Class 10 Science · Chapter 9

Light - Reflection and Refraction

194 Class 10 Science questions on Light - Reflection and Refraction from the NCERT textbook, NCERT Exemplar and CBSE board papers 2023–2026, each with its answer and a step-by-step solution.

CBSE Class 10 Science previous year papers, with answers →

Questions · page 5 of 10

1. Q81 CBSE 2026
   Draw ​the ray ​diagram for ​the ​image ​formation by a ​lens which ​shows ​a magnification of $+2$.
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   A convex lens with the object midway between O and $F_1$ forms a virtual, erect image with $m=+2$.

   Use a convex lens with the object placed between its optical centre $O$ and its principal focus $F_1$ - exactly midway, at $u=-\frac{f}{2}$. In the ray diagram a ray from the top of the object parallel to the principal axis is refracted through $F_2$, and a ray through $O$ passes undeviated, with arrowheads on the rays showing the direction of light; the refracted rays diverge, and their dashed backward extensions meet on the object's side, at $F_1$. The image A′B′ formed there ($v=-f$) is virtual, erect and twice the size of the object: $m=\frac{v}{u}=\frac{-f}{-f/2}=+2$.

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2. Q82 CBSE 2026
   With the help ​of the ​given ​ray ​diagram, ​calculate ​the ​refractive index ​of Medium 2 with respect to Medium 1. (Given that $\sqrt{3}=1.732$)

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   The refractive index of Medium 2 with respect to Medium 1 is $n_{21}=\sqrt{3}=1.732$.

   The angle of incidence is $i=90^{\circ }-30^{\circ }=60^{\circ }$ and the angle of refraction is $r=30^{\circ }$, so $n_{21}=\frac{\sin 60^{\circ }}{\sin 30^{\circ }}=\frac{\sqrt{3}/2}{1/2}=\sqrt{3}=1.732$. The refractive index of Medium 2 with respect to Medium 1 is $1.732$.

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3. Q83 CBSE 2026
   Use the above given ray ​diagram ​to calculate the angle of ​incidence. ​Given that the refractive index of Medium 2 with respect ​to ​the Medium 1 is $\sqrt{3}$.

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   The angle of incidence is $i=60^{\circ }$.

   The printed $60^{\circ }$ is the angle between the refracted ray OB and the boundary, so the angle of refraction is $r=90^{\circ }-60^{\circ }=30^{\circ }$. By Snell's law, $\sin i=n_{21}\sin r=\sqrt{3}\times \frac{1}{2}=\frac{\sqrt{3}}{2}$, so the angle of incidence is $i=60^{\circ }$.

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4. Q84 CBSE 2026
   With the ​help of above given ray diagram, ​calculate the angle of refraction. Given that the refractive index ​of Medium ​2 ​with respect to Medium ​1 ​is $\sqrt{3}$.

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   The angle of refraction is $r=30^{\circ }$.

   By Snell's law, $n_{21}=\frac{\sin i}{\sin r}$. The angle of incidence is $i=90^{\circ }-30^{\circ }=60^{\circ }$, so $\sin r=\frac{\sin 60^{\circ }}{\sqrt{3}}=\frac{\sqrt{3}/2}{\sqrt{3}}=\frac{1}{2}$ and the angle of refraction is $r=30^{\circ }$.

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5. Q85 CBSE 2026
   A ray of ​light enters into medium B from medium A as shown in the given figure :

   1. (a)Which ​one of these ​two media is ​optically ​denser ​than the other medium ? Justify your answer.
   2. (b)If the speed of light in medium A ​is $V_A$ and in medium B is $V_B$, then what ​is the refractive index of medium B with respect to medium A ?

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   Medium A is optically denser because the ray bends away from the normal; $n_{BA}=\frac{V_A}{V_B}$.

   (a) Medium A is optically denser. $N{N}^{\prime }$ is the normal: the refracted ray in medium B makes a larger angle with it than the incident ray in medium A ($r>i$), so the ray bends away from the normal, which happens when light passes from a denser into a rarer medium (it speeds up). (b) The refractive index of medium B with respect to medium A is $n_{BA}=\frac{V_A}{V_B}$.

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6. Q86 CBSE 2026
   Draw the ray diagram ​showing refraction through a glass slab ​and write ​the relation between angle of incidence, angle ​of ​refraction ​and refractive ​index of glass.
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   The ray bends towards the normal on entering and away on leaving, emerging parallel; $\frac{\sin i}{\sin r}=n_g$.

   Ray diagram: see the figure. The incident ray $AO$ bends towards the normal on entering the glass (angle of refraction $r$ less than $i$), and bends away from the normal on leaving it at $O^{\prime }$, emerging along $O^{\prime }B$ at an angle of emergence $e=i$ - parallel to the incident ray but laterally displaced. Relation (Snell's law): $\frac{\sin i}{\sin r}=n_g$, where $i$ is the angle of incidence in air, $r$ the angle of refraction in glass and $n_g$ the refractive index of glass with respect to air.

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7. Q87 CBSE 2026
   1. (a)​Relate the speed of light in ​the given medium with its ​optical density.
   2. (b)Using the information ​given in the table below, arrange the medium A, B and C in the ascending order of their optical ​density.

   Medium	Speed of light (m/s)
   A	2.25 × 10⁸
   B	2 × 10⁸
   C	2.08 × 10⁸

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   Greater optical density means lower speed of light; the order of optical density is $\text{A}<\text{C}<\text{B}$.

   (a) The speed of light in a medium is inversely related to its optical density: the lower the speed of light in a medium, the greater its optical density (and the greater its refractive index, $n=\frac{c}{v}$). Light travels fastest in the optically rarest medium and slowest in the optically densest one. (b) Ascending order of optical density: $\text{A}<\text{C}<\text{B}$ - their speeds are $2.25\times 10^8$, $2.08\times 10^8$ and $2\times 10^8$ m/s, giving refractive indices of about $1.33$, $1.44$ and $1.50$.

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8. Q88 CBSE 2026
   1. (a)​Define absolute ​refractive index of ​an optical ​medium.
   2. (b)​Arrange these material mediums given in ​the table in ​increasing order of speed of light through them.

   Material Medium	Refractive Index
   A	1.50
   B	1.46
   C	1.31
   D	1.77

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   Absolute refractive index is $n=\frac{c}{v}$; increasing order of speed of light is $D<A<B<C$.

   (a) The absolute refractive index of a medium is the ratio of the speed of light in vacuum to the speed of light in that medium, $n=\frac{c}{v}$; being a ratio of two speeds it has no unit. (b) Increasing order of speed of light: $D<A<B<C$, because $v=\frac{c}{n}$ and $n_D=1.77>n_A=1.50>n_B=1.46>n_C=1.31$.

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9. Q89 CBSE 2026
   When an incident ray of light enters in a medium ​'X' from medium 'Y', ​it bends away from ​the normal. Comment about the ​following :

   1. (a)Speed of light in medium 'X' with respect to the speed ​of light in medium of 'Y'.
   2. (b)Optical ​density of ​medium 'X' with respect to the optical density of medium 'Y'. Give reason for your answer in ​each case.
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   Light travels faster in X than in Y, so X is optically rarer than Y.

   (a) The speed of light in medium 'X' is greater than the speed of light in medium 'Y'. A ray turns away from the normal only when it passes into a medium in which light travels faster, and $n=c/v$ makes the faster medium the one with the smaller refractive index. (b) Medium 'X' is optically rarer than medium 'Y'. From $n_Y\sin i=n_X\sin r$ with $r>i$ we get $n_X<n_Y$, and of two media the one with the smaller refractive index is the optically rarer.

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10. Q90 CBSE 2026
    Write any two ​differences between the ​images ​formed ​by ​convex mirror ​and plane mirror.
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    The plane mirror image is the same size as the object; the convex mirror image is diminished and lies between P and F.

    Any two of the following: (1) Size: the image formed by a plane mirror is of the same size as the object ($m=+1$), whereas the image formed by a convex mirror is diminished, smaller than the object ($0<m<1$). (2) Position: the image formed by a plane mirror is as far behind the mirror as the object is in front of it, whereas the image formed by a convex mirror lies behind the mirror between the pole P and the focus F, nearer to the mirror than the object is. (3) Dependence on object distance: the plane mirror image stays the same size wherever the object is, whereas the convex mirror image becomes smaller as the object moves away from the mirror. (Both images are virtual and erect, so those are not differences.)

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11. Q91 CBSE 2026
    The values of refractive indices of two optical media ​X and Y are ​1.44 and ​1.30, respectively. ​Compare the ​values of the following quantities pertaining ​to them :

    1. (a)Speed of light
    2. (b)Angle ​of refraction for same ​angle of incidence Justify your answer.
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    Light travels faster in Y than in X, and for the same angle of incidence the angle of refraction is smaller in X.

    (a) Light travels faster in Y than in X. Since $v=c/n$, $v_X=\frac{3\times 10^8}{1.44}=2.08\times 10^8$ m/s and $v_Y=\frac{3\times 10^8}{1.30}=2.31\times 10^8$ m/s: the larger refractive index belongs to the optically denser medium, X, in which light is slower. (b) For the same angle of incidence the angle of refraction is smaller in X than in Y. By Snell's law $\sin r=\frac{\sin i}{n}$, and $n_X>n_Y$ gives $\sin r_X<\sin r_Y$, so $r_X<r_Y$: light bends more towards the normal in the optically denser medium X.

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12. Q92 CBSE 2026
    Redraw the ​following ​ray diagrams in ​your answer book and show ​the path of reflected/refracted ray :

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    The reflected ray passes through the focus $F$; the refracted ray bends away from the axis and appears to come from $F_1$.

    (i) After reflection from the concave mirror the ray passes through the principal focus $F$: the reflected ray starts where the incident ray meets the mirror, slopes down through $F$ and carries on beyond it. (ii) After refraction through the concave lens the ray bends away from the principal axis and appears to diverge from the principal focus $F_1$ on the same side as the incident ray: produced backwards as a dashed line, the refracted ray passes through $F_1$.

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13. Q93 CBSE 2026
    A mirror always forms ​a ​virtual, erect and diminished ​image. Identify ​the mirror ​and draw ​a labelled ​ray diagram for image formation by this mirror.
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    A convex mirror always gives a virtual, erect, diminished image, formed behind the mirror between P and F.

    The mirror is a convex mirror. In the ray diagram, a ray from the top $A$ of the object travelling parallel to the principal axis is reflected as if it came from the focus $F$, and a ray aimed at the centre of curvature $C$ is reflected back along its own path. Their backward extensions meet at $A^{\prime }$ behind the mirror, so the image $A^{\prime }{B}^{\prime }$ is formed between $P$ and $F$: virtual, erect and diminished.

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14. Q94 CBSE 2026
    When a ​ray of ​light passes from ​water to air (I) will the angle of refraction in air ($r$) be greater than the angle of incidence in water ($i$) or less ​than it ? ​(II) and when we increase the angle ​of incidence in water, will the angle ​of refraction in air increase or decrease ? What is the limiting value for $\angle r$ ?
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    The angle of refraction in air is greater than $i$, it increases with $i$, and its limiting value is $90^{\circ }$.

    (I) The angle of refraction in air is greater than the angle of incidence in water ($r>i$), because the ray passes from an optically denser medium (water) into a rarer one (air), speeds up and bends away from the normal. (II) When the angle of incidence in water is increased, the angle of refraction in air also increases (faster than $i$, since $\sin r=1.33\sin i$). The limiting value of $\angle r$ is $90^{\circ }$: the refracted ray then grazes along the water surface, which for water happens at $i\approx 48.8^{\circ }$.

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15. Q95 CBSE 2026
    (a) Observe the given ray ​diagram and compare :

    1. (i)​the speed of light in medium A and the speed of light ​in medium B.
    2. (ii)the optical densities ​of medium B and ​medium C. (b) Draw a ray diagram for refraction where the optical density of medium ‘A’ becomes equal to the ​optical density of medium ‘C’ and optical density of medium ‘B’ is greater than the optical density of medium ‘A’ and medium ‘C’.

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    Light is faster in A than in B; B and C have equal optical density; in (b) the ray emerges parallel to the incident ray.

    (a) (i) At the A-B face the ray bends towards the normal, that is $r<i$, so medium B is optically denser than medium A. Since $v=c/n$, light travels faster in the optically rarer medium, and therefore the speed of light in medium A is greater than the speed of light in medium B ($v_A>v_B$). (ii) At the B-C face the ray passes on without any change of direction, so $n_B=n_C$: media B and C have the same optical density. (b) See the diagram. With the optical density of ‘A’ equal to that of ‘C’ and the optical density of ‘B’ greater than both, the ray bends towards the normal on entering B and away from the normal by the same amount on entering C, so $e=i$: the emergent ray in medium C is parallel to the incident ray in medium A and is only laterally displaced.

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16. Q96 CBSE 2026
    1. (a)​Write the expression for the magnification produced by ​a lens in terms of object distance and image distance.
    2. (b)A 4 cm ​tall object is placed perpendicular to the ​principal axis of a ​convex lens of ​focal length 20 cm. Calculate the size ​of the image formed, if ​the distance of the object from the lens is 10 cm.
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    $m=v/u$; the image is 8 cm tall, virtual and erect, 20 cm from the lens on the object's side.

    (a) The magnification produced by a lens is $m=\frac{h^{\prime }}{h}=\frac{v}{u}$, where $v$ is the image distance and $u$ the object distance. (b) From the lens formula, $\frac{1}{v}=\frac{1}{f}+\frac{1}{u}=\frac{1}{20}-\frac{1}{10}=-\frac{1}{20}$, so $v=-20$ cm; then $m=\frac{v}{u}=\frac{-20}{-10}=+2$ and $h^{\prime }=m\times h=2\times 4=8$ cm. The image is $8$ cm tall - virtual, erect and magnified, formed $20$ cm from the lens on the same side as the object.

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17. Q97 CBSE 2026
    1. (a)State the ​laws which help to find out the ​angle ​of incidence and angle ​of reflection in the ray diagram given ​above.
    2. (b)Find out the angle of ​incidence and angle of ​reflection in the given ray ​diagram.

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    The angle of incidence equals the angle of reflection; here both are $50^{\circ }$.

    (a) The laws of reflection: the angle of incidence is equal to the angle of reflection; 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 printed $40^{\circ }$ is measured from the mirror surface, so the angle of incidence is $\angle i=90^{\circ }-40^{\circ }=50^{\circ }$, and by the first law the angle of reflection is $\angle r=50^{\circ }$.

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18. Q98 CBSE 2026
    1. (a)Write ​the name of the lens and position of the object in ​front ​of it so that a virtual ​and ​magnified image is formed.
    2. (b)An object is placed ​perpendicular to the principal axis of a convex lens of focal length 20 cm. If object is located at 30 ​cm from this lens, using lens formula find out the position and nature of the image ​formed.
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    Convex lens, object between $F_1$ and $O$; $v=+60$ cm, a real, inverted image twice the size of the object.

    (a) A convex lens, with the object placed between its principal focus $F_1$ and its optical centre $O$ (nearer to the lens than its focal length); the image is then virtual, erect and magnified, on the same side as the object. (b) With $u=-30$ cm and $f=+20$ cm, $\frac{1}{v}=\frac{1}{f}+\frac{1}{u}=\frac{1}{20}-\frac{1}{30}=\frac{1}{60}$, so $v=+60$ cm: the image is formed $60$ cm from the lens on the other side (beyond $2F_2$). It is real and inverted, and twice the size of the object ($m=\frac{v}{u}=-2$).

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19. Q99 CBSE 2026
    The various positions of an object from a ​convex lens and ​the relative sizes of ​the images formed in each position are ​tabulated ​below. Look at the ​data carefully and answer the questions that follow the table :

    1. (a)What is the focal length of the lens ?
    2. (b)What is the nature of the images formed in case of ​observation 3 and 4 ?

    S.No.	Position of object from lens (cm)	Relative size of image (cm)
    1.	45	Smaller than object
    2.	30	Same size as object
    3.	20	Magnified
    4.	10	Magnified

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    $f=15$ cm; observation 3 gives a real, inverted, magnified image and observation 4 a virtual, erect, magnified image.

    (a) The focal length of the lens is $15$ cm. In observation 2 the image is the same size as the object, which happens with a convex lens only when the object is at $2F$; so $2f=30$ cm and $f=15$ cm. (b) Observation 3 (object at $20$ cm, between $F$ and $2F$): the image is real, inverted and magnified, formed beyond $2F$ on the other side of the lens ($v=60$ cm). Observation 4 (object at $10$ cm, between $F$ and the optical centre): the image is virtual, erect and magnified, formed on the same side of the lens as the object ($v=-30$ cm), and it cannot be obtained on a screen.

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20. Q100 CBSE 2026
    1. (a)A ray of light ​after reflection from a concave mirror becomes ​parallel to its principal axis. Draw a labelled ray ​diagram to show this.
    2. (b)(i) ​Differentiate between the virtual images formed by a convex mirror ​and concave mirror. (ii) “The value of magnification ‘$m$’ for a mirror is $-2$.” ​Calculate the height of object if the height of image is 20 cm.
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    A ray through $F$ reflects parallel to the axis; a concave mirror's virtual image is enlarged, a convex mirror's diminished; $h=10$ cm.

    (a) A ray passing through the principal focus $F$ of a concave mirror is reflected parallel to the principal axis: the labelled diagram shows the mirror with $P$, $F$ and $C$ on the principal axis ($PF=FC$), the incident ray drawn through $F$ to the mirror, the normal $CM$ at the point of incidence, and the reflected ray leaving parallel to the principal axis. (b) (i) A concave mirror forms a virtual image only when the object is between $P$ and $F$; that image is erect, enlarged, and lies behind the mirror farther from it than the object. A convex mirror forms a virtual image for every position of the object; that image is erect, diminished, and always lies behind the mirror between $P$ and $F$. (ii) $m=\frac{h^{\prime }}{h}=-2$ is negative, so the image is inverted and $h^{\prime }=-20$ cm; hence $h=\frac{h^{\prime }}{m}=\frac{-20}{-2}=10$ cm. The object is $10$ cm tall.

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