Q6. Which of the following lenses would you prefer to use while reading small letters found in a dictionary?
(a) Convex lens, f = 50 cm
(b) Concave lens, f = 50 cm
(c) Convex lens, f = 5 cm
(d) Concave lens, f = 5 cm

Q7. We wish to obtain an erect image of an object, using a concave mirror of focal length 15 cm. What should be the range of distance of the object from the mirror? What is the nature of the image? Is the image larger or smaller than the object?

Object must be placed between P and F (i.e., u < 15 cm).
Image is virtual, erect, and enlarged.

Q8. Name the type of mirror used in the following situations.
(a) Headlights of a car.
(b) Side/rear-view mirror of a vehicle.
(c) Solar furnace.

(a) Concave mirror — produces powerful parallel beam.
(b) Convex mirror — wider field of view, erect image.
(c) Concave mirror — concentrates sunlight to produce heat.

Q9. One-half of a convex lens is covered with a black paper. Will this lens produce a complete image of the object?

Yes, the lens will still produce a complete image, but with reduced brightness (intensity).

Q10. An object 5 cm in length is held 25 cm away from a converging lens of focal length 10 cm. Find the position, size and nature of the image.

Given: $h = 5 \text{ cm}, u = -25 \text{ cm}, f = +10 \text{ cm}$ .
Lens formula: $\dfrac{1}{v} = \dfrac{1}{f} + \dfrac{1}{u} = \dfrac{1}{10} - \dfrac{1}{25} = \dfrac{3}{50} \Rightarrow v = +16.67 \text{ cm}$ .
Magnification: $m = \dfrac{v}{u} = \dfrac{16.67}{-25} = -0.67$ .
Image height: $h' = m \times h = -3.33 \text{ cm}$ .
Image is real, inverted, diminished, 16.67 cm on the other side.

Q11. A concave lens of focal length 15 cm forms an image 10 cm from the lens. How far is the object placed from the lens?

Given: $f = -15 \text{ cm}, v = -10 \text{ cm}$ (virtual image).
Lens formula: $\dfrac{1}{u} = \dfrac{1}{v} - \dfrac{1}{f} = -\dfrac{1}{10} + \dfrac{1}{15} = -\dfrac{1}{30} \Rightarrow u = -30 \text{ cm}$ .

Q12. An object is placed at a distance of 10 cm from a convex mirror of focal length 15 cm. Find the position and nature of the image.

Given: $u = -10 \text{ cm}, f = +15 \text{ cm}$ .
Mirror formula: $\dfrac{1}{v} = \dfrac{1}{f} - \dfrac{1}{u} = \dfrac{1}{15} + \dfrac{1}{10} = \dfrac{1}{6} \Rightarrow v = +6 \text{ cm}$ .
Image is virtual, erect, and diminished, 6 cm behind the mirror.

Q13. The magnification produced by a plane mirror is +1. What does this mean?

It means the image is virtual, erect, and of the same size as the object.

Q14. An object 5.0 cm in length is placed at a distance of 20 cm in front of a convex mirror of radius of curvature 30 cm. Find the position, nature and size of the image.

Given: $h = 5 \text{ cm}, u = -20 \text{ cm}, R = +30 \text{ cm} \Rightarrow f = +15 \text{ cm}$ .
Mirror formula: $\dfrac{1}{v} = \dfrac{1}{15} + \dfrac{1}{20} = \dfrac{7}{60} \Rightarrow v = +8.57 \text{ cm}$ .
Magnification: $m = -\dfrac{v}{u} = \dfrac{8.57}{20} = +0.43$ .
Image height: $h' = 0.43 \times 5 = 2.15 \text{ cm}$ .
Image is virtual, erect, diminished.

Q15. An object of size 7.0 cm is placed at 27 cm in front of a concave mirror of focal length 18 cm. At what distance from the mirror should a screen be placed to obtain a sharp image? Find the size and nature of the image.

Given: $h = 7 \text{ cm}, u = -27 \text{ cm}, f = -18 \text{ cm}$ .
Mirror formula: $\dfrac{1}{v} = \dfrac{1}{-18} - \dfrac{1}{-27} = -\dfrac{1}{54} \Rightarrow v = -54 \text{ cm}$ .
Magnification: $m = -\dfrac{v}{u} = -\dfrac{-54}{-27} = -2$ .
Image height: $h' = -2 \times 7 = -14 \text{ cm}$ .
Screen at 54 cm in front; image is real, inverted, enlarged.

Q16. Find the focal length of a lens of power –2.0 D. What type of lens is this?

$f = \dfrac{1}{P} = \dfrac{1}{-2} = -0.5 \text{ m} = -50 \text{ cm}$ .
It is a concave (diverging) lens.

Q17. A doctor has prescribed a corrective lens of power +1.5 D. Find the focal length of the lens. Is the prescribed lens diverging or converging?

$f = \dfrac{1}{1.5} = +0.67 \text{ m} = +67 \text{ cm}$ .
It is a convex (converging) lens — used for hypermetropia.

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