Chapter 10 · Light - Reflection and Refraction

NCERT Exemplar Class 10 Science Chapter 10 Short Answer Questions

All 10 questions of NCERT Exemplar Class 10 Science Chapter 10 Short Answer Questions (Light - Reflection and Refraction), in the book's order, each with its answer and a step-by-step solution.

1. Q20
   Identify the device used as a spherical mirror or lens ​in following cases, when the image formed is virtual and erect in ​each case.

   1. (a)Object is placed between device and its focus, image formed is enlarged and behind it.
   2. (b)Object is ​placed between the focus and device, image formed is enlarged and on the same side as that of the object.
   3. (c)Object is placed between infinity and device, image formed ​is diminished ​and between focus and optical centre on the same ​side as that of the object.
   4. (d)Object is placed between infinity ​and device, image formed is diminished and between pole and focus, behind it.
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   The devices are (a) a concave mirror, (b) a convex lens, (c) a concave lens and (d) a convex mirror.

   (a) Concave mirror - with the object between the pole and the focus it forms an enlarged, erect, virtual image behind the mirror ($u=-10$ cm with $f=-20$ cm gives $v=+20$ cm, $m=+2$). (b) Convex lens - with the object between the optical centre and the focus it forms an enlarged, erect, virtual image on the object's own side ($u=-10$ cm with $f=+15$ cm gives $v=-30$ cm, $m=+3$). (c) Concave lens - for every object position it forms a diminished, erect, virtual image between the optical centre and the focus on the object's side ($u=-30$ cm with $f=-15$ cm gives $v=-10$ cm, $m=+\frac{1}{3}$). (d) Convex mirror - for every object position it forms a diminished, erect, virtual image between the pole and the focus, behind the mirror ($u=-30$ cm with $f=+20$ cm gives $v=+12$ cm, $m=+0.4$).

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2. Q21
   Why does a light ​ray incident ​on a rectangular ​glass slab ​immersed ​in any medium emerges parallel ​to itself? ​Explain using a ​diagram.
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   Parallel faces make the two refractions equal and opposite, so $e=i$: the emergent ray is parallel to the incident ray in any medium.

   Because the two faces of a rectangular slab are parallel, so the two refractions are equal and opposite. At the first face $n_1\sin i=n_2\sin r$. The ray then crosses the slab and meets the second face, whose normal is parallel to the first, at the same angle $r$, so there $n_2\sin r=n_1\sin e$. Eliminating $n_2\sin r$ gives $n_1\sin i=n_1\sin e$, hence $\sin i=\sin e$ and $e=i$. Equal angles measured from two parallel normals mean the emergent ray is parallel to the incident ray. The refractive index $n_1$ of the surrounding medium cancels, which is why the result holds for any medium - what matters is only that the **same** medium lies on both sides. The emergent ray is not the same line as the incident ray, though: it is displaced sideways by $d=t\sin (i-r)/\cos r$, which for the $6$ cm slab in water at $i=45^{\circ }$ (so $r=38.8^{\circ }$) is $0.83$ cm.

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3. Q22
   A pencil ​when dipped in water in a glass tumbler ​appears to be bent at the interface of air and water. Will the ​pencil ​appear to be bent to the same extent, if instead of ​water we use liquids like, kerosene or turpentine. Support your answer ​with reason.
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   No: kerosene and turpentine have higher refractive indices than water, so they bend light more and the pencil looks more bent.

   No. The pencil will appear bent to a greater extent in kerosene, and to the greatest extent in turpentine. The apparent bend is caused by refraction where the light leaves the liquid, and the amount of refraction is fixed by the refractive index of the liquid: $n=1.33$ for water, $1.44$ for kerosene and $1.47$ for turpentine oil. Taking the tip of the pencil $8$ cm below the surface, its apparent depth is $8/n$, so the tip appears lifted by $1.98$ cm in water, $2.44$ cm in kerosene and $2.56$ cm in turpentine; equivalently, a ray leaving the liquid at $30^{\circ }$ to the normal is turned through $11.7^{\circ }$, $16.1^{\circ }$ and $17.3^{\circ }$ respectively. Kerosene and turpentine are optically denser than water, so they bend the light more and the pencil looks more sharply broken at the surface.

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4. Q23
   How is ​the ​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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   $n=c/v$, and the refractive index of medium 2 with respect to medium 1 is $n_{21}=v_1/v_2$, the ratio of the speeds.

   The refractive index measures how much a medium slows light down: $n=\frac{c}{v}$, where $c=3\times 10^8$ m/s is the speed of light in vacuum and $v$ is its speed in the medium, so the larger $n$ is, the slower the light travels. For two media, $n_1=\frac{c}{v_1}$ and $n_2=\frac{c}{v_2}$, and 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, $n_{21}$ is the speed of light in medium $1$ divided by the speed of light in medium $2$; likewise $n_{12}=\frac{v_2}{v_1}=\frac{1}{n_{21}}$. For water and glass, $v_1=2.25\times 10^8$ m/s and $v_2=2.00\times 10^8$ m/s give $n_{21}=1.125$.

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5. Q24
   Refractive index of ​diamond ​with respect ​to glass is $1.6$ ​and absolute ​refractive index of glass is $1.5$. Find out the absolute refractive ​index ​of diamond.
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   The absolute refractive index of diamond is $n_d=n_{dg}\times n_g=1.6\times 1.5=2.4$.

   The absolute refractive index of diamond is $2.4$ (a pure number, with no unit). Since $n_{dg}=\frac{n_d}{n_g}$, rearranging gives $n_d=n_{dg}\times n_g=1.6\times 1.5=2.4$.

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6. Q25
   A convex ​lens ​of focal length $20$ cm can produce ​a magnified virtual as well as real image. Is this a correct ​statement? If yes, ​where ​shall the object be ​placed in each case for ​obtaining these images?
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   Yes: an object nearer than 20 cm gives a magnified virtual image; between 20 cm and 40 cm, a magnified real image.

   Yes, the statement is correct. (i) For a magnified **virtual** image the object must be placed between the optical centre and the focus, that is nearer than $20$ cm to the lens. With the object at $10$ cm, $\frac{1}{v}=\frac{1}{20}-\frac{1}{10}=-\frac{1}{20}$, so $v=-20$ cm and $m=+2$: an erect virtual image twice the size, $20$ cm from the lens on the object's own side. (ii) For a magnified **real** image the object must be placed between the focus and twice the focal length, that is between $20$ cm and $40$ cm from the lens. With the object at $30$ cm, $\frac{1}{v}=\frac{1}{20}-\frac{1}{30}=\frac{1}{60}$, so $v=+60$ cm and $m=-2$: an inverted real image twice the size, $60$ cm beyond the lens. Beyond $40$ cm the real image is diminished, and with the object exactly at the focus no image is formed at all.

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7. Q26
   Sudha finds out that the sharp image of the window ​pane ​of her science laboratory ​is formed at a distance of $15$ cm from the lens. She now ​tries to focus the building visible to her outside ​the ​window instead of the window pane without disturbing the lens. In which direction ​will she move the screen to obtain a sharp image ​of the building? What is the approximate focal length of this lens?
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   The screen moves towards the lens, because the building's image forms at the focus; the focal length is approximately 15 cm.

   She must move the screen **towards** the lens. The building is much further away than the window pane, and for a fixed lens a more distant object is imaged closer in: in plain distances $v=\frac{fD}{D-f}$, which falls as the object distance $D$ grows. Since the building is effectively at infinity its rays arrive parallel, so its sharp image forms in the focal plane and the new screen position **is** the focal length. The window pane, only a few metres away, was already far compared with the focal length, so the $15$ cm measured for it is only slightly more than $f$: taking the pane at $3$ m, $\frac{1}{f}=\frac{1}{15}+\frac{1}{300}=\frac{21}{300}$, giving $f=14.3$ cm. The focal length of the lens is therefore approximately $15$ cm, and the screen moves in by less than a centimetre.

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8. Q27
   How are power and focal length ​of a lens related? You are ​provided with ​two ​lenses ​of focal length $20$ cm and $40$ ​cm respectively. ​Which ​lens will you use to obtain more convergent light?
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   $P=\frac{1}{f}$; the 20 cm lens has power $+5$ D, twice that of the 40 cm lens, so it converges light more.

   The power of a lens is the reciprocal of its focal length measured in metres: $P=\frac{1}{f}$, in dioptres (D), where $1$ D is the power of a lens of focal length $1$ m. So a shorter focal length means a greater power. For the two lenses, $f_1=20$ cm $=0.20$ m gives $P_1=\frac{1}{0.20}=+5$ D, and $f_2=40$ cm $=0.40$ m gives $P_2=\frac{1}{0.40}=+2.5$ D. The $20$ cm lens has twice the power, so it converges light more strongly - a ray striking it $2$ cm from the axis is bent through $\arctan (2/20)=5.7^{\circ }$ against only $\arctan (2/40)=2.9^{\circ }$ for the other. Use the lens of focal length $20$ cm.

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9. Q28
   Under what ​condition in an arrangement of two ​plane mirrors, incident ​ray and ​reflected ray will always be parallel to each other, whatever may ​be angle ​of incidence. ​Show the same with the help of diagram.
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   The two plane mirrors must be perpendicular ($90^{\circ }$); the two deviations then add to $180^{\circ }$ for every angle of incidence.

   The two plane mirrors must be placed perpendicular to each other, at $90^{\circ }$. A ray striking the first mirror at an angle of incidence $i$ is deviated by $180^{\circ }-2i$; because the mirrors are at right angles their normals are too, so the ray meets the second mirror at an angle of incidence $90^{\circ }-i$ and is deviated there by $180^{\circ }-2(90^{\circ }-i)=2i$. The total deviation is $(180^{\circ }-2i)+2i=180^{\circ }$, in which the angle of incidence has cancelled - so for every angle of incidence the finally reflected ray is parallel to the incident ray, travelling in the opposite direction. In general two mirrors inclined at $\theta$ deviate a ray by $360^{\circ }-2\theta$ whatever the angle of incidence; besides $\theta =90^{\circ }$ the only other value leaving the ray parallel to itself is $\theta =0^{\circ }$, two mirrors facing each other, which send it on in the **same** direction merely displaced sideways, as in a periscope.

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10. Q29
    Draw a ray diagram showing the ​path of ​rays ​of light when ​it enters with oblique incidence

    1. (i)from ​air into ​water;
    2. (ii)from ​water into air.
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    Air to water: the ray bends towards the normal; water to air: it bends away from the normal.

    (i) Air into water: water is optically denser, so the ray bends **towards** the normal. With $i=45^{\circ }$, $1.00\times \sin 45^{\circ }=1.33\sin r$ gives $\sin r=0.532$ and $r=32.1^{\circ }$, so the refracted ray is drawn on the far side of the normal but closer to it than the incident ray was. (ii) Water into air: air is optically rarer, so the ray bends **away** from the normal. With $i=30^{\circ }$ inside the water, $1.33\times \sin 30^{\circ }=1.00\times \sin r$ gives $\sin r=0.665$ and $r=41.7^{\circ }$. In both diagrams the normal is drawn dashed at the point of incidence, both angles are measured from it, and the incident and refracted rays lie on opposite sides of it; the two are the same path travelled in opposite directions. In case (ii) a refracted ray exists only while the angle of incidence stays below the critical angle $C=48.8^{\circ }$, since $\sin C=1/1.33$.

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