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NCERT Exemplar · Class 10 Science The Human Eye and the Colourful World

30 questions · 30 still being checked

Short Answer Questions 15–24 (part 3 of 4)

  1. Exercise 15

    Draw ray diagrams each showing (i) myopic eye and (ii) hypermetropic eye.

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    NCERT_Solution_Class10_Science_Exemplar_Ch11_Q15_ncert NCERT_Solution_Class10_Science_Exemplar_Ch11_Q15_ncert_2
    Rays are traced from a distant object for the myopic eye and from the normal near point N ($\displaystyle 25$ cm) for the hypermetropic eye.Answer: (i) Myopic eye: parallel rays from a distant object meet in front of the retina. (ii) Hypermetropic eye: rays from N are still converging at the retina and would meet behind it, so the image on the retina is blurred.

  2. Exercise 16

    A student sitting at the back of the classroom cannot read clearly the letters written on the blackboard. What advice will a doctor give to her? Draw ray diagram for the correction of this defect.

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    Hint— The student is suffering from myopia (near sightedness). Doctor advises her to use a concave lens of appropriate power to correct this defect. NCERT_Solution_Class10_Science_Exemplar_Ch11_Q16_ncert
    She cannot see the distant blackboard clearly, so she has myopia. The doctor will advise spectacles with a concave (diverging) lens of suitable power: the lens diverges the parallel rays before they enter the eye, so the eye lens now brings them to focus exactly on the retina instead of in front of it.Answer: Myopia; corrected with concave (diverging) lens spectacles.

  3. Exercise 17

    How are we able to see nearby and also the distant objects clearly?

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    Hint— Human eye is able to see nearby and distant objects clearly by changing the focal length of the eye lens using its power of accommodation
    The eye lens is elastic and its curvature is adjusted by the ciliary muscles (accommodation). Near object: muscles contract \(\displaystyle \Rightarrow\) lens more convex \(\displaystyle \Rightarrow f\downarrow \Rightarrow P\uparrow\). Distant object: muscles relax \(\displaystyle \Rightarrow\) lens flatter \(\displaystyle \Rightarrow f\uparrow \Rightarrow P\downarrow\). Either way the image is brought to focus sharply on the retina.Answer: Accommodation — ciliary-muscle control of the eye lens's curvature (focal length) — focuses both near and distant objects on the retina.

  4. Exercise 18

    A person needs a lens of power -4.5\displaystyle 4.5 D for correction of her vision.
    (a)
    What kind of defect in vision is she suffering from?
    (b)
    What is the focal length of the corrective lens?
    (c)
    What is the nature of the corrective lens?

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    (a)
    Myopia
    (b)
    Hint— \(\displaystyle f=\frac{1}{-4.5}=-\frac{2}{9}=-0.22 \mathrm{~m}\),
    (c)
    Concave lens
    Negative power means a diverging lens, so she has myopia (near-sightedness).\[f = \frac{1}{P} = \frac{1}{-4.5\ \mathrm{D}} = -\frac{2}{9}\ \mathrm{m} \approx -22.2\ \mathrm{cm} \]The negative focal length confirms the lens is concave (diverging).Answer: (a) Myopia (b) \(\displaystyle f \approx -22.2\ \mathrm{cm}\) (c) Concave (diverging) lens.

  5. Exercise 19

    How will you use two identical prisms so that a narrow beam of white light incident on one prism emerges out of the second prism as white light? Draw the diagram.

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    Hint— By using two identical prisms, one placed inverted with respect to the other. NCERT_Solution_Class10_Science_Exemplar_Ch11_Q19_ncert
    Place a second, identical prism right after the first but inverted (upside-down) relative to it, the two almost touching. The first prism disperses the white beam into a spectrum; the inverted second prism bends each colour back by an equal and opposite amount, recombining it into white light again.Answer: Place the second prism inverted relative to the first so it recombines the spectrum back into white light.

  6. Exercise 20

    Draw a ray diagram showing the dispersion through a prism when a narrow beam of white light is incident on one of its refracting surfaces. Also indicate the order of the colours of the spectrum obtained.

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    NCERT_Solution_Class10_Science_Exemplar_Ch11_Q20_ncert
    Inside the glass \(\displaystyle \mu\) increases from red to violet, so deviation follows the same order: \[\mu_{violet} > \mu_{red} \Rightarrow \delta_{violet} > \delta_{red} \] Answer: VIBGYOR — Violet, Indigo, Blue, Green, Yellow, Orange, Red, with red closest to the incident direction and violet farthest.

  7. Exercise 21

    Is the position of a star as seen by us its true position? Justify your answer.

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    No. light from stars undergoes atmospheric refraction which occurs in medium of gradually changing refractive index.
    No. Starlight bends continuously through atmospheric layers whose refractive index falls with height: \[n_{air}\downarrow\ (\text{height}\uparrow) \Rightarrow \text{ray curves toward Earth} \] The eye extrapolates the last straight segment backward, so the star appears slightly higher than its true position; shifting air layers make this apparent position fluctuate (twinkling).Answer: No — atmospheric refraction raises the star's apparent position above its true one.

  8. Exercise 22

    Why do we see a rainbow in the sky only after rainfall?

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    Hint— The water droplets behave like prisms and disperse sunlight.
    \[\text{sunlight} \Rightarrow \text{refraction + dispersion at entry} \Rightarrow \text{internal reflection} \Rightarrow \text{refraction at exit} \Rightarrow \text{colours to the eye} \] Rain leaves countless tiny droplets suspended in air, each acting as a prism. With the Sun behind the observer, together they give the coloured arc; no droplets, no rainbow.Answer: After rain the air holds droplets that disperse and internally reflect sunlight into a coloured arc.

  9. Exercise 23

    Why is the colour of the clear sky blue?

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    Hint— Blue colour gets scattered the maximum.
    Air molecules scatter light following Rayleigh's law, \(\displaystyle I\propto 1/\lambda^4\), so shorter wavelengths scatter far more strongly than longer ones: \[\frac{I_{blue}}{I_{red}} = \left(\frac{\lambda_{red}}{\lambda_{blue}}\right)^{4} = \left(\frac{700\ \mathrm{nm}}{450\ \mathrm{nm}}\right)^{4} \approx 5.9 \] This scattered blue light reaches the eye from every direction in the sky.Answer: Preferential (Rayleigh) scattering of blue light by air molecules makes the clear sky look blue.

  10. Exercise 24

    What is the difference in colours of the Sun observed during sunrise/sunset and noon? Give explanation for each.

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    Hint— During sunrise and sunset the sun appears reddish whereas at noon the sun appears white. Explanation should be given in terms of atmospheric depth travelled by light. Colours are different due to scattering of light by atmospheric particles.
    At noon the Sun is overhead, so its light crosses the least atmosphere; little of any colour is scattered out, and the near-original mix (looking white) reaches us. At sunrise/sunset the Sun is near the horizon, so its light travels a far longer path: \[\ell_{sunset} \gg \ell_{noon} \] \[\ell\uparrow \Rightarrow (I\propto 1/\lambda^4)\ \text{blue/violet scattered away} \Rightarrow \text{mostly red/orange reaches the eye directly} \]Answer: Noon Sun looks white (short atmospheric path, little scattering); sunrise/sunset Sun looks reddish (long path scatters out the blue, leaving red).