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.

CBSE Class 10 Science previous year papers, with answers →

Questions · page 6 of 21

1. Q101 CBSE 2026
   An ​object is placed at a distance of 30 ​cm in front of a convex lens ​of ​focal length 15 ​cm. Use lens formula ​to determine the position of the image. What ​will be the size of the image in this case ?
   Show answer

   The image is at $v=+30$ cm, real and inverted, and the same size as the object ($m=-1$).

   The image is formed at $v=+30$ cm, that is 30 cm from the optical centre on the side of the lens away from the object, and it is real and inverted. Since $m=\frac{v}{u}=\frac{+30}{-30}=-1$, the image is exactly the same size as the object ($|h^{\prime }|=h$); the minus sign only means it is inverted.

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2. Q102 CBSE 2026
   An ​object of ​height 6 cm is placed at a ​distance of ​30 cm from the optical centre of a ​concave ​lens of focal length 15 cm. Use lens formula to determine :

   1. (a)The ​distance of image from the optical centre.
   2. (b)The height of the image ​formed.
   Show answer

   The image is 10 cm from the optical centre, on the object's side, and 2 cm tall (virtual, erect, diminished).

   (a) $v=-10$ cm, so the image is formed $10$ cm from the optical centre, on the same side of the lens as the object. (b) $h^{\prime }=+2$ cm, so the image is $2$ cm tall. It is virtual, erect and diminished.

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3. Q103 CBSE 2026
   By using ​the ​data given in the table, compare ​the properties of images formed ​by ​convex lens and concave mirror in ​terms of the positions and nature of ​images formed.

   Optical device	Object distance (cm)	Focal length (cm)	Height of object (cm)
   Convex lens	20	10	6
   Concave mirror	30	10	6

   Show answer

   Both images are real and inverted, but the lens image is the same size and the mirror image half-size.

   Convex lens ($u=-20$ cm, $f=+10$ cm): $v=+20$ cm and $m=-1$. The image is formed $20$ cm from the lens on the other side, at $2F_2$. It is real, inverted and of the same size as the object ($h^{\prime }=-6$ cm). Concave mirror ($u=-30$ cm, $f=-10$ cm): $v=-15$ cm and $m=-\frac{1}{2}$. The image is formed $15$ cm in front of the mirror, on the same side as the object, between F and C. It is real, inverted and diminished ($h^{\prime }=-3$ cm). Comparison: both images are real and inverted, but the lens forms its image on the side opposite the object and of the same size, while the mirror forms its image on the object's side and half the size of the object.

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4. Q104 CBSE 2026
   A mirror ​used for ​rear-view of ​an automobile ​has a focal length 2.0 m. If a car is located at 8.0 m from the ​mirror -

   1. (a)Find out the position ​of the image of ​this car as seen in the mirror.
   2. (b)Draw the ray-diagram for image formation by the rear-view mirror.
   Show answer

   $v=+1.6$ m: the image is 1.6 m behind the mirror, virtual, erect and diminished.

   (a) A mirror used for the rear view of an automobile is a convex mirror, so under the New Cartesian sign convention its focus lies behind the mirror and $f=+2.0$ m, while the car stands in front of the mirror so $u=-8.0$ m. From the mirror formula, $\frac{1}{v}=\frac{1}{f}-\frac{1}{u}=\frac{1}{2.0}+\frac{1}{8.0}=\frac{5}{8.0}$, giving $v=+1.6$ m. The image of the car is formed $1.6$ m **behind** the mirror. The magnification is $m=-v/u=-(+1.6)/(-8.0)=+0.2$, so the image is virtual, erect and one-fifth the size of the car, and it lies between the pole and the focus - the position a convex mirror always puts its image in, since $0<1.6<2.0$. (b) The ray diagram is drawn in the figure in this solution. From the top of the car one ray is drawn parallel to the principal axis, which after reflection diverges as if coming from the focus $F$ behind the mirror, and a second ray is aimed at the centre of curvature $C$, which strikes the mirror along the normal and is reflected back along its own path. The reflected rays themselves diverge and go on to the driver's eye; their backward extensions, drawn dashed behind the mirror, meet $1.6$ m from the pole, and the erect, diminished virtual image is drawn there.

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5. Q105 CBSE 2026
   The above ​ray diagram is to show image formation by a reflecting telescope. ​Reflecting telescope revolutionized our way of looking into the sky. Reflecting telescopes employ large mirrors to ​gather ​and ​focus light rather ​than relying solely on ​lenses as we do in the refracting telescopes. These telescopes utilize precisely shaped and polished mirrors to capture incoming light and reflect it to a ​focal point, where the telescope forms an image for observation.

   1. (a)What kind of image of a star is seen by the observer in the eye-piece ?
   2. (b)What kind of mirror is used in this reflecting telescope ?
   3. (c)Explain with the reason, what kind of optical device (type of lens or mirror) is used at the eye-piece.
   4. (c)OR : What is the role of plane mirror in image formation by this telescope ?
   Show answer

   Virtual, magnified, inverted image; a concave mirror with a plane mirror; a convex lens as eye-piece; or the plane mirror redirects the light.

   (a) A virtual, magnified image, inverted with respect to the star. The curved mirror first forms a real, inverted image of the star at its focus, and the eye-piece, used as a magnifying glass on that image, gives the enlarged virtual image the observer actually sees. (b) A concave (converging) mirror - the large curved mirror that brings the parallel rays from a distant star to a real focus - together with a small plane mirror that turns the converging beam towards the eye-piece. (c) A convex (converging) lens is used at the eye-piece. It magnifies the image formed by the curved mirror: that small real image lies at or just within the focal length of the lens, so the lens acts as a magnifying glass and gives a much enlarged virtual image of it; a concave lens would only diminish the image, and a mirror there would send the light back into the tube instead of to the eye. OR (c) The plane mirror redirects (reflects) the light rays coming from the curved mirror towards the eye-piece. Held at $45^{\circ }$ in the path of the converging beam, it turns the beam through $90^{\circ }$ and brings the real image out through the side of the tube, so that the eye-piece and the observer's head do not block the starlight falling on the curved mirror; it forms no image of its own and does not change the size of the image.

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6. Q106 CBSE 2026
   Read the following passage and answer the questions given : Lenses can form different types of images ​depending upon their focal length and position of object. A convex lens can create real, ​inverted or virtual, erect images, while ​a concave lens forms only virtual ​and diminished images. The focal length determines the ​power of ​lens. Convex lenses have positive focal length while concave lenses ​have negative focal length by convention. When lenses are placed together, their combined power is determined ​by the sum of their individual powers. Ray diagrams help to visualize how light converges or diverges through lens to form an image.

   1. (a)A convex lens of focal length 20 cm is used to form an image. If an object is placed at 40 cm from the lens, what will be the position and nature of image ?[1]
   2. (b)Illustrate the formation of image with the help of ray diagram, when the object is placed between the optical centre and principal focus of concave lens.[1]
   3. (c) (i)A lens combination consists of a convex lens of focal length 30 cm and a concave lens of focal length 15 cm placed together. Find the equivalent focal length and power of this lens combination.[2]
   4. (c) (ii)OR : Two lenses are placed in contact. One is a concave lens with focal length 2 m and the other is a convex lens with focal length 1.5 m. What type of lens will the combination behave as (convex or concave) ? Give reason.[2]
   Show answer

   Image at $v=+40$ cm is real, inverted, same size; a concave lens gives a virtual, diminished image; combination $F=-30$ cm, $P\approx -3.33$ D.

   (a) The image is formed at $v=+40$ cm, that is $40$ cm on the other side of the lens (at $2F_2$); it is real, inverted and of the same size as the object ($m=-1$). (b) The ray parallel to the axis diverges as if it came from $F_1$ and the ray through $O$ goes straight on; they meet (by backward extension) between $O$ and $F_1$, so the image is formed between the optical centre and the focus on the object's side, nearer the lens than the object - virtual, erect and diminished, as shown in the ray diagram. (c) (i) $\frac{1}{F}=\frac{1}{30}-\frac{1}{15}=-\frac{1}{30}$, so the equivalent focal length is $F=-30$ cm and the power is $P=\frac{1}{-0.30\text{m}}\approx -3.33$ D; the combination acts as a concave (diverging) lens. OR (c) (ii) $P=-\frac{1}{2}+\frac{1}{1.5}=-0.5+0.67=+0.17$ D (focal length $+6$ m). The total power is positive, so the combination behaves as a convex (converging) lens, because the convex lens has the shorter focal length and hence the greater power.

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7. Q107 CBSE 2026
   The ​variation of ​image distance ($v$) with the object distance ($u$) for a convex lens ​is given in ​the ​following observation table. Analyse it and ​answer the ​questions ​given below :

   S.No.	Object distance ($u$) cm	Image distance ($v$) cm
   1	$-150$	$+30$
   2	$-75$	$+37.5$
   3	$-50$	$+50$
   4	$-37.5$	$+75$
   5	$-30$	$+150$
   6	$-15$	$+37.5$

   1. (a)Without calculation, find out the focal length of the given convex lens. Justify your answer.[1]
   2. (b)Which one of the observations given in the table is not correct and why ?[1]
   3. (c)(i)Find out the value of magnification for $u=-30$ cm. Write the nature of the image formed. Give reason for your answer.[2]
   4. (c)(ii)OR : Analyse the data given in the table above and draw the labelled ray diagram for $u=-30$ cm and $v=150$ cm.[2]
   Show answer

   $f=25$ cm; observation 6 is incorrect because $v$ must be negative; $m=-5$, so the image is real, inverted and magnified.

   (a) $f=25$ cm. In observation 3 the object is at $u=-50$ cm and the image at $v=+50$ cm, the same distance on the other side of the lens, which happens only when the object stands at $2F_1$; hence $2f=50$ cm and $f=25$ cm. (b) Observation 6 ($u=-15$ cm, $v=+37.5$ cm) is not correct. The object is only $15$ cm from the lens, inside its focal length of $25$ cm, and a convex lens then forms a virtual, erect, magnified image on the same side as the object, so $v$ must be negative - the lens formula gives $v=-37.5$ cm, not $+37.5$ cm. (c)(i) $m=\frac{v}{u}=\frac{+150}{-30}=-5$. The image is real, inverted and magnified, five times the size of the object, because $m$ is negative (inverted, and so real) and $|m|=5>1$ (enlarged); the object at $30$ cm lies between $F_1$ and $2F_1$, which is the position that gives such an image beyond $2F_2$. OR (c)(ii) With $f=25$ cm, the object $AB$ stands at $30$ cm, between $F_1$ and $2F_1$. The ray from $B$ parallel to the axis passes through $F_2$ after refraction and the ray from $B$ through $O$ goes on undeviated; they meet at $B^{\prime }$, $150$ cm behind the lens and beyond $2F_2$, giving a real, inverted image $A^{\prime }{B}^{\prime }$ five times the height of the object (see the ray diagram).

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8. Q108 CBSE 2026
   1. (i)In which mirror we can always see the full image of a distant object ? Support your ​answer by a ray diagram for it.
   2. (ii)A convex mirror forms an erect ​image of an object of ​size, one-third of the size ​of the object. Find the position of ​the object ​if the radius of curvature of convex mirror is ​30 cm.
   Show answer

   A convex mirror always shows the full image (virtual, erect, diminished); the object is 30 cm in front of the mirror.

   (i) A convex mirror. Whatever the position of the object, a convex mirror always forms a virtual, erect and diminished image lying between the pole P and the focus F, so the whole of even a large or distant object is compressed into the mirror and its full image can always be seen (this, with its wide field of view, is why a convex mirror is used as a rear-view mirror). The ray diagram is drawn with a ray from the tip of the object parallel to the principal axis, reflected as though it came from F, and a second ray aimed at C, reflected back along itself; their dashed backward extensions meet at the image tip behind the mirror. (ii) $u=-30$ cm - the object is $30$ cm in front of the convex mirror. Working: $f=R/2=+15$ cm; $m=+\frac{1}{3}=-\frac{v}{u}$ gives $v=-\frac{u}{3}$; substituting in $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$ gives $-\frac{3}{u}+\frac{1}{u}=\frac{1}{15}$, so $-\frac{2}{u}=\frac{1}{15}$ and $u=-30$ cm. The image is then at $v=+10$ cm, i.e. virtual and erect, $10$ cm behind the mirror.

   Step-by-step solution →
9. Q109 CBSE 2026
   1. (i)With the help of given ray diagram, calculate ​the refractive ​index of medium ​2 with respect to medium 1.
   2. (ii)If speed of ​light in ​medium-1 is $3\times 10^8$ m/s, then ​calculate the speed of light in medium-2.
   3. (iii)Compare the optical density of ​medium-1 and medium-2 and justify your ​answer.

   Show answer

   $n_{21}=\sqrt{3}\approx 1.73$; $v_2\approx 1.73\times 10^8$ m/s; medium-2 is optically denser than medium-1.

   The arcs in the figure are drawn from the boundary, so $i=90^{\circ }-30^{\circ }=60^{\circ }$ and $r=90^{\circ }-60^{\circ }=30^{\circ }$. (i) $n_{21}=\frac{\sin i}{\sin r}=\frac{\sin 60^{\circ }}{\sin 30^{\circ }}=\frac{\sqrt{3}/2}{1/2}=\sqrt{3}\approx 1.73$. (ii) $v_2=\frac{v_1}{n_{21}}=\frac{3\times 10^8}{\sqrt{3}}=\sqrt{3}\times 10^8\approx 1.73\times 10^8$ m/s. (iii) Medium-2 is optically denser than medium-1, because $n_{21}=\sqrt{3}>1$: the ray bends towards the normal on entering medium-2 ($r<i$) and light travels more slowly in it ($1.73\times 10^8$ m/s against $3\times 10^8$ m/s).

   Step-by-step solution →
10. Q110 CBSE 2026
    1. (i)Which mirror will always form ​a ​full length image of a distant object ? Support your answer by a ray diagram for it.
    2. (ii)A convex lens forms an image with magnification $m=+2$. If ​the object is kept ​at $15$ cm from the lens, find out the focal length ​of ​convex ​lens and nature of image formed.
    Show answer

    (i) A convex mirror; (ii) $f=+30$ cm, and the image is virtual, erect and twice the size of the object.

    (i) A convex mirror. For every position of the object it forms a virtual, erect and diminished image between its pole and its focus, and its outward bulge gives it a much wider field of view than a plane or a concave mirror of the same size, so the whole of a distant object is always taken in - which is exactly why a convex mirror is used as a rear-view mirror. The ray diagram shows a parallel ray reflected as though it came from $F$ and a ray to the pole reflected at an equal angle on the other side of the axis, their dashed extensions meeting at the small erect image between $P$ and $F$. (ii) Focal length $f=+30$ cm. Since $m=\frac{v}{u}=+2$ and $u=-15$ cm, the image distance is $v=-30$ cm, and $\frac{1}{f}=\frac{1}{-30}-\frac{1}{-15}=\frac{1}{30}$. The image is virtual, erect and magnified - twice the size of the object - and it is formed $30$ cm from the lens on the same side as the object.

    Step-by-step solution →
11. Q111 CBSE 2026
    1. (i)Define the optical centre of a lens.
    2. (ii)The light from an object placed on a meter scale at 8 ​cm mark, was focused on a white screen placed at ​92 cm mark, with the help of a converging lens placed on the scale at 50 cm mark ​: (I) Find the ​focal length ​of converging lens. (II) Find the position ​of image if the object is shifted towards the lens ​and kept at 29 cm mark. (III) What will be the nature of the image formed if the object is further shifted towards ​the lens ?
    Show answer

    $f=+21$ cm; at the 29 cm mark the image forms at infinity; nearer the lens it is virtual, erect and magnified.

    (i) The optical centre of a lens is the central point of the lens, lying on its principal axis midway between its two surfaces, such that a ray of light directed towards it passes through the lens without suffering any deviation. (ii) (I) Measuring from the lens at the $50$ cm mark, $u=8-50=-42$ cm and $v=92-50=+42$ cm, so $\frac{1}{f}=\frac{1}{v}-\frac{1}{u}=\frac{1}{42}+\frac{1}{42}=\frac{1}{21}$, giving $f=+21$ cm. (II) At the $29$ cm mark, $u=29-50=-21$ cm $=-f$, so the object is exactly at the principal focus. Then $\frac{1}{v}=\frac{1}{f}+\frac{1}{u}=\frac{1}{21}-\frac{1}{21}=0$, so $v=\infty$: the refracted rays emerge parallel to one another and the image is formed at infinity (real, inverted and highly magnified); it cannot be caught on the screen at any finite distance. (III) Brought still nearer, the object lies between the optical centre and the focus, and the image is then virtual, erect and magnified, formed on the same side of the lens as the object and not obtainable on a screen - the magnifying-glass case.

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12. Q112 CBSE 2026
    1. (i)Define ​the focus of a convex lens.
    2. (ii)An object placed on a meter scale at 8 cm mark, was focused ​on a white ​screen placed at 92 cm mark, by using a converging lens placed on the scale at 50 cm mark. (I) Find the focal length of converging lens. (II) Find the position of object if the image forms at ​71 ​cm mark on meter scale. (III) ​State the nature of image formed if the object is ​kept on 36 cm ​mark of meter scale.
    Show answer

    $f=+21$ cm, the object for the 71 cm image is at infinity, and the 36 cm mark gives a virtual, erect, enlarged image.

    (i) The principal focus of a convex lens is the point on its principal axis at which rays of light coming parallel to the principal axis meet after refraction through the lens. A convex lens has two such foci, one on each side, each at a distance $f$ from the optical centre. (ii)(I) With the lens at the $50$ cm mark, $u=8-50=-42$ cm and $v=92-50=+42$ cm, so $\frac{1}{f}=\frac{1}{42}+\frac{1}{42}=\frac{1}{21}$ and $f=+21$ cm. (II) An image at the $71$ cm mark is $21$ cm from the lens, which is exactly $f$, so $\frac{1}{u}=\frac{1}{21}-\frac{1}{21}=0$ and the object must be at infinity; no mark on the metre scale will give this image, because only a beam of light arriving parallel to the principal axis is brought to a focus at $F$. (III) An object at the $36$ cm mark is $14$ cm from the lens, which is less than $f=21$ cm, so it lies between the optical centre and the focus: $\frac{1}{v}=\frac{1}{21}-\frac{1}{14}=-\frac{1}{42}$, giving $v=-42$ cm and $m=\frac{-42}{-14}=+3$. The image is virtual, erect and enlarged three times, formed $42$ cm from the lens on the same side as the object, that is, at the $8$ cm mark.

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13. Q113 CBSE 2026
    1. (i)What ​is meant by refraction of ​light ? Write the laws of refraction.
    2. (ii)If two lenses of focal length $+20$ cm and $-10$ ​cm are kept in contact, ​find out the net power of the lens combination. Will the combination behave as ​converging ​lens or a diverging lens ​?
    Show answer

    The combination has net power $-5$ D, so it behaves as a diverging lens of focal length $-20$ cm.

    (i) Refraction of light is the change in the direction of a ray of light when it passes obliquely from one transparent medium into another; it occurs because the speed of light is different in the two media, so the ray bends towards the normal on entering a denser medium and away from the normal on entering a rarer one. The laws of refraction are: (1) the incident ray, the refracted ray and the normal to the surface at the point of incidence all lie in the same plane; (2) Snell's law - for light of a given colour and a given pair of media, $\frac{\sin i}{\sin r}=n_{21}$, a constant, which is the refractive index of the second medium with respect to the first. (ii) $P_1=\frac{1}{+0.20\text{m}}=+5$ D and $P_2=\frac{1}{-0.10\text{m}}=-10$ D, so the net power is $P=P_1+P_2=+5-10=-5$ D. The power is negative, so the combination behaves as a diverging (concave) lens, of equivalent focal length $f=\frac{100}{-5}=-20$ cm.

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14. Q114 CBSE 2025
    To ​get ​an image of magnification $-1$ on a ​screen using a lens of focal length $20$ cm, ​the object ​distance must be :

    1. (a)Less than $20$ cm
    2. (b)$30$ cm
    3. (c)$40$ cm
    4. (d)$80$ cm
    Show answer

    (c) $40$ cm

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15. Q115 CBSE 2025
    If the ​absolute ​refractive indices ​of two media X and ​Y are $\frac{6}{5}$ and $\frac{4}{3}$ respectively, then the refractive index of ​Y with ​respect to ​X will be :

    1. (a)$\frac{10}{9}$
    2. (b)$\frac{9}{10}$
    3. (c)$\frac{9}{8}$
    4. (d)$\frac{8}{9}$
    Show answer

    (a) $\frac{10}{9}$

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16. Q116 CBSE 2025
    Absolute refractive index of water ​and glass is $\frac{4}{3}$ ​and $\frac{3}{2}$ respectively. If the speed of ​light in glass is $2\times 10^8$ m/s, the speed of ​light in water ​is :

    1. (a)$\frac{9}{4}$ m/s
    2. (b)$\frac{7}{3}$ m/s
    3. (c)$\frac{16}{9}$ m/s
    4. (d)$\frac{9}{8}$ m/s
    Show answer

    (a) $\frac{9}{4}$ m/s

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17. Q117 CBSE 2025
    A 5 cm ​long ​rod is placed vertically ​in front of a ​concave mirror. If its 5 cm long image is formed at ​20 cm ​in front of ​the ​mirror, the focal length of the mirror is :

    1. (a)$-20$ cm
    2. (b)$-10$ cm
    3. (c)$-5$ cm
    4. (d)$20$ cm
    Show answer

    (b) $-10$ cm

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18. Q118 CBSE 2025
    An ​optical device ​‘X’ is ​placed obliquely in the path of a narrow ​parallel beam of light. If ​the ​emergent beam gets displaced laterally, ​the device ‘X’ is :

    1. (a)plane mirror
    2. (b)convex lens
    3. (c)glass slab
    4. (d)glass prism
    Show answer

    (c) glass slab

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19. Q119 CBSE 2025
    A candle flame is placed in front of the ​reflecting surface ​of a convex mirror ​of focal length $f$. If the distance of the flame from the ​pole ​of the mirror is ‘$f$’, its ​image is formed :

    1. (a)at infinite distance from the mirror
    2. (b)behind the mirror at the principal focus
    3. (c)behind the mirror at a distance $2f$
    4. (d)behind the mirror at a distance $\frac{f}{2}$
    Show answer

    (d) behind the mirror at a distance $\frac{f}{2}$

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20. Q120 CBSE 2025
    An object is placed at a distance of ​30 cm from the pole of a ​concave mirror. If its real and inverted image is ​formed ​at 60 ​cm in front of the mirror, the ​focal length of the ​mirror ​is :

    1. (a)$-15$ cm
    2. (b)$-20$ cm
    3. (c)$+20$ cm
    4. (d)$+15$ cm
    Show answer

    (b) $-20$ cm

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