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 8 of 21

1. Q141 CBSE 2025
   List ​four properties of ​the ​images ​formed ​by plane ​mirrors.
   Show answer

   The image is virtual, erect, the same size, as far behind the mirror as the object is in front, and laterally inverted.

   (i) The image is virtual and erect. (ii) It is of the same size as the object, so the magnification is 1. (iii) It is formed as far behind the mirror as the object is in front of it. (iv) It is laterally inverted, that is, its left and right are interchanged.

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2. Q142 CBSE 2025
   A ray of light ​after refraction from a convex lens emerges parallel to its principal axis.

   1. (i)Draw a labelled ​ray diagram to show ​it.
   2. (ii)In ​this ​case, the ​incident ​ray before refraction from the lens passes through a point on ​its principal axis. Name the point.
   Show answer

   A ray passing through the principal focus emerges parallel to the axis after a convex lens; the point is the principal focus.

   (i) The ray diagram shows a ray that crosses the principal axis at the principal focus on the incident side, strikes the convex lens and emerges parallel to the principal axis. (ii) The point is the principal focus of the convex lens (the focus on the side from which the light comes).

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3. Q143 CBSE 2025
   List four ​properties ​of the images formed by convex ​mirrors when objects ​are placed anywhere in front of ​their reflecting ​surfaces.
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   A convex mirror always forms a virtual, erect, diminished image behind the mirror, between the pole and the focus.

   For an object anywhere in front of a convex mirror the image is (i) always virtual, (ii) always erect, (iii) always diminished, that is smaller than the object, and (iv) always formed behind the mirror, between the pole and the principal focus.

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4. Q144 CBSE 2025
   An ​object of size 5 cm is placed at a ​distance of ​30 ​cm from the optical centre of a converging lens ​of ​focal length 20 cm. Use lens formula to determine the position and ​size of the image ​formed.
   Show answer

   The image is real, 60 cm from the optical centre on the other side, and 10 cm tall (inverted).

   Position: $v=+60$ cm — the image is formed $60$ cm from the optical centre on the other side of the lens, and it is real. Size: $h^{\prime }=-10$ cm — the image is $10$ cm tall and inverted, twice the size of the object ($m=-2$).

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5. Q145 CBSE 2025
   Draw ray diagrams to show the nature, ​position and relative size of the image formed ​by ​a convex ​mirror when the object ​is placed

   1. (i)at infinity and ​
   2. (ii)between ​infinity and ​pole P of the mirror.
   Show answer

   A convex mirror gives a virtual, erect, diminished image: at F for an object at infinity, and between P and F otherwise.

   (i) Object at infinity: the parallel beam is reflected so that the rays appear to diverge from the principal focus, so the image is formed **at** $F$, behind the mirror. It is virtual, erect and highly diminished (point-sized). (ii) Object between infinity and the pole $P$: the image is formed between $P$ and $F$, behind the mirror. It is virtual, erect and diminished. Both results follow from the mirror formula, which gives $v=\frac{f|u|}{f+|u|}$ (always positive and less than $f$) and $m=\frac{f}{f+|u|}$ (always positive and less than $1$), so a convex mirror always forms a virtual, erect, diminished image between $P$ and $F$.

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6. Q146 CBSE 2025
   A convex lens ​forms an $8.0$ cm long image ​of a $2.0$ cm long object which is kept at a distance of $6.0$ cm from the optical centre of the ​lens. If ​the object and ​the image are on the same side of the lens, find

   1. (i)the nature of the image,
   2. (ii)the position of the image, and
   3. (iii)the focal ​length of ​the lens.
   Show answer

   Virtual, erect and enlarged image at $v=-24$ cm on the object's side; focal length $f=+8$ cm.

   (i) The image is virtual, erect and enlarged (four times the size of the object), formed on the same side of the lens as the object. (ii) It is at $v=-24$ cm, that is $24$ cm from the optical centre on the same side as the object. (iii) The focal length is $f=+8$ cm.

   Step-by-step solution →
7. Q147 CBSE 2025
   If we want ​to obtain a virtual and magnified image of ​an object by using a concave mirror of focal ​length ​18 cm, where ​should the object be placed? ​Use ​mirror formula to determine the object distance for an image of magnification $+2$ produced by this mirror to justify your ​answer.
   Show answer

   Place the object between the pole and the focus; for $m=+2$, $u=-9$ cm, so it is 9 cm from the mirror and $v=+18$ cm.

   The object must be placed between the pole and the principal focus of the mirror, that is at a distance less than the focal length of $18$ cm. Taking $f=-18$ cm and $m=+2$: $m=-v/u$ gives $v=-2u$, and substituting in $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$ gives $\frac{1}{-2u}+\frac{1}{u}=\frac{1}{-18}$, so $\frac{1}{2u}=-\frac{1}{18}$ and $u=-9$ cm. The object must therefore be placed $9$ cm from the mirror, and since $9$ cm is less than the focal length $18$ cm this position lies between the pole and the focus, which justifies the answer. The image is then formed at $v=+18$ cm, that is $18$ cm behind the mirror, and is virtual, erect and twice the size of the object.

   Step-by-step solution →
8. Q148 CBSE 2025
   The ​power of a lens is $-0.25$ D. Based on this ​information, find out

   1. (a)The type of lens and its focal length.
   2. (b)The eye ​defect for which ​it may be ​used as a ​corrective lens.
   3. (c)The nature and size of the ​image formed by this lens when an object is ​placed between F and 2F from the optical centre of this lens.
   Show answer

   Concave lens, $f=-4$ m; used for myopia; the image is virtual, erect and diminished, between $F$ and $O$.

   (a) $f=\frac{1}{P}=\frac{1}{-0.25}=-4$ m $=-400$ cm. The focal length is negative, so the lens is a concave (diverging) lens of focal length $4$ m. (b) It can be used to correct myopia (short-sightedness); a person needing this lens has a far point of about $4$ m. (c) The image is virtual, erect and diminished, and is formed between the focus $F$ and the optical centre $O$, on the same side of the lens as the object. For example, with $u=-6$ m (between $F$ and $2F$), $v=-2.4$ m and $m=+0.4$, so the image is $0.4$ times the size of the object.

   Step-by-step solution →
9. Q149 CBSE 2025
   A student placed a candle ​flame at ​different ​distances from a convex lens and focused its image ​on a screen. ​He recorded his observation in tabular form as given below : Analyse the observation table and on the basis of your analysis ​only, answer the following questions (without doing any calculations) :

   1. (a)What ​is the focal length of the convex lens used ? Give reason to justify your answer.
   2. (b)Which one of the sets of observations ​is not correct and why ?
   3. (c)Draw ray diagram to show image formation for any correct set of observation.

   Observations of the candle flame and its image for a convex lens

   S.No.	Distance of flame from the lens (cm)	Distance of the image from the lens
   1	$-90$	$+18$
   2	$-60$	$+20$
   3	$-40$	$+24$
   4	$-30$	$+30$
   5	$-24$	$+40$
   6	$-20$	$+60$
   7	$-18$	$+90$
   8	$-12$	$+120$

   Show answer

   $f=15$ cm, as object and image are both at 30 cm ($2F$); set 8 is wrong because the flame is inside the focal length.

   (a) $f=15$ cm. In set 4 the flame and its image are both 30 cm from the lens; for a convex lens the two distances are equal only when the object is at 2F, so $2f=30$ cm and $f=15$ cm. (b) Set 8 is not correct. The flame there is 12 cm from the lens, which is less than the focal length of 15 cm, so it lies between the optical centre and the focus. A convex lens then gives a virtual, erect and magnified image on the same side as the flame, which cannot be caught on a screen, so an image distance of $+120$ cm is impossible. (c) For set 4 the ray diagram shows the object at 2F and a real, inverted image of the same size at 2F on the other side, with $v=+30$ cm and $m=-1$.

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10. Q150 CBSE 2025
    An object of ​height 4.0 cm is placed at a distance of 30 cm ​from the optical centre of ​a convex ​lens of focal length 20 cm. Use lens formula to determine

    1. (i)the distance ​of the image from the optical ​centre ​and ​
    2. (ii)height of the image formed.
    Show answer

    The image is 60 cm from the optical centre, on the opposite side, and 8.0 cm tall (real, inverted, magnified).

    (i) The image is formed at $v=+60$ cm, that is 60 cm from the optical centre on the side of the lens opposite to the object. (ii) The magnification is $m=\frac{v}{u}=\frac{60}{-30}=-2$, so the image height is $h^{\prime }=mh=-2\times 4.0=-8.0$ cm. The image is real, inverted and magnified, and it is $8.0$ cm tall.

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11. Q151 CBSE 2025
    An ​object is placed at a distance of $20$ cm from the ​optical centre of a concave lens and its image is formed on the same side of the ​lens ​as the object. If the ​distance of the image from optical centre of the lens is $10$ cm, use lens formula to determine

    1. (i)​focal ​length, and
    2. (ii)power of the lens in ​new Cartesian sign conventions.
    Show answer

    The focal length is $-20$ cm and the power is $-5$ D, showing a diverging lens.

    (i) Focal length $f=-20$ cm; the minus sign shows a diverging lens. (ii) Power $P=\frac{1}{f}=\frac{1}{-0.20\text{m}}=-5$ D.

    Step-by-step solution →
12. Q152 CBSE 2025
    A convex mirror used for ​rear view on an automobile has a focal length of $1.5$ m. If a $3$ m high bus is ​located at $6.0$ m from ​the mirror, ​use ​mirror formula to ​determine the position and size ​of the image of ​the bus as seen in the mirror.
    Show answer

    The image of the bus is 1.2 m behind the mirror and 0.6 m high: virtual, erect and diminished.

    Position: $v=+1.2$ m, that is, the image is $1.2$ m behind the mirror. Size: $h\prime =+0.6$ m, so the image of the bus is $0.6$ m high, one-fifth of the height of the bus ($m=+0.2$). The image is virtual, erect and diminished.

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13. Q153 CBSE 2025
    If we want to obtain a real and ​magnified image ​of an object by using a concave ​mirror of ​focal length 18 cm. Where should ​the object be placed? Use mirror ​formula to determine the object distance for an image of magnification $-2$ by this mirror to ​justify your answer.
    Show answer

    The object goes between F and C; for $m=-2$ it is 27 cm from the mirror.

    To get a real and magnified image, the object must be placed between the principal focus F and the centre of curvature C of the mirror, that is, between $18$ cm and $36$ cm from the mirror. Justification: with $f=-18$ cm and $m=-2$, $m=-v/u$ gives $v=2u$; then $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$ gives $\frac{1}{2u}+\frac{1}{u}=\frac{3}{2u}=-\frac{1}{18}$, so $u=-27$ cm. The object is $27$ cm in front of the mirror, between F ($18$ cm) and C ($36$ cm); the image forms at $v=-54$ cm, beyond C, and is real, inverted and twice the size of the object.

    Step-by-step solution →
14. Q154 CBSE 2025
    1. (i)The power of a lens 'X' is $-2.5$ D. Name the lens and determine its focal length in cm. For which eye defect of vision will an optician prescribe this type of lens as a corrective lens ?
    2. (ii)"The value ​of magnification $m$ for a lens is $-2$." Using new Cartesian Sign Convention and considering that an object is placed at a distance of $20$ cm from the optical centre of this lens, state : (I) ​the nature of the image formed; (II) size of ​the image compared to the ​size of ​the object; (III) position of the image, and (IV) sign of the height of the image.
    3. (iii)The numerical values ​of the focal lengths of two lenses ​A and B are $10$ cm and $20$ cm respectively. Which one of the two will show higher degree of convergence/divergence ? Give reason to justify your answer.
    Show answer

    (i) Concave lens, $f=-40$ cm, for myopia; (ii) real, inverted, magnified image at $v=+40$ cm; (iii) lens A, the larger power.

    (i) The power is negative, so lens X is a concave (diverging) lens; its focal length is $f=1/P=1/(-2.5\text{D})=-0.40$ m $=-40$ cm, and an optician prescribes this type of lens to correct myopia (short-sightedness). (ii) With $u=-20$ cm and $m=-2$, $v=mu=+40$ cm. (I) The image is real and inverted. (II) It is magnified — twice the size of the object. (III) It is formed $40$ cm from the optical centre, on the side of the lens opposite the object. (IV) The height of the image is negative, since $h^{\prime }=mh=-2h$. (The lens of this part must be a convex lens of focal length $+40/3\approx +13.3$ cm, because a concave lens can never give a magnification of $-2$.) (iii) Lens A, of focal length $10$ cm, shows the higher degree of convergence or divergence, because power is the reciprocal of focal length: $|P_A|=1/0.10=10$ D against $|P_B|=1/0.20=5$ D, so lens A bends the rays twice as strongly.

    Step-by-step solution →
15. Q155 CBSE 2025
    1. (i)​“In refraction of light through a rectangular glass slab, the emergent ray is always ​parallel to the direction of the incident ray.” Why ? Explain with the help of a ray ​diagram. What happens when a ray of light falls normally on one of the faces of a rectangular ​glass prism ? Draw diagram.
    2. (ii)An object is placed at a ​distance of 30 cm from ​the optical centre of a concave lens of focal length 20 cm. Use Lens formula to determine the position of the image formed in ​this ​case.
    Show answer

    Parallel faces give $r_2=i_1$, so the emergent ray is parallel but shifted; a normal ray goes straight; $v=-12$ cm.

    (i) Let the ray meet the first face at $i_1$ and be refracted at $r_1$; then $\sin i_1/\sin r_1=n$. The two faces of a rectangular slab are parallel, so their normals are parallel and the ray inside the glass makes equal angles with them: the angle of incidence at the second face is $i_2=r_1$. At the second face the light goes from glass to air, so $\sin i_2/\sin r_2=1/n$, giving $\sin r_2=n\sin r_1=\sin i_1$, that is $r_2=i_1$. The emergent ray therefore makes the same angle with its normal as the incident ray does with its own, and the two normals are parallel, so the emergent ray is parallel to the incident ray. It is only shifted sideways, by the lateral displacement. When a ray falls **normally** on a face, $i=0^{\circ }$, so $\sin r=0$ and $r=0^{\circ }$: the ray is not bent at all, it passes straight through the glass and emerges along the same line, undeviated and with no lateral displacement. (ii) Here $u=-30$ cm and $f=-20$ cm. From $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$, $\frac{1}{v}=\frac{1}{-20}+\frac{1}{-30}=-\frac{5}{60}=-\frac{1}{12}$, so $v=-12$ cm. The image is formed $12$ cm from the optical centre on the same side as the object; with $m=v/u=+0.4$ it is virtual, erect and diminished.

    Step-by-step solution →
16. Q156 CBSE 2025
    (a) Observe the following diagram and compare

    1. (i)speed of light and
    2. (ii)optical densities of the ​three media A, B ​and C. Also give justification for your answer of any one of the ​two cases in ​terms of refractive indices of A, ​B and C. (b) Redraw the ​path of ​a ray of light through the three media, if the ray of ​light starting from medium A falls on the medium B (i) Obliquely and the optical density of medium B is made more than that of A and C. (ii) The ray falls normally from medium A to medium B.

    Show answer

    Light is fastest in C ($v_A=v_B$ less than $v_C$); A and B are equally dense and C is the rarest.

    In the given diagram the ray goes straight on at the A|B surface and bends away from the normal at the B|C surface. (a) (i) Speed of light: it is the **same** in A and in B, and greatest in C, that is $v_A=v_B<v_C$. (ii) Optical density: A and B are equally dense, and C is the rarest of the three, that is density of $A$ = density of $B$ > density of $C$. Justification in terms of refractive indices: at the A|B surface the angle of refraction equals the angle of incidence, so $n_A\sin i=n_B\sin i$ gives $n_A=n_B$. At the B|C surface the ray bends away from the normal, so the angle of refraction is greater than the angle of incidence and $n_C=n_B\frac{\sin i}{\sin r}<n_B$. Hence $n_A=n_B>n_C$, and since $v=c/n$, $v_A=v_B<v_C$. (b) (i) With B made optically denser than A and C, the ray bends **towards** the normal on entering B and **away** from the normal on entering C; because C remains rarer than A, the emergent ray in C makes a larger angle with the normal than the incident ray in A. (ii) At normal incidence the angle of incidence is $0^{\circ }$, so the angle of refraction is $0^{\circ }$ at every surface: the ray passes straight through A, B and C without any bending, though its speed changes in each medium.

    Step-by-step solution →
17. Q157 CBSE 2025
    1. (i)​Draw a ray diagram to show the refraction of a ray of ​light through a ​rectangular glass slab when it falls obliquely from air into glass.
    2. (ii)State Snell's law ​of refraction of light.
    3. (iii)​Differentiate between the virtual images formed by a convex lens and a concave lens on the basis of : (I) object distance, and (II) ​magnification.
    Show answer

    Emergent ray is parallel to the incident ray; a convex lens gives a virtual image only inside $F$ ($m>1$), a concave lens always ($m<1$).

    (i) The ray diagram is given above: the incident ray strikes the upper face of the slab, bends towards the normal inside the glass so that the angle of refraction $r$ is less than the angle of incidence $i$, travels straight to the lower face, and there bends away from the normal as it re-enters air, emerging at an angle of emergence $e=i$; the emergent ray is therefore parallel to the incident ray but shifted sideways by the lateral displacement. (ii) Snell's law: when light passes obliquely from one medium to another, the ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant for a given pair of media and for light of a given colour, that is $\frac{\sin i}{\sin r}=n_{21}$, the refractive index of the second medium with respect to the first. (iii)(I) Object distance: a convex lens forms a virtual image only when the object is placed between its optical centre and its principal focus, whereas a concave lens forms a virtual image for every position of the object, from infinity right up to the optical centre. (II) Magnification: the virtual image formed by a convex lens is enlarged, with $m>1$, while that formed by a concave lens is always diminished, with $m<1$; both images are erect and lie on the same side of the lens as the object.

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18. Q158 CBSE 2025
    1. (i)A student wishes ​to study the image formation by a concave mirror using candle flame ​as object. State the type of the image formed by the mirror and mention the change in ​the image formed, if any, ​that he observes when the candle flame is gradually moved away from the pole of the mirror. Draw a ray ​diagram to show the image formation when the object distance is nearly equal to the radius of curvature of the mirror.
    2. (ii)A convex mirror used for rear-view on an automobile has a focal length of $3.0$ m. If a bus is located at $6.0$ m from this mirror, use mirror formula ​to find the position ​of the image of the bus ​as seen in the mirror.
    Show answer

    Real, inverted image that shrinks towards F as the flame moves away; the bus's image is 2.0 m behind the mirror.

    (i) The image formed by a concave mirror depends on where the flame is. With the flame between the pole $P$ and the focus $F$ the image is virtual, erect and enlarged, and lies behind the mirror. As the flame is gradually moved away from the pole, the image changes: at $F$ it is formed at infinity, highly enlarged, real and inverted; between $F$ and $C$ it is real, inverted and enlarged and lies beyond $C$; at $C$ it is real, inverted and of the same size as the flame and lies at $C$; and beyond $C$ it is real, inverted and diminished and lies between $F$ and $C$, shrinking towards $F$ as the flame goes further away. So the image turns from virtual and erect to real and inverted as the flame passes $F$, and the real image then travels in from infinity towards $F$, getting steadily smaller. The ray diagram for an object distance nearly equal to the radius of curvature shows the object at $C$: a ray parallel to the axis reflects through $F$, a ray through $F$ reflects parallel to the axis, and the two meet at $C$ to give a real, inverted image of the same size. (ii) Here $f=+3.0$ m (convex) and $u=-6.0$ m. From $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$, $\frac{1}{v}=\frac{1}{3.0}+\frac{1}{6.0}=\frac{1}{2}$, so $v=+2.0$ m. The image is formed $2.0$ m behind the mirror; with $m=-v/u=+1/3$ it is virtual, erect and one-third the size of the bus.

    Step-by-step solution →
19. Q159 CBSE 2025
    Analyse the following observation table showing variation of ​image distance (v) with object distance (u) in case of a ​convex lens and answer the questions that follow without doing any calculations :

    1. (a)Determine the focal length of the lens. Give reason for your ​answer.
    2. (b)​Find ​magnification of the image formed in ​Observation No. 3.
    3. (c)The numerical value of magnifications in cases ​of observation 1 and 2 is same. List two differences in the images formed ​in these two cases.

    Variation of image distance (v) with object distance (u) for a convex lens

    Observation Number	Object distance (u) in cm	Image distance (v) in cm
    1	$-15$	$-60$
    2	$-25$	$+100$
    3	$-30$	$+60$
    4	$-40$	$+40$
    5	$-60$	$+30$
    6	$-100$	$+25$

    Show answer

    (a) $f=+20$ cm; (b) $m=-2$; (c) observation 1 gives a virtual, erect image, observation 2 a real, inverted one.

    (a) $f=+20$ cm. Reason: in observation 4 the image distance and the object distance are numerically equal ($40$ cm each) and lie on opposite sides of the lens. For a convex lens that happens only when the object is placed at $2F$, when the image is formed at $2F$ on the other side, the same size and inverted. Hence $2f=40$ cm, so $f=20$ cm. (b) For observation 3, $m=\frac{v}{u}=\frac{+60}{-30}=-2$. The image is real, inverted and twice the size of the object. (c) Observation 1 gives $m=+4$ and observation 2 gives $m=-4$, so the sizes are the same but the images differ in two ways: (i) The image in observation 1 is **virtual** and **erect**, while the image in observation 2 is **real** and **inverted**. (ii) The image in observation 1 is formed on the **same** side of the lens as the object and cannot be obtained on a screen; the image in observation 2 is formed on the **opposite** side of the lens and can be obtained on a screen.

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20. Q160 CBSE 2025
    A highly ​polished surface which reflects most of the light falling on it is called a mirror. ​Mirrors may be plane as well as spherical. The ​reflecting surface of a spherical mirror ​may be curved ​inwards or outwards. A spherical mirror where ​reflecting surface is ​curved inwards is ​called concave mirror, whereas spherical mirrors where reflecting surfaces are curved outwards are known as convex mirrors.

    1. (a)State the laws of reflection of light.[1]
    2. (b)Define the term ‘Principal axis of a concave mirror.’[1]
    3. (c)(i)Write the path of the following light rays after reflection from a concave mirror: (I) A ray of light moving parallel to the principal axis of a concave mirror. (II) A ray incident obliquely to the principal axis towards the pole of the concave mirror making angle $\theta$.[2]
    4. (c)(ii)OR : An object is placed at 18 cm in front of a concave mirror of focal length 12 cm. Use mirror formula to determine the position of the image.[2]
    Show answer

    $\angle i=\angle r$; the principal axis joins P and C; a parallel ray reflects through F; the image is real, at $v=-36$ cm.

    (a) The angle of incidence is equal to the angle of reflection, $\angle i=\angle r$; and the incident ray, the normal to the mirror at the point of incidence and the reflected ray all lie in the same plane. (b) The principal axis of a concave mirror is the straight line passing through its pole and its centre of curvature; it is normal to the mirror at the pole. (c)(i) (I) A ray parallel to the principal axis is reflected through the principal focus F of the concave mirror. (II) A ray incident obliquely at the pole, making an angle $\theta$ with the principal axis, is reflected obliquely at the same angle $\theta$ on the other side of the principal axis, the axis being the normal at the pole. OR (c)(ii) With $u=-18$ cm and $f=-12$ cm, $\frac{1}{v}=\frac{1}{f}-\frac{1}{u}=-\frac{3}{36}+\frac{2}{36}=-\frac{1}{36}$, so $v=-36$ cm. The image is formed 36 cm in front of the mirror, on the same side as the object; it is real, inverted and, since $m=-2$, twice the size of the object.

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