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 14 of 21

1. Q261 CBSE 2020
   1. (a)A concave mirror of focal length 10 cm can produce a magnified real as well as virtual image of an object placed in front of it. Draw ray ​diagrams ​to justify this statement.
   2. (b)An object is placed perpendicular to the principal axis of a convex mirror ​of focal ​length ​10 cm. The distance of the object from the pole of the ​mirror is 10 cm. Find the position of ​the image formed.
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   Concave mirror: object between F and C gives a magnified real image, between P and F a magnified virtual image; convex mirror: $v=+5$ cm.

   (a) For the concave mirror, F is $10$ cm and C is $20$ cm from the pole. (i) With the object between F and C (an object distance between $10$ cm and $20$ cm), the reflected rays meet beyond C: the image is real, inverted and magnified. (ii) With the object between P and F (less than $10$ cm), the reflected rays spread apart and appear to meet behind the mirror: the image is virtual, erect and magnified. The two ray diagrams justify the statement. (b) For the convex mirror $f=+10$ cm and $u=-10$ cm, so $\frac{1}{v}=\frac{1}{f}-\frac{1}{u}=\frac{1}{10}+\frac{1}{10}=\frac{1}{5}$ and $v=+5$ cm: the image is formed $5$ cm behind the mirror (virtual, erect and diminished, $m=+0.5$).

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2. Q262 CBSE 2020
   1. (a)​A ​security mirror used in a big showroom has radius ​of curvature 5 m. If a customer is standing at a distance of 20 m from the cash counter, find the position, ​nature and size of the image formed in the security mirror.
   2. (b)Neha visited a dentist in his clinic. She observed that the dentist was holding an instrument fitted with ​a ​mirror. State the nature of this mirror and reason for its use in the instrument used ​by dentist.
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   (a) Virtual, erect, diminished image 2.22 m behind the convex mirror; (b) a concave mirror gives an enlarged image of the tooth.

   (a) A security mirror is convex, so $R=+5$ m and $f=+2.5$ m, with $u=-20$ m. From $\frac{1}{v}=\frac{1}{f}-\frac{1}{u}=\frac{1}{2.5}+\frac{1}{20}=\frac{9}{20}$, $v=\frac{20}{9}\approx +2.22$ m. Position: $2.22$ m behind the mirror, between P and F. Nature: virtual and erect. Size: diminished; $m=-\frac{v}{u}=+\frac{1}{9}$, so the image is one-ninth the size of the customer. (b) The mirror is concave. With the tooth held between its pole and its focus, it forms an erect and enlarged (virtual) image of the tooth, so the dentist sees the tooth magnified.

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3. Q263 CBSE 2020
   1. (a)Draw a ​labelled ray diagram to show the ​path of a ray of light incident obliquely on one face of a glass slab.
   2. (b)Calculate the refractive index of the material of a glass slab. Given that the speed of ​light through the glass slab ​is $2\times 10^8$ m/s and ​in air is $3\times 10^8$ m/s.
   3. (c)Calculate the focal length ​of a ​lens, if its power ​is $-2.5$ D.
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   The ray bends towards the normal in glass and emerges parallel but shifted; $n=1.5$; $f=-40$ cm (concave lens).

   (a) The ray diagram shows a ray AO incident obliquely at O on one face of the glass slab, with the normal NN′ at O. On entering the glass the ray bends towards the normal and travels along OO′, the angle of refraction $r$ being smaller than the angle of incidence $i$. At O′ on the opposite face it bends away from the normal MM′ and emerges along O′B at the angle of emergence $e=i$, so the emergent ray is parallel to the incident ray but laterally displaced. Arrows mark the direction of light on the incident, refracted and emergent rays. (b) $n_{ga}=\frac{\text{speed of light in air}}{\text{speed of light in glass}}=\frac{3\times 10^8\text{m/s}}{2\times 10^8\text{m/s}}=1.5$. (c) $f=\frac{1}{P}=\frac{1}{-2.5\text{D}}=-0.4$ m $=-40$ cm; the negative sign shows a concave (diverging) lens.

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4. Q264 CBSE 2020
   (a) Define the ​following terms :

   1. (i)Power ​of a lens
   2. (ii)Principal focus of a ​concave mirror (b) Write the relationship among the object ​distance ​($u$), image distance ($\nu$) and the focal length ($f$) of a (i) Spherical lens (ii) ​Spherical mirror (c) An object is placed at a distance of 10 cm from optical centre of a convex lens of focal ​length 15 cm. Draw a labelled ray diagram to show the formation of image in this case.
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   Power is the reciprocal of focal length; lens $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$, mirror $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$; the image for $u=-10$ cm is virtual, erect and magnified.

   (a) (i) Power of a lens is the degree of convergence or divergence of light rays achieved by the lens; it is the reciprocal of its focal length, $P=\frac{1}{f}$ ($f$ in metres), measured in dioptres (D). (ii) The principal focus of a concave mirror is the point on the principal axis at which rays of light incident parallel to the principal axis meet after reflection from the mirror. (b) (i) Spherical lens: $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$. (ii) Spherical mirror: $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$. (c) The object lies between $F_1$ and the optical centre O ($BO=10$ cm, $O{F}_1=15$ cm). A ray from A parallel to the principal axis passes through $F_2$ after refraction, and a ray through O goes straight on; their backward extensions meet at $A^{\prime }$ on the same side as the object. The image $A^{\prime }{B}^{\prime }$ is virtual, erect and magnified, $30$ cm from the lens ($v=-30$ cm, $m=+3$), as shown in the ray diagram.

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5. Q265 CBSE 2020
   Rishi went ​to a palmist to show his ​palm. The palmist used a ​special lens for this purpose.

   1. (i)State the nature of the lens and reason for its use.
   2. (ii)Where should the palmist place/hold the lens so as to have a real ​and magnified image of an object ?
   3. (iii)If the focal length of this lens is 10 cm ​and the ​lens ​is held at a distance ​of 5 cm from the palm, use lens formula to find the position and size of the image.
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   Convex lens; palm between $F_1$ and $2F_1$; image at $v=-10$ cm, virtual, erect and twice the size ($m=+2$).

   (i) A convex lens (a magnifying glass). With the palm held within its focal length, it forms a virtual, erect and magnified image, so the lines on the palm look bigger and are seen clearly. (ii) The palm should lie between $F_1$ and $2F_1$ of the lens, that is, the lens should be held at a distance between $f$ and $2f$ from the palm (between $10$ cm and $20$ cm for this lens); the image is then real, inverted and enlarged, formed beyond $2F_2$. Holding the palm at $F_1$ gives a real, highly enlarged image at infinity, which is also accepted. (iii) With $f=+10$ cm and $u=-5$ cm, $\frac{1}{v}=\frac{1}{f}+\frac{1}{u}=\frac{1}{10}-\frac{1}{5}=-\frac{1}{10}$, so $v=-10$ cm: the image is $10$ cm from the lens, on the same side as the palm. $m=\frac{v}{u}=\frac{-10}{-5}=+2$: the image is virtual, erect and twice the size of the object.

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6. Q266 CBSE 2020
   1. (a)A person suffering from myopia (near-sightedness) was advised to ​wear corrective lens of power $-2.5$ D. A spherical lens ​of same focal length was taken in the laboratory. ​At what distance should a student place an object from this lens so that it ​forms ​an image at a distance of 10 cm ​from the lens ?
   2. (b)Draw a ray diagram ​to ​show the position and nature of the image formed in the above case.
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   $u\approx -13.3$ cm: the object is about 13.3 cm from the concave lens; the image is virtual, erect and diminished.

   (a) $f=\frac{1}{P}=\frac{1}{-2.5\text{D}}=-0.4$ m $=-40$ cm; the negative power shows a concave (diverging) lens. A concave lens forms a virtual image on the object's side, so $v=-10$ cm. From $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$: $\frac{1}{u}=\frac{1}{-10}-\frac{1}{-40}=-\frac{3}{40}$, so $u=-\frac{40}{3}$ cm $\approx -13.3$ cm. The object must be placed $\approx 13.3$ cm in front of the lens. (b) Since the power is negative, the lens is concave (diverging). In the ray diagram, a ray from the top of the object parallel to the axis diverges after refraction as if coming from $F_1$ (its backward extension is dashed), and a ray through the optical centre passes undeviated; they meet 10 cm from the lens on the object's side. The image is virtual, erect and diminished ($m=\frac{v}{u}=+0.75$), between the optical centre and the focus $F_1$.

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7. Q267 CBSE 2019
   Draw ​a labelled ray diagram to ​show the path of the reflected ray ​corresponding to an incident ray of light parallel to the ​principal ​axis of a convex mirror. Mark ​the angle ​of incidence ​and angle of reflection on it.
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   The ray parallel to the axis reflects along MB, appearing to come from the principal focus F; $\angle i=\angle r$.

   Ray diagram (see the figure): the ray $AM$, parallel to the principal axis, strikes the convex mirror at $M$ and is reflected along $MB$, away from the axis; the dotted backward extension of $MB$ passes through the principal focus $F$ behind the mirror. $CM$ extended to $N$ is the normal at $M$. The angle of incidence $\angle AMN=i$ and the angle of reflection $\angle NMB=r$ are marked, with $\angle i=\angle r$. The mirror, $P$, $F$, $C$ and the arrows on both rays are shown.

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8. Q268 CBSE 2019
   Draw a labelled ray diagram ​to show the path of ​the ​reflected ray corresponding to the ​ray which is directed towards ​the principal focus of a ​convex mirror. Mark the angle of incidence and angle of reflection ​on it.
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   The ray directed towards the focus is reflected parallel to the principal axis, with $\angle i=\angle r$.

   Ray diagram (see the figure): the ray $AM$, directed towards the principal focus $F$ behind the convex mirror, meets the mirror at $M$ and is reflected along $MB$, parallel to the principal axis. $CM$ produced to $N$ is the normal at $M$. The angle of incidence $\angle i=\angle AMN$ and the angle of reflection $\angle r=\angle NMB$ are marked, with $\angle i=\angle r$. The part of $AM$ beyond the mirror, towards $F$, is dashed; arrows show the direction of both rays; $P$, $F$ and $C$ are labelled, with $F$ midway between $P$ and $C$.

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9. Q269 CBSE 2019
   Draw a labelled ray diagram ​to show the ​path of the ​reflected ray ​corresponding to the ray which is incident obliquely to the principal ​axis, towards the pole of a convex mirror. Mark the angle of incidence and angle ​of ​reflection ​on it.
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   The ray towards the pole is reflected to the other side of the axis with $\angle r=\angle i$, both measured from the axis.

   The ray aimed at the pole $P$ is reflected obliquely to the other side of the principal axis. At the pole the principal axis is the normal, so the reflected ray makes the same angle with the axis as the incident ray: $\angle r=\angle i$. In the diagram, $\angle i$ is marked between the incident ray and the principal axis, and $\angle r$ between the reflected ray and the principal axis (see the figure).

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10. Q270 CBSE 2019
    State ​laws of reflection ​of light.
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    The incident ray, normal and reflected ray lie in one plane, and the angle of incidence equals the angle of reflection.

    (i) The incident ray, the normal to the reflecting surface at the point of incidence and the reflected ray all lie in the same plane. (ii) The angle of incidence is equal to the angle of reflection, $\angle i=\angle r$.

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11. Q271 CBSE 2019
    State laws ​of refraction ​of light.
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    The incident ray, refracted ray and normal lie in one plane; $\frac{\sin i}{\sin r}$ is constant for a given pair of media (Snell's law).

    (i) First law: The incident ray, the refracted ray and the normal to the interface of the two media at the point of incidence all lie in the same plane. (ii) Second law (Snell's law): For light of a given colour and a given pair of media, the ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant, $\frac{\sin i}{\sin r}=\text{constant}$. This constant is the refractive index of the second medium with respect to the first, $n_{21}$.

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12. Q272 CBSE 2019
    List ​four characteristics ​of ​the image formed by ​a convex lens of focal length 20 ​cm when the object is placed in front of it at a distance of 10 cm ​from ​its optical centre.
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    The image is virtual, erect, magnified twice ($m=+2$) and formed on the object's side, 20 cm from the lens, at $F_1$.

    The object ($10$ cm from the lens) is inside the focal length ($20$ cm), so the image is: (i) Virtual: it cannot be obtained on a screen. (ii) Erect. (iii) Magnified: twice the size of the object ($m=+2$). (iv) On the same side of the lens as the object, $20$ cm from the lens ($v=-20$ cm), that is, at the focus $F_1$.

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13. Q273 CBSE 2019
    Define ​absolute ​refractive ​index and ​express ​it ​mathematically.
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    Absolute refractive index is the ratio of the speed of light in vacuum to its speed in the medium, $n_m=\frac{c}{v}$.

    The absolute refractive index of a medium is the refractive index of that medium with respect to vacuum (or air), that is, the ratio of the speed of light in vacuum to the speed of light in the medium. Mathematically, $n_m=\frac{c}{v}=\frac{\text{speed of light in vacuum (or air)}}{\text{speed of light in the medium}}$.

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14. Q274 CBSE 2019
    List ​four characteristics of the image formed by ​a ​concave mirror of focal length 40 cm when ​the ​object is ​placed in front of ​it at a distance of 20 ​cm from its pole.
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    The image is virtual, erect, magnified twice ($m=+2$) and formed 40 cm behind the mirror.

    The image is: (i) Virtual (it cannot be obtained on a screen); (ii) Erect; (iii) Magnified, twice the size of the object ($m=+2$); (iv) Formed behind the mirror, $40$ cm from its pole ($v=+40$ cm).

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15. Q275 CBSE 2019
    Define refractive index of ​a transparent medium. ​The ​speed of ​light in a medium of absolute refractive index $1.5$ is $2\times 10^8{\text{ms}}^{-1}$. What is ​the speed of ​light in vacuum ​?
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    Refractive index is the ratio of the speed of light in vacuum to that in the medium; $c=3\times 10^8$ m/s.

    Refractive index: The absolute refractive index of a transparent medium is the ratio of the speed of light in vacuum (or air) to the speed of light in the medium, $n_m=\frac{c}{v}$. Speed of light in vacuum: $c=n_m\times v=1.5\times 2\times 10^8=3\times 10^8$ m s${}^{-1}$.

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16. Q276 CBSE 2019
    The absolute ​refractive index of Ruby is 1.7. Find the ​speed ​of light ​in Ruby. The speed of light in ​vacuum is $3\times 10^8$ ​m/s.
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    The speed of light in ruby is $v\approx 1.76\times 10^8$ m/s.

    Speed of light in ruby: $v=\frac{c}{n}=\frac{3\times 10^8}{1.7}\approx 1.76\times 10^8$ m/s.

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17. Q277 CBSE 2019
    The power of ​a lens is $+5$ diopters. What is ​the nature and focal length of this lens ​? ​At what distance from this lens ​should an object be ​placed so as to get its ​inverted ​image of the same size ?
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    The lens is convex with $f=+20$ cm; the object must be 40 cm in front of it, at $2F_1$.

    Focal length: $f=\frac{1}{P}=\frac{1}{+5}$ m $=+0.2$ m $=+20$ cm. Nature: the focal length is positive, so the lens is a convex (converging) lens. Object distance: an inverted image of the same size is formed when the object is at $2F_1$, that is, $2f=40$ cm in front of the lens ($u=-40$ cm). The image is then formed at $2F_2$, $40$ cm behind the lens ($v=+40$ cm), real, inverted and of the same size.

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18. Q278 CBSE 2019
    State Snell’s law ​of ​refraction ​of light. ​Write an expression for the absolute ​refractive ​index ​of a medium in terms ​of speed of light.
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    Snell's law: $\frac{\sin i}{\sin r}$ is constant for a given pair of media; the absolute refractive index is $n_m=\frac{c}{v}$.

    Snell's law: the ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant, for light of a given colour and a given pair of media: $\frac{\sin i}{\sin r}=\text{constant}=n_{21}$. Absolute refractive index of a medium: $n_m=\frac{c}{v}$, where $c$ is the speed of light in vacuum and $v$ is the speed of light in the medium.

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19. Q279 CBSE 2019
    How is refractive index of a ​medium related to ​the speed of light ​? ​Obtain an expression ​for refractive index of ​a medium ​with respect to ​another in terms of speed of light in these two media.
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    Refractive index is inversely related to the speed of light, and $n_{21}=\frac{v_1}{v_2}$.

    The refractive index of a medium is inversely related to the speed of light in it: the absolute refractive index is $n_m=\frac{c}{v}$, the speed of light in vacuum divided by the speed of light in the medium, so the higher the refractive index, the slower light travels in that medium. For medium 1 (speed $v_1$) and medium 2 (speed $v_2$): $n_1=\frac{c}{v_1}$ and $n_2=\frac{c}{v_2}$, so the refractive index of medium 2 with respect to medium 1 is $n_{21}=\frac{n_2}{n_1}=\frac{c/v_2}{c/v_1}=\frac{v_1}{v_2}$, that is, the speed of light in medium 1 divided by the speed of light in medium 2.

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20. Q280 CBSE 2019
    State two ​laws ​on the basis of which ​the ​refraction ​of light occurs.
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    The incident ray, refracted ray and normal lie in one plane; $\frac{\sin i}{\sin r}$ is constant (Snell's law).

    (i) The incident ray, the refracted ray and the normal to the interface of the two transparent media at the point of incidence all lie in the same plane. (ii) The ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant for light of a given colour and for a given pair of media: $\frac{\sin i}{\sin r}=\text{constant}$ (Snell's law of refraction). This constant is the refractive index of the second medium with respect to the first.

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