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NCERT Exemplar · Class 10 Science Light - Reflection and Refraction

38 questions · 38 still being checked

Short Answer Questions 20–29 (part 3 of 4)

  1. Exercise 20

    Identify the device used as a spherical mirror or lens in following cases, when the image formed is virtual and erect in each case.
    (a)
    Object is placed between device and its focus, image formed is enlarged and behind it.
    (b)
    Object is placed between the focus and device, image formed is enlarged and on the same side as that of the object.
    (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.
    (d)
    Object is placed between infinity and device, image formed is diminished and between pole and focus, behind it.

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    NCERT’s answer
    (a)
    concave mirror
    (b)
    convex lens
    (c)
    concave lens
    (d)
    convex mirror
    (a)
    Concave mirror — object between P and F.
    (b)
    Convex lens — object between F and O.
    (c)
    Concave lens — image is always diminished, whatever the object distance.
    (d)
    Convex mirror — image is always diminished, whatever the object distance.
    Answer: (a) concave mirror (b) convex lens (c) concave lens (d) convex mirror

  2. Exercise 21

    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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    NCERT’s answer
    Hint— Draw the diagram and explain using laws of refractions at both the interfaces.
    Parallel faces, same medium $\displaystyle 1$ on both sides of the glass slab ($\displaystyle 2$).First face ($\displaystyle 1$ \(\displaystyle \to\) $\displaystyle 2$): \[n_1 \sin i = n_2 \sin r \] Normals are parallel, so the angle of incidence at the second face is again \(\displaystyle r\) (alternate angles). Second face ($\displaystyle 2$ \(\displaystyle \to\) $\displaystyle 1$): \[n_2 \sin r = n_1 \sin e \] \[n_1 \sin i = n_1 \sin e \Rightarrow i = e \] Answer: Same medium on both sides of parallel faces gives \(\displaystyle e = i\), so the emergent ray is parallel to the incident ray, shifted sideways by \(\displaystyle d\).

  3. Exercise 22

    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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    NCERT’s answer
    Hint— No. Bending will be different in different liquids since velocity of light at the interface separating two media depends on the relative refractive index of the medium.
    The bending is produced by refraction, so it depends on the liquid's refractive index. \[n_{\text{liquid}}\uparrow \Rightarrow \text{bending at the interface}\uparrow \] Answer: No. Kerosene (\(\displaystyle n \approx 1.44\)) and turpentine (\(\displaystyle n \approx 1.47\)) are optically denser than water (\(\displaystyle n \approx 1.33\)), so the pencil looks more bent in each, by a different amount.

  4. Exercise 23

    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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    Hint— \(\displaystyle \mathrm{n}=\frac{\mathrm{c}}{v}\) \[\mathrm{n}_{21}=\frac{\mathrm{v}_1}{\mathrm{v}_2} \]
    Absolute refractive index of a medium: \[n = \dfrac{c}{v} \] \(\displaystyle c\) = speed of light in vacuum, \(\displaystyle v\) = speed in the medium; \(\displaystyle v\downarrow \Rightarrow n\uparrow\). Medium $\displaystyle 2$ relative to medium $\displaystyle 1$, light travelling \(\displaystyle 1 \to 2\): \[n_{21} = \dfrac{n_2}{n_1} = \dfrac{c/v_2}{c/v_1} = \dfrac{v_1}{v_2} \] Answer: \(\displaystyle n = c/v\) (\(\displaystyle n\) is inversely proportional to the speed of light in the medium); \(\displaystyle n_{21} = v_1/v_2\).

  5. Exercise 24

    Refractive index of diamond with respect to glass is 1.6\displaystyle 1.6 and absolute refractive index of glass is 1.5. Find out the absolute refractive index of diamond.

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    NCERT’s answer
    Hint— \(\displaystyle \mathrm{n}_{\mathrm{dg}}=\frac{\mathrm{v}_{\mathrm{g}}}{\mathrm{v}_{\mathrm{d}}}=1.6, \mathrm{n}_{\mathrm{g}}=\frac{\mathrm{c}}{\mathrm{v}_{\mathrm{g}}}\), and \(\displaystyle \mathrm{n}_{\mathrm{d}}=\frac{\mathrm{c}}{\mathrm{v}_{\mathrm{d}}}\) Therefore, \(\displaystyle \frac{\mathrm{v}_{\mathrm{g}}}{\mathrm{v}_{\mathrm{d}}} \times \frac{\mathrm{c}}{\mathrm{v}_{\mathrm{g}}}=\mathrm{n}_{\mathrm{d}}=1.6 \times 1.5=2.40\).
    Given: \[{}_{g}n_d = 1.6, \quad n_g = 1.5 \] Relative refractive index: \[{}_{g}n_d = \dfrac{n_d}{n_g} \] Substituting: \[n_d = {}_{g}n_d \times n_g = 1.6 \times 1.5 \] \[n_d = 2.4 \] Answer: The absolute refractive index of diamond is \(\displaystyle 2.4\).

  6. Exercise 25

    A convex lens of focal length 20\displaystyle 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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    NCERT’s answer
    Hint— Statement is correct if the object is placed within $\displaystyle 20$ cm from the lens in the first case and between $\displaystyle 20$ cm and $\displaystyle 40$ cm in the second case.
    Yes — a convex lens forms a magnified real image for an object between \(\displaystyle F\) and \(\displaystyle 2F\), and a magnified virtual image for an object between the lens and \(\displaystyle F\).Case $\displaystyle 1$ (real, magnified), \(\displaystyle f=20\) cm, take \(\displaystyle u=-30\) cm: \[\frac{1}{v}-\frac{1}{u}=\frac{1}{f} \] \[\frac{1}{v}=\frac{1}{20\ \text{cm}}-\frac{1}{30\ \text{cm}}=\frac{1}{60\ \text{cm}} \] \[v=60\ \text{cm},\qquad m=\frac{v}{u}=\frac{60}{-30}=-2 \]Case $\displaystyle 2$ (virtual, magnified), take \(\displaystyle u=-10\) cm: \[\frac{1}{v}=\frac{1}{20\ \text{cm}}-\frac{1}{10\ \text{cm}}=-\frac{1}{20\ \text{cm}} \] \[v=-20\ \text{cm},\qquad m=\frac{v}{u}=\frac{-20}{-10}=2 \]Answer: real, inverted, magnified image for object distance \(\displaystyle 20\ \text{cm}<|u|<40\ \text{cm}\) (between \(\displaystyle F\) and \(\displaystyle 2F\)); virtual, erect, magnified image for \(\displaystyle |u|<20\ \text{cm}\) (between the lens and \(\displaystyle F\)).

  7. Exercise 26

    Sudha finds out that the sharp image of the window pane of her science laboratory is formed at a distance of 15\displaystyle 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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    NCERT’s answer
    Hint— Sudha should move the screen towards the lens so as to obtain a clear image of the building. The approximate focal length of this lens will be $\displaystyle 15$ cm.
    Lens formula, object distance \(\displaystyle x\) (\(\displaystyle u=-x\)): \[\frac{1}{v}=\frac{1}{f}-\frac{1}{x} \] The building is farther than the window pane: \[x\uparrow\ \Rightarrow\ \frac{1}{x}\downarrow\ \Rightarrow\ v\downarrow \] so its sharp image lies nearer the lens.Both objects are far away, so \(\displaystyle \frac{1}{x}\approx 0\) and \(\displaystyle v\approx f\): \[f\approx 15\ \text{cm} \]Answer: move the screen towards the lens; \(\displaystyle f\approx 15\) cm.

  8. Exercise 27

    How are power and focal length of a lens related? You are provided with two lenses of focal length 20\displaystyle 20 cm and 40\displaystyle 40 cm respectively. Which lens will you use to obtain more convergent light?

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    NCERT’s answer
    \(\displaystyle \mathrm{P}=\frac{1}{\mathrm{f}}, \mathrm{P} \propto \frac{1}{\mathrm{f}}\). Power of a lens is inversely proportional to its focal length therefore lens having focal length of $\displaystyle 20$ cm will provide more convergence.
    Power is the reciprocal of focal length, in metres: \[P=\frac{1}{f} \] For \(\displaystyle f_1=20\ \text{cm}=0.20\ \text{m}\): \(\displaystyle P_1=\dfrac{1}{0.20}=+5\ \text{D}\). For \(\displaystyle f_2=40\ \text{cm}=0.40\ \text{m}\): \(\displaystyle P_2=\dfrac{1}{0.40}=+2.5\ \text{D}\). Shorter focal length means higher power and bends light more sharply.Answer: The $\displaystyle 20$ cm lens (higher power, \(\displaystyle +5\) D) gives more convergent light.

  9. Exercise 28

    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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    NCERT’s answer
    When two plane mirrors are placed at right angle to each other then the incident and reflected rays will always be parallel to each other. NCERT_Solution_Class10_Science_Exemplar_Ch10_Q28_ncert
    Deviation produced by one mirror reflection is \(\displaystyle 180^\circ-2\theta\). When the two mirrors are at \(\displaystyle 90^\circ\) to each other, their normals are also perpendicular, so the two angles of incidence satisfy \(\displaystyle \theta_1+\theta_2=90^\circ\). Total deviation: \[(180^\circ-2\theta_1)+(180^\circ-2\theta_2)=360^\circ-2(\theta_1+\theta_2)=360^\circ-180^\circ=180^\circ \] This is independent of \(\displaystyle \theta_1\): a \(\displaystyle 180^\circ\) deviation means the emergent ray is parallel to the incident ray, in the opposite direction, for every angle of incidence.Answer: The two plane mirrors must be inclined at \(\displaystyle 90^\circ\) to each other.

  10. Exercise 29

    Draw a ray diagram showing the path of rays of light when it enters with oblique incidence (i) from air into water; (ii) from water into air.

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    NCERT’s answer
    Hint— NCERT_Solution_Class10_Science_Exemplar_Ch10_Q29_ncert NCERT_Solution_Class10_Science_Exemplar_Ch10_Q29_ncert_2
    By Snell's law, \(\displaystyle n_1\sin\theta_i=n_2\sin\theta_r\).Air \(\displaystyle \to\) water (\(\displaystyle n_{water}>n_{air}\)): \[\sin\theta_r=\frac{n_{air}}{n_{water}}\sin\theta_i<\sin\theta_i \] so the ray bends towards the normal on entering water.Water \(\displaystyle \to\) air: \[\sin\theta_r=\frac{n_{water}}{n_{air}}\sin\theta_i>\sin\theta_i \] so the ray bends away from the normal on entering air.Answer: entering water the ray bends towards the normal; entering air from water it bends away from the normal.