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These NCERT Solutions for Class 10 Science Chapter 11 The Human Eye and the Colourful World Questions and Answers are prepared by our highly skilled subject experts to help students while preparing for their exams.

The Human Eye and the Colourful World NCERT Solutions for Class 10 Science Chapter 11

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

In-text Questions (Page 190)

Question 1.
What is meant by power of accommodation of the eye?
Answer:
The property of the eye lens to change its focal length is called its power of accomodation.

Question 2.
A person with a myopic eye cannot see objects beyond 1.2 m distinctly. What should be the nature of the corrective lens used to restore proper vision ?
Answer:
Given v = -1.2 m
u = ∞

So, if the far point of myopic eye is 1.2 m, focal length of corrective lens is -1.2 m. And it will be concave.

Question 3.
What is the far point and near point of the human eye with normal vision?
Answer:
The near point is 25 cm far point is infinity of the human eye with normal vision.

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:
Myopia : It can be corrected by using concave lens. A concave lens of suitable power will bring the image back on to the retina and thus defect is corrected.

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

Page no. 197 & 198

Question 1.
The human eye can focus on objects at different distances by adjusting the focal length of the eye lens. This is due to
(a) 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) cornea
(b) iris
(c) pupil
(d) retina.
Answer:
(d) Retina

Question 3.
The least distance of distinct vision 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
(c) ciliary muscles
(d) iris
Answer:
(c) Ciliary muscles.

Question 5.
A person needs a lens of power -5.5 dioptres for correcting his distant vision. For correcting his near vision he needs a lens of power +1.5 dioptre. What is the focal length of the lens required for correcting (i) distant vision, and (ii) near vision?
Answer:
Given p = -5.5 dioptres

Question 6.
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?
Answer:
When object is at 80 cm, focal length of eye lens of f and image distance i.e., distance between eye lens and retina is v., By using lens formula :
\(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\)
\(\frac{1}{f}=\frac{1}{v}-\frac{1}{-80 \mathrm{~cm}}\) ………(i)
When the object is at infinity, the object is not distinctly visible to short sighted person so he will use corrective lens of focal length f’ in front of eye lens. By applying lens formula, we have

So the lens will be concave because its focal length is negative.

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 the lens required to correct this defect? Assume that the near point of the normal eye is 25 cm.
Answer:
Let the focal length of eye lens is ‘f’
Given u = – 1m
By lens formula, we get

When the object is placed at normal eye point, then the focal length of corrective lens be f’. By using lens formula we get

By substituting (i) and (ii) we get

Question 8.
Why is a normal eye not able to see clearly the objects placed closer than 25 cm?
Answer:
The eye lens is composed of a fibrous, jelly-like material. Its curvature can be modified to some extent by the ciliary muscles. The change in the curvature of the eye lens can thus chage its focal length. When the muscles are released the lens becomes thin. Thus, its focal length increases This enables us to see distant object clearly. When we are looking at objects closer to the eye lens decreases. However, the focal length of eye lens cannot be decreased below a cetain limit, minimum limit i.e., below 25 cm. In this situation image is formed behind the retina so a normal eye is not able to see clearly the objects placed closer than 25 cm.

Question 9.
What happens to the image distance in the eye when we increase the distance of an object from the eye ?
Answer:
Image distance in the eye also increases when we increase The distance of an object from the eye.

Question 10.
Why do stars twinkle ?
Answer:
Light emitted by star passes through the atmosphere of the earth before reaching our eyes. The atmosphere of the earth is not uniform but consists of many layers of different densities.

The layers close to the surface of the earth are denser. As we go higher the density of layers and refractive index decreases progressively. As the light from a star enters the upper-most layer of the atmosphere, it bends towards the normal as it enters the next layer. This process continues till the light enters our eyes. So due to refraction of light, the apparent position of the star is different from the actual position of the star.

Moreover, the different layers of the atmosphere are mobile and the temperature and the density of layers of atmosphere changes continuously. Hence the apparent position of the star changes continuously. This change in the apparent position of the star continuously leads to the twinkling of a star.

Question 11.
Explain why the planets do not twinkle.
Answer:
Planets are very dose to the earth as compared to the stars. So they appear bigger due to their comparatively smaller distances and stars appear smaller due to their very large distances. Because of this difference planet is considered as a collection of many point sources of light while a star is considered as a point source of light. So, a planet is made up of a number of point sources of light, each twinkling in a random fashion.

Random twinkling of each individual point light source, nullify the twinkling effect of other source and as a result, there is null twinkling effect as a whole. Due to this, the variations in the atmospheric conditions are unable to produce twinkling effect of the planet in the eye of the observer.

Question 12.
Why does the sun appear reddish early is the morning?
Answer:
Light of lower frequencies such as yellow, orange, red is scattered the least by oxygen and nitrogen molecules of the atmosphere. Thus, the red, orange and yellow lights are transmitted through the atmosphere much more than violet and blue. Red which is scattered the least due to long wavelength, passes through more atmosphere than other colours. Therefore, most of the light with high frequencies, violet, indigo, blue, green are scattered away in the atmosphere at sunrise. Only red and a little orange light which are least scattered enter our eyes, and appear reddish early in the morning.

Question 13.
Why does the sky appear dark instead of blue to an astronaut ?
Answer:
At very high altitude earth has no atmosphere, so the astronaut would not have been any scattering of light. Then the sky would look dark. The sky appear dark to astronaut flying at very high altitudes, as scattering is not prominent at such heights.

Class 10 Science Chapter 11 The Human Eye and the 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.
• Look for the images of the pins, fixed at P and Q, through the other face AC.
• Fix two more pins, at points R and 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.). 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.

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

Here PE is the incident ray, EF is the refracted ray and FS is the emergent ray. You may note that a ray of light is entering from air to glass at the first surface AB. The light rayon 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. Compare the angle of incidence and the angle of refraction at each refracting surface of the prism. Is this 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 ∠D is the angle of deviation.

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..
• Turn the prism slowly until the light that comes out of it appears on a nearby screen.

Question 1.
What do you observe? You will find a beautiful band of colours. Why does this happen?
Answer:
Observations : The prism has probably split the incident white light into a band of colours.

Note the colours that appear at the two ends of the colour band. What is the sequence of colours that you see on the screen? The various colours seen are Violet, Indigo, Blue, Green, Yellow, Orange and Red, as shown in Fig. The acronym VIBGYOR will help you to remember the sequence of colours. The band of the coloured components of a light beam is called its spectrum. You might not be able to see all the colours separately. Yet something makes each colour distinct from the other. The splitting of light into its component colours is called dispersion.

Different colours of light bend 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.

Activity 11.3 (Page 196)

• Place a strong source (S) of white light at the focus of a converging lens (Ly. This lens provides a parallel beam of light.
• Allow the light beam to pass through a transparent glass tank (T) containing clear water.
• Allow the beam of light to pass through a circular hole (c) made in a cardboard. Obtain a sharp image of the circular hole on a screen (MN) using a second converging lens (L₂), as shown in Fig.

Question 1.
Dissolve about 200 g of sodium thiosulphate (hypo) in about 2L of clean water taken in the tank. Add about 1 to 2 mL of concentrated sulphuric acid to the water. What do you observe?
Answer:
Observation : We will find fine microscopic sulphur particles precipitating in about 2 to 3 minutes. As the sulphur particles begin to form, you can observe the blue light from the three sides of the glass tank. This is due to scattering of short wavelengths by minute colloidal sulphur particles. Observe the colour of the transmitted light from the fourth side of the glass tank facing the circular hole. It is interesting to observe at first the orange-red colour and then bright crimson red colour on the screen.

Class 10 Science Chapter 11 The Human Eye and the Colourful World Additional Important Questions and Answers

Very Short Answer Type Questions

Question 1.
What is the near point off a normal eye ?
Answer:
25 cm.

Question 2.
What is Datonism ?
Answer:
The defect of vision due to which a person is not able to distinguish between the red and green colours is called datonism.

Question 3.
What is dispersion of white light ?
Answer:
The phenomenon of splitting white light into seven colours when it passes through a glass prism is called dispersion.

Question 4.
What is white light spectrum ?
Answer:
A band of seven colours of white light arranged according to the increasing wavelength is called spectrum of white light spectrum.

Question 5.
Name the constituent colours of white light.
Answer:
VIBGYOR i.e., Violet, Indigo, Blue, Green, Yellow, Orange Red.

Short Answer Type Questions

Question 1.
Why does it take some time to see objects in a dimly Lightroom when you enter the room from outside which is bright?
Answer:
When we are in bright light, the size of our pupil becomes very small and the rod shaped cells on retina also comes in excited state. Thus, when we enter in a dimly light room the pupil of our eye tries some time to expand and rod shaped cells also take sometime to come back normal state and we are not able to see the objects immediately.

Question 2.
Why colours of an object are not visible in dim light?
Answer:
We are able to see the colours of an object with the help of cone shaped cells. But these cells are not sensitive to dim light due to which colours of an object are not visible in dim light.

Question 3.
What are rods and cones cells ? Give their functions.
Answer:
The retina of our eye contains larger number of cells which are either rod-shaped or cone-shaped. The rod-shaped cells are known as rods and cone-shaped cells are known as cones. The rod-shaped cells respond to the intensity of light while cone shaped cells respond to colours.

Long Answer Type Questions

Question 1.
Explain the formation of a rainbow.
Answer:
Rainbow is one of the most A highly example of spectrum formed due to the dispersion of light raindrop in nature. The is produced due to the dispersion of sunlight by tiny droplets of water suspended in air, just after the rain.

The suspended tiny droplets of water act as innumerable small prisms. When the sunlight is incident on the side ‘A of the tiny droplet of water, it gets refracted as well as dispersed. The dispersed rays on striking the surface ‘B’ of the tiny waterdrop suffer total internal reflection and hence moves on towards surface ‘A’. At the surface ‘A’, the rays further suffer refraction and emerge out in the form of band of colours in the form of a circular arc along the horizon. The red colour appears on the upper arc of rainbow and violet colour on the innermost arc. We can also see rainbow on a bright sunny day, in the mist created by a water fall or water fountain.

Question 2.
Draw a labelled diagram of eye and give brief description of its different parts with function.
Answer:
Parts of an eye:
1. Sclerotic : It is the outermost covering of the eye. It consists of white tough fibrous tissue. Its function is to protect and contain the vital internal parts of the eye.

2. Cornea: It is the front bulging portion of the eye. It allow the light to enter the eye ball.

3. Choroid: It is a grey membrane attached to the sclerotic from the inner side. Its function is to darken the eye from inside, and hence, prevent any internal reflection.

4. Optic nerve: It is a bundle of approximately 70000 nerves originating from the brain and entering the eye ball from posterior side. Its function is to carry the optical messages to the brain.

5. Retina: The optic nerve on entering the eyeball, spreads like a canopy and each nerve attaches itself to the choroids. The nerve ending from a kind of hemispherical screen called retina. The function of the retina is to receive the optical image of an object and then convert it into electrical pulses which are finally sent to the brain through the optic nerve.

6. Yellow spot: It is situated at the Centre of the retina and is slightly raised. It has a little depression called Fovea-Centralist, which is extremely sensitive to light. Its function is to form an extremely clear image.

7. Blindspot: The region on the retina, where the optic nerve enters the eyeball is called the blind spot. Apparently it has no function. Any image formed at this spot is not visible.

8. Crystalline lens: It is a double convex lens, more bulging the posterior side. It is made of transparent, flexible in position by a ring of muscles, commonly called ciliary muscles. Its function is to focus the image of the objects at different distances, clearly on the retina.

9. Ciliary muscle: It is a ring of muscles, which along with the suspensory ligament, holds the crystalline lens in position when these muscles contract, they decrease the focal length of the which along with crystalline lens. When these muscles are relaxed they increase the focal length of the crystalline lens.

10. Iris: It is a circular contractile diaphragm suspended in front of the crystalline lens. It has a tiny hole in the middle, commonly called the pupil Its function is to control the amount of light entering the eye.

11. Viterous humour: It is a dense, jelly like fluid, slightly grey in colour, filling the posterior part of the eye ball. It prevents the eye ball from collapsing, due to the changes in the atmospheric pressure.

12. Aqueous humour : It is a watery, saline fluid, filling the anterior portion of the eye. It prevents the anterior portion of the eye from collapsing, due to the changes in the atmospheric pressure.

Multiple Choice Questions

Question 1.
Splitting of white light into seven colours is known as:
(a) refraction
(b) reflaction
(c) interference
(d) dispersion
Answer:
(d) dispersion

Question 2.
Which of the following colours has the least wavelength ?
(a) Red
(b) Orange
(c) Violet
(d) Blue
Answer:
(c) Violet

Question 3.
A white light falls on the glass prism. The least deviated Colour is
(a) Violet
(b) Orange
(c) Yellow
(d) Red
Answer:
(d) Red

Question 4.
Persistence of vision is
(a) 1/10 sec
(b) 1/16 sec
(c) 1/20 sec
(d) 1/32 sec
Answer:
(b) 1/16 sec

Question 5.
The colour having maximum wavelength is:
(a) Violet
(b) Yellow
(c) Red
(d) blue
Answer:
(c) Red

These NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.1 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Exercise 8.1

Question 1.
In ∆ABC right angled at B, AB = 24 cm, BC = 7 cm. Determine:
(i) sin A, cos A
(ii) sin C, cos C
Solution:
We have given that,
∆ABC is a right angle ∆, right angled at B and AB = 24 cm, BC = 7 cm
∴ By Pythagoras theorem

Question 2.
In given figure, find tan P – cot R.
Solution:
We have given that ∆PQR is a right angled A, right angle at Q. and PQ = 12 cm, PR = 13 cm
∴ By Pythagoras theorem, we know that

Question 3.
If sin A = \(\frac { 3 }{ 4 }\), calculate cos A and tan A.
Solution:
Given

Question 4.
Given 15 cot A = 8, find sin A and sec A.
Solution:

Question 5.
Given sec θ = \(\frac { 13 }{ 12 }\) , calculate all other trigonometric ratios.
Solution:

Question 6.
If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠A = ∠B.
Solution:

Question 7.
If cot θ = \(\frac { 7 }{ 8 }\), evaluate:
(i) \(\frac { \left( 1+sin\theta \right) \left( 1-sin\theta \right) }{ \left( 1+cos\theta \right) \left( 1-cos\theta \right)}\)
(ii) cot²θ
Solution:

Question 8.
If 3 cot A = 4, check whether \(\frac { 1-tan^{ 2 }A }{ 1+tan^{ 2 }A }\) = cos² A – sin² A or not.
Solution:

Question 9.
In triangle ABC, right angled at B, if tan A = \(\frac { 1 }{ \surd 3 }\), find the value of:
(i) sin A cos C + cos A sin C
(ii) cos A cos C – sin A sin C
Solution:

Question 10.
In ΔPQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P.
Solution:

Question 11.
State whether the following statements are true or false. Justify your answer.
(i) The value of tan A is always less than 1.
(ii) sec A = \(\frac { 12 }{ 5 }\) for some value of angle A.
(iii) cos A is the abbreviation used for the cosecant of angle A.
(iv) cot A is the product of cot and A.
(v) sin θ = \(\frac { 4 }{ 3 }\) for some angle.
Solution:
(i) tan 60° = √3 , Since √3 > 1. (False)
(ii) sec A is always ≥ 1. (True)
(iii) cos A is the abbreviation for cosine A. (False)
(iv) cot without ∠A is meaningless. (False)
(v) sin θ can never be greater than 1.
∴ sin θ = \(\frac { P }{ H }\), hypotenuse is always greater than other two sides. (False)

These NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.2 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Exercise 8.2

Question 1.
Evaluate the following:

Solution:

Question 2.
Choose the correct option and justify your choice:
(i) \(\frac{2 \tan 30^{\circ}}{1+\tan ^{2} 30^{\circ}}\) =
(A) sin 600 (B) cos 60°
(C) tan 60° (D) sin 300

(ii) \(\frac{1-\tan ^{2} 45^{\circ}}{1+\tan ^{2} 45^{\circ}}\) =
(A) tan 90° (B) 1
(C) sin 450 (D) O

(iii) sin 2A = 2 sin A is true when A =
(A) 00 (B) 30°
(C) 45° (D) 60°

(iv) \(\frac{2 \tan 30^{\circ}}{1-\tan ^{2} 30^{\circ}}\) =
(A) cos 60° (B) cos 60°
(C) tan 60° (D) sin 300
Solution:
(i) We have given,

(ii) We have given,

(iii) We have given,

Therefore,
sin 2 A = 2 sin A is true only when A = 0°
∴ correct option is (A)

(iv) We have given,

Question 3.
If tan (A + B) = √3 and tan (A – B) = \(\frac { 1 }{ \surd 3 }\); 0° < A + B ≤ 90°; A > B, find A and B.
Solution:
We have given
tan (A + B) = √3
⇒ tan (A + B) = tan 60°
⇒ A + B = 60° … (i)
tan (A – B) = \(\frac { 1 }{ \surd 3 }\)
⇒ tan (A – B) = tan 30°
⇒ A – B = 30° … (ii)
Adding equation (i) and (ii), we get
2A = 90° ⇒ A = 45°
Putting the velue of A in equation (i), we get
45° + B = 60° ⇒ B = 60° – 45° = 15°
Therefore A = 45° and B = 15°

Question 4.
State whether the following statements are true or false. Justify your answer.
(i) sin (A + B) = sin A + sin B.
(ii) The value of sin θ increases as θ increases.
(iii) The value of cos θ increases as θ increases.
(iv) sin θ = cos θ for all values of θ.
(v) cot A is not defined for A = 0°.
Solution:
(i) False, because if A = 60° and B = 30° then
sin (A + B) = sin (60° – 30°)
= sin 90° = 1
sin A + sin B = sin 60° + sin 30°
= \(\frac{\sqrt{3}}{2}+\frac{1}{2}\) = \(\frac{\sqrt{3}+1}{2}\)
∴ sin (A + B) ≠ sin A + sin B, when A = 60° and B = 30°

(ii) True, because the value of sin θ increases as θ increases from θ to 90°, but when θ increases from 90° to 180° then the value of sin 0 decreases.

(iii) False, because the value of cos θ decreases as θ increases from 0 to 90°.

(iv) False, because sin θ = cos θ is true only when θ = 45°. It is not true for all values of θ,

(iv) True, because cot 0° = \(\frac { 1 }{ 0 }\) = not defined.