Human Eye and The Colourful World For Class 10 Physics Important Questions
Q 1 – A person cannot see distinctly objects kept beyond 2 m. This defect can be corrected by using lens of power
(a) +0.5 D
(b) -0.5 D
(c) +0.2 D
(d) -0.2 D
Ans. (b) -0.5 D
Q 2 – A student sitting on the last bench can read the letters written on the blackboard but is not able to read / the letters written in his textbook. Which of the following statements is correct?
(a) The near point of his eyes has receded away.
(b) The near point of his eyes has come closer to him.
(c) The far point of his eyes has come closer to him.
(d) The far point of his eyes has receded away.
Ans. (a) The near point of his eyes has receded away.
Q 3 – At noon the sun appears white as
(a) light is least scattered.
(b) all the colours of the white light are scattered away.
(c) blue colour is scattered the most.
(d) red colour is scattered the most.
Ans. (a) light is least scattered.
Q 4 – Twinkling of stars is due to atmospheric
(a) dispersion of light by water droplets
(b) refraction of light by different layers of varying refractive indices
(c) scattering of light by dust particles
(d) internal reflection of light by clouds
Ans. (b) refraction of light by different layers of varying refractive indices
Q 5 – The clear sky appears blue because
(а) blue light gets absorbed in the atmosphere.
(b) ultraviolet radiations are absorbed in the atmosphere.
(c) violet and blue lights get scattered more than lights of all other colours by the atmosphere.
(d) light of all other colours is scattered more than the violet and blue colour lights by the atmosphere.
Ans. (c) violet and blue lights get scattered more than lights of all other colours by the atmosphere.
Q 6 – Which of the following statements is correct regarding the propagation of light of different colours of white light in air?
(a) Red light moves fastest.
(b) Blue light moves faster than green light.
(c) All the colours of the white light move with the same speed.
(d) Yellow light moves with the mean speed as that of the red and the violet light.Ans. (c) All the colours of the white light move with the same speed.
Ans. (c) All the colours of the white light move with the same speed.
Q 7 – The danger signals installed at the top of tall buildings are red in colour. These can be easily seen from a distance because among all other colours, the red light
(a) is scattered the most by smoke or fog.
(b) is scattered the least by smoke or fog.
(c) is absorbed the most by smoke or fog.
(d) moves fastest in air.
Ans. (b) is scattered the least by smoke or fog.
Q 8 – When light rays enter the eye, most of the refraction occurs at the [1]
(a) crystalline lens
(b) outer surface of the cornea
(c) iris
(d) pupil
Ans. (b) outer surface of the cornea
Q 9 – The focal length of the eye lens increases when eye muscles
(a) are relaxed and lens becomes thinner
(b) contract and lens becomes thicker
(c) are relaxed and lens becomes thicker
(d) contract and lens becomes thinner
Ans. (a) are relaxed and lens becomes thinner
Q 10 – A student traces the path of a ray through a glass prism for four different values of angle of incidence. On analysing the diagrams he is likely to conclude that the emergent ray
(a) is always parallel to the incident ray.
(b) is always perpendicular to the incident ray.
(c) is always parallel to the refracted ray.
(d) always bends at an angle to the direction of incident ray.
Ans. (d) always bends at an angle to the direction of incident ray.
Q 11 – What is Presbyopia ?
Q 12 – The far point of a myopic person is 80 cm in front of the eye. What is the nature and power of the lens required to correct the problem?
Ans. The person is suffering from myopia and hence need a concave lens to correct the defect. The lens should be such that an object at infinity must form its image at the far point.
Hence, f = −80 cm = − 0.8 m
The power of the lens can be obtained as:
P= 1/ f
P= 1 / (−0.8) = −1.25 D
Q 13 – A person wants to read a book placed at 20 cm, whereas near point of his eye is 30 cm. calculate the power of the lens required.
Ans. Let x be the distance near point and d be the distance of the object.
Now :
f = xd / (x – d)
In this case :
x = 30 cm
d = 20 cm
Doing the substitution we have :
f = (20 × 30) / (30 – 20)
f = 600/10
We now have f as follows.
f = 60
The power formula is given as follows :
P = 100/f
Doing the substitution we have :
P = 100/60
= 1.67D
This person is hypermetropic which means he is long-sighted .
Q 14 – How can change of size of eyeball be one of the reason for: (a) Myopic eye (b) Hypermetropic eye. Compare the size of eyeball with that of a normal eye in each case. How does this change of size affect the position of image in each case?
Ans. (i) The eye suffering from myopia, has long eyeball than that of normal eye due to which the retina is at a larger distance from the eye lens. This results in the formation of the image in front of the retina.
(ii) The eye suffering from hypermetropia has short eyeball than that of normal eye due to which the retina is at smaller distance from the eye lens. This results in the formation of the image behind the retina.
Q 15 – (a) Explain how a normal human eye is able to see distinctly the object placed at a distance as well as those placed at a nearer distance.
(b) What is the far point and near point of a normal human eye?
Q 16 – (a) Write the function of each of the following parts of human eye: (i) Cornea (ii) Iris
(iii) Crystalline lens (iv) Ciliary muscles (b) Why does the sun appear reddish early in the morning? Will this phenomenon be observed by an astronaut on the Moon?
Ans. (a) The boy is suffering from myopia.
This defect is caused :
(i) due to increase in length of eyeball, and
(ii) decrease in focal length of eye lens, when the eye is fully relaxed.
Correction : The image of a distant object (i.e., at infinity) is formed in front of the retina of eye suffering from myopia as shown in figure
(a) As the image of the object lying at infinity is not formed on the retina of the eye, so such object can not be seen clearly by the myopic eye. The far point of such an eye is near to the eye as shown in fig.
(b) This defect can be corrected by using a concave lens of suitable focal length. So, a man suffering from this defect wears spectacles having concave lens of suitable focal length. The concave lens diverges the rays of light entering the eye from infinity. Hence this lens makes the rays of light appear come from the far point (O’) of the defective eye as shown in figure (c).
(b) The twinkling of a star is due to atmospheric refraction of starlight. The atmospheric refraction occurs in a medium of gradually changing refractive index. The atmosphere of the earth is made of different layers. It is affected by winds, varying temperatures, and different densities as well. When light from a distant source (a star) passes through our atmosphere, it undergoes refraction many times. When we finally perceive this light from a star, it appears to be twinkling. This is because some of the light rays reach us directly and some bends away from and toward us. It happens so fast that it gives a twinkling effect.
OR
(a) (i) Cornea : It is a thin membrane, covering the surface of eyeball, through which light enters. It acts as a primary lens, which provides the refraction for light rays that enters the eye.
(ii) Iris : It is a dark muscular diaphragm that controls the size of the pupil and is located just behind the cornea in the eye.
(iii) Crystalline lens : Crystalline lenses are converging in nature, made by the jelly-like proteinaceous material. The focal length of the crystalline lens is changed by the ciliary muscles. Its function is to focus the incoming light rays from the object on the retina using its refractive property.
(iv) Ciliary muscles : It modifies the curvature and thereby the focal length of the eye lens by contracting or relaxing itself to focus the image of an object on the retina according to the distance of the object. It also holds the eye lens in position.
(b) The wavelength of red light is greater than that of wavelength of blue light. More the wavelength of the light, less the light scatters. During sunrise, the Sun looks almost reddish because only red colour λb<λr which is least scattered, is transmitted through the Earth’s atmosphere more quickly than blue. Hence, the appearance of the sun at the sunrise, near the horizon looks almost reddish.
This phenomenon will not be observed by an astronaut on Moon, since there is no atmosphere so no scattering of light takes place, thus the Sun appears dark.
Q 17 – With the help of a diagram, explain the formation of a rainbow in the sky.
Ans. A collection of suspended water droplets in the atmosphere serves as a refractor of light. The water represents a medium with a different optical density than the surrounding air. Light waves refract when they cross over the boundary from one medium to another. The decrease in speed upon entry of light into a water droplet causes a bending of the path of light towards the normal. And upon exiting the droplet, light speeds up and bends away from the normal. The droplet causes a deviation in the path of light as it enters and exits the drop.
There are countless paths by which light rays from the sun can pass through a drop, Each path is characterized by this bending towards and away from the normal. One path of great significance in the discussion of rainbows is the path in which light refracts into the droplet, internally reflects, and then refracts out of the droplet. The diagram at the right depicts such a path.
A light ray from the sun enters the droplet with a slight downward traiectory. Upon refracting twice and reflecting once, the light ray is dispersed and bent downward towards an observer on earth’s surface. Other entry locations into the droplet may result in similar paths or even in light continuing through the droplet and out the opposite side without significant internal reflection. But for the entry location shown in the diagram at the right, there is an optimal concentration of light exiting the airborne droplet at an angle towards the ground.
As in the case of the refraction of light through prisms with nonparallel sides, the refraction of light at two boundaries of the droplet results in the dispersion of light into a spectrum of colors. The shorter wavelength blue and violet light refract a slightly greater amount than the longer wavelength red light. Since the boundaries are not parallel to each other, the double refraction results in a distinct separation of the sunlight into component colors.
Q 18 – What is meant by scattering of light? Mention the factor on which it depends. Explain why the colour of the clear sky is blue? An Astronaut in space finds sky to be dark. Explain reason for this observation.
Ans. Scattering of light
When white light from the sun enters the earth’s atmosphere, light disperses i.e., light disperses in all directions by the dust particles, free water molecules, and the molecules of the gases present in the atmosphere. This phenomenon is called the scattering of light.
Factors on Which the Scattering of Light depends are-
1. The scattering of light depends on the wavelength of the incident light and also on the size of the particle.
2. The sky appears dark in space because there is no particles to scatter the light due to the absence of atmosphere in space.
Hence, the Sky Appear Dark in Space due to the absence of atmosphere in space.