Bihar Board Class 10 Science Solutions Chapter 11 The Human Eye and Colourful World

 

Bihar Board Class 10 Science Chapter 11 The Human Eye and Colourful World InText Questions and Answers

Intext Questions (Page 190)

Question 1.
What is meant by power of accommodation of the eye ?
Answer:
A healthy human eye can see the distant objects as well as nearby objects (not less than 25 cm from eye) distinctly by decreasing or increasing the focal length. This ability of human eye is called power of accommodation of the eye.

Question 2.
A person with a myopic eye cannqt see objects beyond 1.2 m distinctly. What should be the type of the corrective lens used to restore proper vision ?
Answer:
The person who cannot see object beyond 1.2 m from eye is suffering from myopia. This defect of vision is corrected by using a concave length of suitable power.

Question 3.
What is the far point and near point of the human eye with normal vision ?
Answer:
A healthy human eye can see the objects at infinity distinctly. So far point is infinity. The eye can also see the nearby objects beyond 25 cm from eye. So near point is 25 cm.

Question 4.
A student has difficulty in reading the blackboard while sitting in the last row. What could be the defect the child is suffering from ? How can it be corrected ?
Answer:
The student is suffering from myopia. This eye defect can be corrected by using a concave lens of suitable focal length.
(380)

Bihar Board Class 10 Science Chapter 11 The Human Eye and Colourful World Textbook Questions and Answers

Question 1.
The human eye can focus objects at different distances by adjusting the focal length of the eye lens. This is due to :
(а) presbyopia.
(b) accommodation.
(c) near-sightedness.
(d) far-sightedness.
Answer:
(b) accommodation.

Question 2.
The human eye forms the image of an object at its
(a) corijeja.
(b) iris.
(c) pupil
(d) retina.
Answer:
(d) retina.

Question 3.
The least distance of distinct visión for a young adult with normal vision is about
(a) 25 m.
(b) 2.5 cm.
(c) 25 cm.
(d) 2.5 m.
Answer:
(c) 25 cm.

Question 4.
The change in focal length of an eye lens is caused by the action of the
(a) pupil.
(b) retina.
(e) ciliary musçes.
(d) iris.
Answer:
(c) ciliary musdes.

Question 5.
A person needs a lens of power – 5.5 dioptres for coriecting his distant vision. Fôr correcting his near vision he needs a lens of power + 1. dioptre. What is the focal length of the lens re4uired for ‘correcting (i) distant vision, and (ii) near vision? .
Answer:
(i) For distant vision:

P = 1f (f in metre)
f = 1p = – 15.5 = 1055 = –211 m
f = – 211 x 100 cm
f = – 211 cm
f= – 18.2 cm = – 0.18m

(ii) For near vision:
f = 1p = 11.5

f = 2003 cm
f = 66.67cm = + 0.67 m

Question 6.
The far point of a myopic person is 80 cm in front of the eye. What is the nature and power of lens required to correct the problem ?
Answer:
The lens used for correcting this disease is concave lens.

Question 7.
Make a diagram to show how hypermetropia is corrected. The near point of a hypermetropic eye is 1 m. What is the power of lens required to correct this defect ? Assume that the near point of the normhl eye is 25 cm.
Answer:
For the convex lens used for correction.
u = l metre = – 100 cm, u = – 25 cm

We know lens formula :
1v = 1u = 1f
1−100 – 1−25 = 1f
−1+4100 = 1f
f = 1003 cm

f = 13 m

P = 3 dioptres = + 3.0 D

Question 8.
Why is a normal eye not able to see clearly the objects placed closer than 25 cm ?
Answer:
The rays coming from the objects which are at a distant less than 25 cm are not properly focused at retina and the image of object appears blanked. So the image cannot be seen clearly.

Question 9.
What happehs to the image distance in the eye when we increase the distance of an object from the eye ?
Answer:
The image distance remains unchanged. It is because as distance of object increases the focal length of eye lens is adjusted by ciliary muscle such that image is always formed at retina.

Question 10.
Why do stars twinkle?
Answer:
The refractive index of atmosphere is changing continuously. As starlight enters into atmosphere, it undergoes several times refraction. Hence, the apparent position of stars are changing cpntinuously. As the path of rays of light coming rom star goes on varying slighting, starlight entering to the ;yes flickers. Hence, stars twinkle.

Question 11.
Explain why the planets do not twinkle.
Answer:
Planets are much closer to the earth and are extended source of light. As planets are large number of point-sized sources of light, the total variation in the amount of light entering our eyes will average out to zero. Hence, planets do hot twinkle.

Question 12.
Why does the Sun appear reddish early in the morning ?
Answer:
Due to the presence of fine particles in atmosphere, scattering of light is caused. Scattering of light is reciprocal to wave-length of light. So red light is scattered less and spread in the sky early in the morning and evening. Other light rays are much more scattered and not visible in the sky. So, the Sun appears reddish early in the morning.

Question 13.
Why does the sky appear dark instead of blue to an astronaut?
Answer:
There is nearly no atmosphere for the astronauts as they are flying high in sky. So, there is no scattering of light. This is why the sky appears dark to the astronauts.

Bihar Board Class 10 Science Chapter 11 The Human Eye and Colourful World Textbook Activities

Activity 11.1 (Page 192)

• Fix a sheet of white paper on a drawing board using drawing pins.
• Place a glass prism on it in such a way that it rests on its triangular base. Trace the outline of the prism using a pencil.
• Draw a straight line PE inclined to one of the refracting surfaces, say AB, of the prism.
• Fix two pins, say at points P and Q, on the line PE as shown in Fig 11.8

Fig. 11.8. Refraction of light through a triangular glass prism.

PE — Incident ray
EF — Refracted ray
FS — Emergent ray
∠i — Angle of incidence
∠r — Angle of refraction
∠e — Angle of emergence
∠A — Angle of the prism
∠D — Angle of deviation

• Look for the images of the pins, fixed at P and Q, through the other face AC.
• Fix two more pins, at points R anc( S, such that the pins at R and S and the images of the pins at P and Q lie on the same straight line.
• Remove the pins and the glass prism.
• The line PE meets the boundary of the prism at point E (See Fig. 11.8). Similarly, join and produce the points R and S. Let these lines meet the boundary of the prism at E and F, respectively. Join E and F.
• Draw perpendiculars to the refracting surfaces AB and AC of the prism at points E and F, respectively.
• Mark the angle of incidence (∠i), the angle of refraction (∠r) and the angle of emergence'(∠e) as shown in Fig. 11.8.

Questions Based on Activity

Question 1.
Draw a diagram to show the incident ray, refracted ray, emergent ray and angle of deviation.
Answer:
See Fig. 11.8.
PE = incident ray,
EF = refracted ray
FG = emergent ray
ZD = angle of deviation

Question 2.
How many times a light ray undergoes refraction and what is the direction of refracted ray in each refraction?
Answer:
The light is refracted twice when it passes through a prism. Once when it enters to glass from air and next time when it enters to air from glass.
In both the cases the refracted ray bends toward base of prism.

Question 3.
What is angle of deviation?
Answer:
The angle between (direction of incident ray and direction of emergent ray is called angle of deviation.

Question 4.
For which kind of light the angle of deviation is least and for which kind of light it is maximum ?
Answer:
For red light it is minimum and for blue light it is maximum.

Activity 11.2 (Page 193)

• Take a thick sheet of cardboard and make a small hole or narrow slit in its middle.
• Allow sunlight to fall on the narrow slit. This gives a narrow beam of white light.
• Now, take a glass prism and allow the light from the slit to fall on one of its faces as shown in Fig. 11.5 at Textbook Page No. 193.
• Turn the prism slowly until the light that comes out of it appears on a nearby screen.

Question 1.
What do you observe ?
Answer:
Seven coloured spectrum of white ight is obtained. Their names are as — Violet, Indigo, Blue, Green, Yellow, Orange and Red. ~

Question 3.
Why does this happen ?
Answer:
This happens due to the dispersion of light. Dispersion of white light occurs due to different velocities of different colours of light in glass. Due to different velocities the ponstituent colours of white light undergoes deviation through different angles.

Question 3.
Which colour is at the top of spectrum ?
Answer:
In the spectrum of sunlight red colour is at the top.

Question 4.
Which colour is at the bottom of spectrum of white light ?
Answer:
Violet colour is at the bottom of spectrum of white light.

Activity 11.3 (Page 196)

Questions Based on Activity
Question 1.
Name the lens used to observe scattering of light experimentally.
Answer:
Convex lens.

Question 2.
How does scattering of light occur ?
Answer:
Scattering of light occurs due to striking of light to very minute suspended particles of colloidal solution.

Question 3.
Which kind of light is scattered more and why ?
Answer:
Scattering of light is reciprocal to the wavelength so violet light is scattered more as its wavelength is relatively less.

Question 4.
Why do we observe blue light from three sides of the glass tank ?
Answer:
We observe blue light from three sides of the glass tank due to scattering of short wavelength by minute colloidal sulphur particles.

Bihar Board Class 10 Science Chapter 11 The Human Eye and Colourful World NCERT Exemplar Problems

Short Answer Type Questions

Question 1.
Draw ray diagrams each showing (i) myopic eye and (ii) hypermetropic eye.
Answer:

Hypermetropic eye

Question 2.
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.
Answer:
The student is suffering from myopia (near sightedness). Doctor advises her to use a concave lens of appropriate power to correct this defect.

Question 3.
How are we able to see nearby and also the distant objects clearly?
Answer:
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.

Question 4.
A person needs a lens of power – 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?
Answer:
(a) Myopia v
(b) f = 1−4.5 = 29 = -0.22 m
(c) Concave lens

Question 5.
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.
Answer:
By using two identical prisms, one placed inverted with respect to the other.

Question 6.
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.
Answer:

Question 7.
Is the position of a star as seen by us its true position? Justify your answer.
Answer:
No. Light from stars undergoes atmospheric refraction which occurs in medium of gradually changing refractive index.

Question 8.
Why do we see a rainbow in the sky only after rainfall?
Answer:
The water droplets suspend in the atmosphere behave like prisms and disperse sunlight. Thus, we see a rainbow in the sky only after rainfall.

Question 9.
Why is the colour of the clear sky blue?
Answer:
We know that out of the seven constituents of white light, blue colour gets scattered the maximum. Thus, the colour of the clear sky appears blue.

Question 10.
What is the difference in colours of the Sun observed during sunrise/sunset and noon? Give explanation for each-
Answer:
Colours are different due to scattering of light by atmospheric particles. Red light is scattered less and spread in the sky early in the morning and evening. Thus, sun appears reddish during sunrise and sunset. However at noon, sun appears white as only a little of the blue and violet colours are scatered.

Long Answer Type Questions

Question 11.
Explain the structure and functioning of human eye. How are we able to see nearby as well as distant objects? .
Answer:
Give explanation of each part and discuss power of accommodation.

The human eye is like a camera. Its lens system formt an image on a light-sensitive screen called the retina. Light enters the eye through a thin membrane called the cornea. It forms the transparent bulge on the front surface of the eyeball as shown in the figure The eyeball is approximately spherical in shape with a diameter of about 2.3 cm. Most of the refraction for the light rays entering the eye occurs at the outer surface of the cornea.

The crystalline lens merely provides the finer adjustment of focal length required to focus objects at different distances on the retina. We find a structure called iris behind the cornea. Iris is a dark muscular diaphragm that controls the size of the pupil. The pupil regulates and controls the amount of light entering the eye. The eye lens forms an inverted real image of the object on the retina. The retina is a delicate membrane having enormous number of light-sensitive cells.

The light-sensitive cells get activated upon illumination and generate electrical signals. These signals are sent to the’brain via the optic nerves. The brain interprets these signals, and finally, processes the information so that we perceive objects as they are.
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.

Question 12.
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?
Answer:
When a person is not able to see distant objects clearly but can see nearby objects clearly then he is considered to be myopic. Whereas a hypermetropic person can see distant objects clearly but cannot see nearby objects distinctly.
Myopic vision can be corrected by using a concave lens of suitable power while a hypermetropic vision can be corrected by using a convex lens of suitable power.

Question 13.
Explain the refraction of light through a triangular glass prism using a labelled ray diagram. Hence define the angle of deviation.
Answer:
To explain the phenomenon of refraction of light through a prism, a figure is drawn.

Here PE is the incident ray, EF is the refracted ray and FS is the emergent ray. A ray of light is entering from air to the glass at the first surface AB. The light ray on refraction has bent towards the normal. At the second surface AC, the light ray has entered from glass to air. Hence, it has bent away from normal. The angle of incidence and the angle of refraction at each refracting surface of the prism is similar to the kind of bending that occurs in a glass slab. The peculiar shape of the prism makes the emergent ray bend at an angle to the direction of the incident ray. This angle is called the angle of deviation. In this case ZD is the angle of deviation.

Question 14.
How can we explain the reddish appearance of sun at sunrise or sunset? Why does it not appear red at noon?
Answer:
Sun appears reddish at sunrise or sunset as blue light gets scattered away. At the time of sunset or sunrise, the Sun is near the horizon and light rays pass through a longer distance in atmosphere before reaching our eyes. So blue and violet colours are scattered away and only longer wavelength light suish as red approaches to our eye. Thus Sun appears reddish in colour.

At noon the Sun’s position is just above the atmosphere and so it travel shorter distance and hence very less blue and violet colours are scattered away. So Sun appears white as most of visible radiation approach to our eye.

Question 15.
Explain the phenomenon of dispersion of white light through a glass prism, using suitable ray diagram.
Answer:
When light passes through prism, it split into its seven constituents colours. This optical phenomenon is called dispersion. The various colours seen are Violet, Indigo, Blue, Green, Yellow, Orange and Red as shown in the figure below. The band of the coloured components of a light beam is called its spectrum. This happens because different colours of light bend through different angles with respect to the incident ray, as they pass through a prism. The red light bends the least while the_violet the most. Thus the rays of each colour emerge along different paths and thus become distinct. It is the band of distinct colours that we see in a spectrum.

Question 16.
How does refraction take place in the atmosphere? Why do stars twinkle but not the planets?
Answer:
The hotter air present in atmosphere is lighter (less dense) than the cooler air above it, and has a refractive index slightly less than that of the cooler air. Since the physical conditions of the object, as seen through the hot air, fluctuates. This wavering is thus an effect of atmospheric refraction (refraction of light by the earth’s atmosphere) on a small scale in our local environment.  The refractive index of atmosphere is changing continuously.

As starlight enters into atmosphere, it undergoes several times refraction. Hence, the apparent position of stars are changing continuously. As the path of rays of light coming from star goes on varying slighting, starlight entering to the eyes flickers. Hence, stars twinkle. Planets are much closer to the earth and are extended source of light. As planets are large number of point-sized sources of light, the total variation in the amount of light entering our eyes will average out to zero. Hence, planets do not twinkle.