Class 10 Science · Chapter 9

Light - Reflection and Refraction

409 Class 10 Science questions on Light - Reflection and Refraction from the NCERT textbook, NCERT Exemplar and CBSE board papers 2017–2026, each with its answer and a step-by-step solution.

NCERT exercises

Exercises (17) Intext Questions (Page 142) (4) Intext Questions (Page 145) (2) Intext Questions (Page 150) (5) Intext Questions (Page 158) (3)

NCERT Exemplar exercises

Multiple Choice Questions (19) Short Answer Questions (10) Long Answer Questions (9)

CBSE Class 10 Science previous year papers, with answers →

Questions · page 11 of 21

Q201 CBSE 2023
A light ray enters from medium A to medium B as shown in the figure.
1. (a)Which one of the two media is denser w.r.t. other medium? Justify your answer.
2. (b)If the speed of light in medium A is $v_{a}$ and in medium B is $v_{b}$ , what is the refractive index of B with respect to A.
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Medium B is denser because the ray bends towards the normal; $n_{B A} = \frac{v _{a}}{v _{b}}$ .
(a) Medium B is denser than medium A. On entering B the ray bends **towards** the normal, so the angle of refraction is smaller than the angle of incidence, and that happens only when light passes into an optically denser medium. (b) $n_{B A} = \frac{n _{B}}{n _{A}} = \frac{c / v _{b}}{c / v _{a}} = \frac{v _{a}}{v _{b}}$ , that is, the speed of light in A divided by the speed of light in B.

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Q202 CBSE 2023
1. (a)A ray of light starting from diamond is incident on the interface separating diamond and water. Draw a labelled ray diagram to show the refraction of light in this case.
2. (b)Absolute refractive indices of diamond and water are 2.42 and 1.33 respectively. Find the value of refractive index of water w.r.t. diamond.
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The ray bends away from the normal as it passes from diamond into water, and $n_{w d} = 0.55$ .
(a) Diamond is optically denser than water, so the ray bends **away** from the normal as it passes into the water: the angle of refraction is larger than the angle of incidence (see the labelled diagram). (b) $n_{w d} = \frac{n _{w}}{n _{d}} = \frac{1.33}{2.42} = 0.55$ .

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Q203 CBSE 2023
A student wants to project the image of a candle flame on a screen 60 cm in front of a mirror by keeping the candle at a distance of 15 cm from its pole.
1. (a)Name the type of mirror used.
2. (b)Also calculate : (i) Magnification of the image produced (ii) Distance between object and its image (c) Draw a ray diagram to show the image formation.
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A concave mirror; $m = - 4$ ; the object and image are 45 cm apart.
(a) A concave mirror, because only a concave mirror forms a real image of a real object, and only a real image can be caught on a screen. (b)(i) With $u = - 15$ cm and $v = - 60$ cm, $m = - \frac{v}{u} = - \frac{- 60}{- 15} = - 4$ : the image is real, inverted and four times as large as the candle flame. (ii) The candle and the image are on the same side of the mirror, so the distance between them is $60 - 15 = 45$ cm. (c) Here $f = - 12$ cm, so F is at $12$ cm and C at $24$ cm, and the candle lies between F and C. A ray parallel to the axis is reflected through F, and a ray through F is reflected parallel to the axis; they meet beyond C, at $60$ cm, forming a real, inverted, enlarged image on the screen, as the ray diagram shows.

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Q204 CBSE 2023

List two differences in the characteristic properties of the virtual images formed by the two types of spherical lenses (concave and convex). How are these characteristics of the two lenses used in the correction of the two common defects of vision namely myopia and hypermetropia ?

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A convex lens gives a magnified virtual image and a concave lens a diminished one; concave corrects myopia and convex corrects hypermetropia.
Differences between the virtual images: (1) Size: the virtual image formed by a convex lens is magnified (enlarged), while that formed by a concave lens is diminished. (2) Position: the virtual image of a convex lens forms on the same side as the object but farther from the lens than the object, while that of a concave lens forms between the principal focus $F_{1}$ and the optical centre, nearer to the lens than the object. (Also, a convex lens forms a virtual image only when the object is between $F_{1}$ and $O$ , whereas a concave lens forms one for every position of the object. Both images are erect.) Myopia: the far point of a myopic eye is nearer than infinity, and the image of a distant object forms in front of the retina. A concave lens of suitable power forms a virtual, erect, diminished image of a distant object at the far point of the eye, nearer to the eye than the object, so the eye sees it clearly; the diverging lens brings the image back onto the retina. Hypermetropia: the near point of a hypermetropic eye is farther than 25 cm, and light from a nearby object is focused behind the retina. A convex lens of suitable power forms a virtual, erect, magnified image of an object held at 25 cm at the near point of the eye, farther from the eye than the object, so the eye can focus it; the converging lens supplies the extra focusing power needed to form the image on the retina.

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Q205 CBSE 2023

The magnification produced when an object is placed at a distance of 20 cm from a spherical mirror is $+ 1/2$ . Where should the object be placed to reduce the magnification to $+ 1/3$ .

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The object should be placed 40 cm in front of the mirror, which is 20 cm farther away than before.
The object should be placed $40$ cm in front of the mirror ( $u = - 40$ cm), that is, $20$ cm farther away than before. Working: a positive magnification less than 1 means an erect, diminished image, so the mirror is convex. With $u = - 20$ cm and $m = + \frac{1}{2}$ , $v = - m u = + 10$ cm, and $\frac{1}{f} = \frac{1}{10} - \frac{1}{20} = \frac{1}{20}$ , so $f = + 20$ cm. For $m = + \frac{1}{3}$ , $v = - \frac{u}{3}$ , and $- \frac{3}{u} + \frac{1}{u} = \frac{1}{20}$ gives $u = - 40$ cm; the image is then $13.3$ cm behind the mirror.

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Q206 CBSE 2023
1. (a)Complete the following ray diagram to show the formation of image :
2. (b)Mention the nature, position and size of the image formed in this case.
3. (c)State the sign of the image distance in this case using the Cartesian sign convention.
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The image is virtual, erect and diminished, behind the mirror between P and the focus, and the image distance $v$ is positive.
(a) Ray 1, being parallel to the principal axis, is reflected away from the axis so that its backward extension passes through the principal focus (the point marked $R$ ). Ray 2, being directed towards the centre of curvature $C$ , strikes the mirror along the normal and is reflected back along the same path. The two reflected rays diverge, so they are produced backwards; their dashed extensions meet behind the mirror, between $P$ and $R$ , and that is where the image $A^{'} B^{'}$ is formed. (b) Nature: virtual and erect. Position: behind the mirror, between the pole $P$ and the principal focus (the point marked $R$ ). Size: diminished, i.e. smaller than the object. (c) The image distance $v$ is **positive**, because the image lies behind the mirror, on the same side as the direction in which the incident light travels.

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Q207 CBSE 2023
A student wants to draw a ray diagram to show the formation of image due to a concave mirror when an object is placed in between its focus and centre of curvature.
1. (a)Write the minimum number of incident rays required.
2. (b)State the paths of the incident rays.
3. (c)State the paths of the reflected rays.
4. (d)Write the sign of magnification according to the New Cartesian Sign Convention.
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Two rays are enough; the image is real, inverted and magnified beyond C, so the magnification is negative.
(a) Two incident rays are required, the minimum needed to fix the image point by their intersection. (b) Ray 1 travels from the top of the object parallel to the principal axis; ray 2 travels from the top of the object through the principal focus F of the mirror. (c) Ray 1 is reflected so as to pass through the principal focus F; ray 2 is reflected parallel to the principal axis. The two reflected rays meet beyond the centre of curvature C, where a real, inverted and magnified image is formed. (d) The magnification is **negative**. For example, with $f = - 10$ cm and $u = - 15$ cm the mirror formula gives $v = - 30$ cm and $m = - v / u = - 2$ .

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Q208 CBSE 2023
A candle flame is placed at a distance of 18 cm from the optical centre of a convex lens. If its image is formed on a screen placed at a distance of 36 cm from the lens, answer the following questions :
1. (a)What is the magnification of the image formed ?
2. (b)What is the focal length of the lens ? Use the “Lens Formula” to determine it.
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The magnification is $- 2$ (real, inverted, twice the size) and the focal length is $+ 12$ cm.
(a) Magnification $m = \frac{v}{u} = \frac{+ 36}{- 18} = - 2$ . The image is real, inverted and twice the size of the candle flame. (b) Using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ with $v = + 36$ cm and $u = - 18$ cm: $\frac{1}{f} = \frac{1}{36} + \frac{1}{18} = \frac{3}{36} = \frac{1}{12}$ , so $f = + 12$ cm.

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Q209 CBSE 2023
(a) The image of an object formed by a lens is of same size but inverted. If the object distance is 30 cm, calculate
1. (i)The distance between the object and its image.
2. (ii)Focal length of the lens. (b) Draw a ray diagram to show the image formed in above case.
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The object and image are 60 cm apart and the focal length is $+ 15$ cm (convex lens).
(a)(i) Same size and inverted means $m = - 1$ , so $v = m u = (- 1) (- 30) = + 30$ cm: the image is $30$ cm from the lens on the other side. The distance between the object and its image is $30 + 30 = 60$ cm. (a)(ii) $\frac{1}{f} = \frac{1}{v} - \frac{1}{u} = \frac{1}{30} + \frac{1}{30} = \frac{1}{15}$ , so $f = + 15$ cm; the lens is convex. (b) The object is at $2 F_{1}$ ( $30$ cm $= 2 f$ ) and the image forms at $2 F_{2}$ on the other side of the lens: real, inverted and of the same size (see the ray diagram).

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Q210 CBSE 2023
Define the following terms in the context of a diverging mirror :
1. (i)Principal focus
2. (ii)Focal length Draw a labelled ray diagram to illustrate your answer.
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The principal focus is the point behind the mirror from which parallel rays appear to diverge; focal length is the pole-to-focus distance, $f = \frac{R}{2}$ .
(i) Principal focus: the point on the principal axis of a diverging (convex) mirror from which a beam of light rays parallel to the principal axis appears to diverge after reflection from the mirror. It lies behind the mirror and is a virtual focus, denoted $F$ . (ii) Focal length: the distance between the pole of the mirror and its principal focus, $f = P F$ . For a spherical mirror of small aperture it equals half the radius of curvature, $f = R /2$ , so $F$ lies midway between the pole $P$ and the centre of curvature $C$ . The ray diagram shows a parallel beam striking the convex mirror, the reflected rays diverging, and their backward extensions meeting at $F$ behind the mirror.

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Q211 CBSE 2023
(i) An object of 5 cm height is placed at a distance of 20 cm from the optical centre of a concave lens of focal length 18 cm. Calculate
1. (1)image distance and
2. (2)the magnification in this case. (ii) Compare the values of magnification obtained by a concave lens and a convex lens when both the lenses form virtual images.
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$v = - 9.47$ cm and $m = + 0.47$ (image 2.37 cm tall); a concave lens gives $m < 1$ , a convex lens forming a virtual image gives $m > 1$ .
(i) (1) Image distance $v = - \frac{180}{19} = - 9.47$ cm, so the image is $9.47$ cm from the lens on the same side as the object and is therefore virtual. (2) Magnification $m = \frac{v}{u} = \frac{9}{19} = + 0.47$ , so the image is erect and diminished; its height is $2.37$ cm. (ii) Both magnifications are positive, so both images are erect. For a concave lens the magnification is always less than $1$ , so its virtual image is always diminished; for a convex lens forming a virtual image the magnification is always greater than $1$ , so that image is always magnified.

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Q212 CBSE 2023
An object is kept at a distance of 1m from a lens of power +2D :
1. (i)Identify the type of lens.
2. (ii)Calculate its focal length and distance of the image formed.
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Convex lens; $f = + 50$ cm; $v = + 100$ cm, so the image is real, inverted and the same size, 100 cm from the lens.
(i) The power is positive, so the lens is a convex (converging) lens. (ii) Focal length: $f = 1/ P = 1/ (+ 2)$ m $= + 0.5$ m $= + 50$ cm. With $u = - 100$ cm, the lens formula gives $\frac{1}{v} = \frac{1}{50} - \frac{1}{100} = \frac{1}{100}$ , so $v = + 100$ cm. The image is formed 100 cm (1 m) from the lens on the other side, at $2 F_{2}$ ; it is real, inverted and of the same size as the object.

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Q213 CBSE 2023

An object of height 10 cm is placed 25 cm away from the optical centre of a converging lens of focal length 15 cm. Calculate the image-distance and height of the image formed.

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$v = + 37.5$ cm and $h^{'} = - 15$ cm: the image is real, inverted and magnified, 15 cm tall.
The image distance is $v = + 37.5$ cm, so the image is formed $37.5$ cm from the optical centre on the side of the lens opposite the object. The magnification is $m = v / u = 37.5/ (- 25) = - 1.5$ , so the image height is $h^{'} = - 15$ cm: the image is $15$ cm tall, real, inverted and magnified.

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Q214 CBSE 2023
A convex lens can form a
1. (i)real, inverted and magnified image as well as
2. (ii)virtual, erect and magnified image of an object. If the focal length of the lens is 10 cm, what should be the range of the object distance in both cases ? Draw ray diagrams to justify your answer.
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Real, inverted, magnified image: object distance between 10 cm and 20 cm; virtual, erect, magnified image: object distance between 0 and 10 cm.
(i) Real, inverted and magnified: the object must lie between $F$ and $2 F$ , that is the object distance must be between $10$ cm and $20$ cm (both ends excluded). For example at $u = - 15$ cm, $v = + 30$ cm and $m = - 2$ . (ii) Virtual, erect and magnified: the object must lie between the optical centre and $F$ , that is the object distance must be between $0$ and $10$ cm (both ends excluded). For example at $u = - 6$ cm, $v = - 15$ cm and $m = + 2.5$ .

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Q215 CBSE 2023
Define the following terms in the context of a diverging lens :
1. (i)Principal focus,
2. (ii)Focal length. Draw a labelled ray diagram to illustrate your answer.
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The principal focus is the point from which parallel rays appear to diverge after refraction, and the focal length is the distance $O F_{1}$ .
(i) Principal focus: the point on the principal axis of a diverging (concave) lens from which rays of light, incident parallel to the principal axis, appear to diverge after refraction through the lens. It lies on the same side of the lens as the incoming light and is virtual; a lens has two principal foci, $F_{1}$ and $F_{2}$ . (ii) Focal length: the distance between the optical centre $O$ of the lens and its principal focus, $f = O F_{1}$ . By the sign convention the focal length of a concave lens is negative. Ray diagram: rays parallel to the principal axis diverge after passing through the concave lens, and their backward extensions meet at the principal focus $F_{1}$ ; the distance $O F_{1}$ is marked as the focal length.

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Q216 CBSE 2023

The power of a lens is +4D. Find the focal length of this lens. An object is placed at a distance of 50 cm from the optical centre of this lens. State the nature and magnification of the image formed by the lens and also draw a ray diagram to justify your answer.

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$f = + 25$ cm; the image is real, inverted and the same size, $m = - 1$ , formed 50 cm from the lens at 2F.
The focal length is $f = 1/ P = 1/4$ m $= 0.25$ m $= + 25$ cm, and the lens is converging because the power is positive. With $u = - 50$ cm, the lens formula gives $\frac{1}{v} = \frac{1}{25} - \frac{1}{50} = \frac{1}{50}$ , so $v = + 50$ cm. The magnification is $m = v / u = 50/ (- 50) = - 1$ . The image is therefore real, inverted and of the same size as the object, formed 50 cm from the optical centre on the other side of the lens, that is at $2 F$ .

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Q217 CBSE 2023
(a) Define the following terms in the context of spherical mirrors :
1. (i)Pole
2. (ii)Centre of curvature
3. (iii)Principal axis
4. (iv)Principal focus (b) An object is placed at a distance of 60 cm from the pole of a concave mirror. If its real and inverted image is formed on a screen placed at a distance of 60 cm in front of the mirror, what is the focal length of the mirror ? Give justification for your answer.
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The four terms are defined; the focal length is $f = - 30$ cm, since the object is at the centre of curvature.
(a) (i) Pole: the centre of the reflecting surface of a spherical mirror; it lies on the surface of the mirror and is denoted by P. (ii) Centre of curvature: the centre of the sphere of which the reflecting surface of the mirror forms a part; it is denoted by C and lies in front of a concave mirror. (iii) Principal axis: the straight line passing through the pole and the centre of curvature of the mirror; it is normal to the mirror at its pole. (iv) Principal focus: the point F on the principal axis at which rays parallel to the principal axis meet after reflection from a concave mirror, or appear to diverge from after reflection from a convex mirror. (b) $f = - 30$ cm, that is, the focal length of the mirror is $30$ cm. Justification: the image is real, inverted and formed at the same distance ( $60$ cm) as the object, which happens only when the object is placed at the centre of curvature. Hence $R = 60$ cm and $f = R /2 = 30$ cm. The mirror formula gives the same result: $\frac{1}{- 60} + \frac{1}{- 60} = \frac{1}{f}$ , so $f = - 30$ cm.

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Q218 CBSE 2023
A student has focussed the image of an object of height 3 cm on a white screen using a concave mirror of focal length 12 cm. If the distance of the object from the mirror is 18 cm, find the values of the following :
1. (i)Distance of the image from the mirror
2. (ii)Height of the image
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(i) $v = - 36$ cm: a real image 36 cm in front of the mirror; (ii) $h^{'} = - 6$ cm: 6 cm tall and inverted.
(i) $v = - 36$ cm, so the image is formed $36$ cm from the mirror, on the same side as the object, and it is real. (ii) $m = - \frac{v}{u} = - 2$ , so $h^{'} = m \times h = - 6$ cm: the image is $6$ cm tall and inverted (twice the size of the object).

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Q219 CBSE 2023
A candle flame and a screen are placed 1 m apart. When a lens is placed exactly in the middle of the two, a distinct image of the candle flame is observed on the screen.
1. (i)Name the type of the lens used.
2. (ii)How is focal length of the lens determined in this case? State the method and determine its value.
3. (iii)Write the power of the lens used.
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A convex lens; with the object at $2 F$ the focal length is 25 cm, so the power is $+ 4$ D.
(i) A convex (converging) lens, because the image is obtained on a screen and is therefore real. (ii) The flame and the screen are each $50$ cm from the lens, so $u = - 50$ cm and $v = + 50$ cm. Since $∣ u ∣ = ∣ v ∣$ , the magnification is $- 1$ and the image is real, inverted and the same size as the object, which for a convex lens happens only when the object is at $2 F$ . Hence $2 f = 50$ cm, that is, the focal length is one quarter of the flame-to-screen separation. By the lens formula $\frac{1}{f} = \frac{1}{50} - \frac{1}{- 50} = \frac{2}{50}$ , so $f = 25$ cm $= 0.25$ m. (iii) $P = \frac{1}{f ( m )} = \frac{1}{0.25} = + 4$ D.

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Q220 CBSE 2023

Define power of a lens. The focal length of a lens is $- 10$ cm. Write the nature of the lens and find its power. If an object is placed at a distance of 20 cm from the optical centre of this lens, according to the New Cartesian Sign Convention, what will be the sign of magnification in this case ?

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Power is the reciprocal of focal length in metres; the lens is concave with $P = - 10$ D, and the magnification is positive ( $+ \frac{1}{3}$ ).
The power of a lens is the degree to which it converges or diverges a beam of light falling on it, and it is measured as the reciprocal of its focal length in metres, $P = 1/ f$ ; its SI unit is the dioptre (D). Since $f = - 10$ cm is negative, the lens is a **concave** (diverging) lens. Power: $f = - 0.10$ m, so $P = 1/ (- 0.10) = - 10$ D. For the object at $u = - 20$ cm, the lens formula gives $\frac{1}{v} = \frac{1}{- 10} + \frac{1}{- 20} = - \frac{3}{20}$ , so $v = - \frac{20}{3} \approx - 6.7$ cm, and $m = \frac{v}{u} = \frac{- 20/3}{- 20} = + \frac{1}{3}$ . The sign of the magnification is therefore **positive**, which means the image is virtual, erect and diminished.

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