Chapter 11 · The Human Eye and the Colourful World

NCERT Exemplar Class 10 Science Chapter 11 Long Answer Questions

All 6 questions of NCERT Exemplar Class 10 Science Chapter 11 Long Answer Questions (The Human Eye and the Colourful World), in the book's order, each with its answer and a step-by-step solution.

1. Q25
   Explain ​the ​structure ​and ​functioning of Human ​eye. How ​are we able ​to ​see nearby as well as distant objects?
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   The eye forms a real, inverted image on the retina; the ciliary muscles change the lens focal length to see near and distant objects.

   The eye is a nearly spherical ball about $2.3$ cm across. Light enters through the transparent **cornea**, where most of the refraction takes place; it crosses the **aqueous humour**, passes through the **pupil** whose size is controlled by the coloured **iris**, and is finely focused by the **crystalline lens**, whose curvature is set by the **ciliary muscles**. It then crosses the **vitreous humour** and forms a real, inverted, diminished image on the **retina**, whose rods and cones send signals along the **optic nerve** to the brain; where that nerve leaves the eyeball is the **blind spot**, which has no receptors. We can see near and distant objects alike because of the power of accommodation: the retina is fixed $2.5$ cm behind the lens, so the eye varies the focal length of the lens instead. For a distant object the ciliary muscles are relaxed, the lens is thin and $f=2.50$ cm ($40$ D); for an object at the near point they contract, the lens thickens and $f=2.27$ cm ($44$ D). A normal eye therefore sees clearly from $25$ cm out to infinity.

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2. Q26
   When ​do we consider a person ​to be myopic or ​hypermetropic? Explain using diagrams ​how the defects associated ​with myopic and hypermetropic eye can ​be ​corrected?
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   Myopia, with the image in front of the retina, needs a concave lens; hypermetropia, with the image behind it, needs a convex lens.

   A person is **myopic** when he can see nearby objects clearly but not distant ones: his far point has come in from infinity, so the image of a distant object falls in **front** of the retina (here at $2.27$ cm against a retina at $2.5$ cm). It is corrected by a **concave** (diverging) lens whose focal length equals minus the far-point distance, which for a far point of $25$ cm is $f=-25$ cm, that is $P=-4$ D; the lens forms a virtual image of the distant object at the far point and the eye then focuses it on the retina. A person is **hypermetropic** when he can see distant objects clearly but not nearby ones: his near point lies beyond $25$ cm, so the image of a book at $25$ cm would fall **behind** the retina (here at $2.70$ cm). It is corrected by a **convex** (converging) lens which forms a virtual image of the book at his own near point; for a near point of $1$ m this needs $f=+33.3$ cm, that is $P=+3$ D.

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3. Q27
   Explain the refraction of light through a triangular ​glass prism using ​a ​labelled ray ​diagram. Hence define ​the angle ​of ​deviation.
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   The ray bends towards the normal at the first face and away at the second, turning towards the base; $\delta =i+e-A$.

   A ray $PQ$ falling on the face $AB$ of the prism $ABC$ passes from air into the denser glass at $Q$ and so bends **towards** the normal: $\sin i=n\sin r_1$ gives $r_1=28.1^{\circ }$ for $i=45^{\circ }$ and $n=1.50$. The geometry of the prism forces $r_1+r_2=A$, so $r_2=31.9^{\circ }$. At $R$ on the face $AC$ the ray passes from glass into the rarer air and bends **away** from the normal: $n\sin r_2=\sin e$ gives $e=52.4^{\circ }$. Both bends turn the ray in the same sense, towards the base, so the total turning is $\delta =(i-r_1)+(e-r_2)=i+e-A=45^{\circ }+52.4^{\circ }-60^{\circ }=37.4^{\circ }$. The **angle of deviation** is defined from this figure as the angle between the incident ray produced forward and the emergent ray produced backward, that is the angle through which the prism turns the ray away from its original direction, always towards the base.

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4. Q28
   How can ​we ​explain the ​reddish ​appearance of ​sun at sunrise or sunset? Why does it not appear red ​at ​noon?
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   The long slant path at sunrise and sunset scatters out the blue, leaving red; at noon the short path leaves the Sun white.

   The Sun sends the same white light all day; what changes is the thickness of atmosphere its light must cross. At sunrise and sunset the Sun is near the horizon, so the light travels a slant path about $\sqrt{2R/H}\approx 39$ times longer than the overhead path. Because scattering by air molecules varies as $1/{\lambda }^4$, and $(700/400{)}^4=9.4$, the short-wavelength blue and violet is scattered sideways out of the beam far more strongly than red. Over so long a path nearly all of it is removed, so the light that still travels straight to the eye is mainly red, and the Sun looks reddish. At noon the Sun is almost overhead and the light crosses the least thickness of atmosphere; comparatively little of any colour is scattered away, all the colours reach the eye in nearly their original proportions, and the Sun appears white.

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5. Q29
   Explain the phenomenon of dispersion of white ​light ​through ​a ​glass ​prism, using suitable ray ​diagram.
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   White light splits into seven colours because glass has a different refractive index for each, so violet is deviated most and red least.

   Dispersion is the splitting of white light into its seven constituent colours. It happens because white light is a mixture and the refractive index of the glass is different for each colour, being largest for violet ($n=1.532$) and smallest for red ($n=1.513$). Since a prism deviates a ray by $\delta =i+e-A$ and both $r_1$ and $e$ depend on $n$, each colour is turned through a different angle. Tracing a beam incident at $45^{\circ }$ on a $60^{\circ }$ prism gives ${\delta }_{\text{red}}=38.6^{\circ }$ and ${\delta }_{\text{violet}}=40.4^{\circ }$, an angular dispersion of $1.8^{\circ }$. Violet, being deviated most, emerges nearest the base and red, deviated least, furthest from it, so with the base down the band on the screen runs red, orange, yellow, green, blue, indigo, violet from the top downwards. That band is the spectrum, and Newton's inverted second prism recombines it into white light, proving the prism only separates colours that were already present.

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6. Q30
   How ​does refraction ​take place ​in the atmosphere? Why do stars twinkle but ​not the planets?
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   Air density increases downwards, so light bends along a curved path; stars twinkle as point sources, while a planet's many points average out.

   The atmosphere is not uniform: its density, and hence its refractive index, increases steadily downwards. Applying Snell's law layer by layer gives $n\sin z=$ constant, so as light descends into denser air it bends **towards** the vertical and travels along a curved path; in the worked model the zenith angle falls from $62^{\circ }$ to $50^{\circ }$. We see a body along the tangent to that path at the eye, so objects near the horizon appear raised, and the sun is seen about two minutes before actual sunrise and two minutes after actual sunset. **Stars twinkle** because a star is so distant that it is a point source: the layers of air are constantly moving, so the bending keeps changing, the apparent position of that point shifts and the amount of starlight entering the eye keeps varying. **Planets do not twinkle** because a planet is far nearer and is seen as a small disc, that is as a large collection of point sources. Each point flickers independently, and the total light entering the eye is their sum, so the variations average out and the planet shines steadily.

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