Chapter 11 : Human Eye and The Colourful World - Ncert Solutions for Class 10 Physics CBSE

In CBSE Class 10, Physics is a combination of theoretical and practical knowledge. Physics can be one of the toughest subjects to understand. In the CBSE Class 10 Physics syllabus, there are crucial topics such as electricity, magnetic effects of electric current, refraction of light, etc. Through the Physical Practical Class 10 CBSE sessions, you dive deeper into how things work. TopperLearning presents study materials for CBSE Class 10 Physics which will help you to effectively prepare for your final examination. Our Physics study materials are prepared by subject experts and include video lessons, revision notes, question banks, sample papers and past years' question papers.

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Chapter 11 - Human Eye and The Colourful World Excercise 190

Solution 1

When the ciliary muscles are relaxed, the eye lens becomes thin, the focal length increases, and the distant objects are clearly visible to the eyes. To see the nearby objects clearly, the ciliary muscles contract making the eye lens thicker.
Thus, the focal length of the eye lens decreases and the nearby objects become visible to the eyes. Hence, the human eye lens is able to adjust its focal length to view both distant and nearby objects. This ability is called the power of accommodation of the eye.

Solution 2

The person is able to see nearby objects clearly, but he is unable to see objects beyond 1.2 m. This happens because the image of an object beyond 1.2 m is formed in front of the retina and not at the retina, as shown in the given figure.

To correct this defect of vision, he must use a concave lens. The concave lens will bring the image back to the retina as shown in the given figure.

  

Solution 3

For human eye with normal vision, the far point is at infinity and near point is at 25 cm from the eye.

Solution 4

A student has difficulty in reading the blackboard while sitting in the last row. It shows that he is unable to see distant objects clearly. He is suffering from myopia.
This defect can be corrected by using a concave lens.

Chapter 11 - Human Eye and The Colourful World Excercise 197

Solution 1

(b) accommodation

Concept Insight: Human eye can change the focal length of the eye lens to see the objects situated at various distances from the eye. This property of light is called accommodation.

Chapter 11 - Human Eye and The Colourful World Excercise 198

Solution 1

(d) Retina

The human eye forms the image of an object at its retina.

Solution 2

(c) 25 cm

Concept Insight: The least distance of distinct vision is the minimum distance from the eye where an object can be seen clearly and distinctly. It is 25 cm for a young adult with normal vision.

Solution 3

(c) ciliary muscles

Concept Insight: The relaxation or contraction of ciliary muscles changes the curvature of the eye lens. The change in curvature of the eye lens changes the focal length of the eyes. Hence, the change in focal length of the eye lens is caused by the action of ciliary muscles.

Solution 4

The power P of a lens of focal length f is given by the relation

begin mathsize 14px style straight P equals fraction numerator 1 over denominator straight f space left parenthesis in space metres right parenthesis end fraction end style

(i) Power of the lens used for correcting distant vision = -5.5 D

Focal length of the required lens, begin mathsize 14px style straight f equals 1 over straight P end style

begin mathsize 14px style straight f equals fraction numerator 1 over denominator negative 5.5 end fraction equals negative 0.181 space straight m end style

The focal length of the lens for correcting distant vision is -0.181 m.

 

(ii) Power of the lens used for correcting near vision = +1.5 D

Focal length of the required lens, begin mathsize 14px style straight f equals 1 over straight P end style

begin mathsize 14px style straight f equals fraction numerator 1 over denominator 1.5 end fraction equals 0.667 space straight m end style

The focal length of the lens for correcting near vision is 0.667 m.


Concept Insight: Concave lens is used for correcting short sightedness. Convex lens is used for correcting long sightedness.

Solution 5

Object distance, u = ∞

Image distance, v = -80 cm (at far point)

Focal length = f

According to the lens formula,

begin mathsize 14px style 1 over straight v minus 1 over straight u equals 1 over straight f
minus 1 over 80 minus 1 over infinity equals 1 over straight f
1 over straight f equals negative 1 over 80
straight f equals negative 80 space cm equals negative 0.8 space straight m end style


We know,

Power, begin mathsize 14px style straight P equals fraction numerator 1 over denominator straight f space left parenthesis in space metres right parenthesis end fraction end style

Syntax error from line 1 column 49 to line 1 column 90. Unexpected 'mathvariant'.

A concave lens of power -1.25 D is required by the person to correct his defect.

Concept Insight :80 cm distance corresponds to the distance at which the image of a far off object should be formed by the corrective lens, because that corresponds to the maximum distance at which if an object is placed, can be seen clearly. 

Solution 6

A person suffering from hypermetropia can see distant objects clearly but faces difficulty in seeing nearby objects. It happens because the eye lens focuses the incoming rays from the object lying at normal near point beyond the retina. This defect of vision can be corrected by using a convex lens. A convex lens of suitable power converges the incoming light in such a way that the image is formed on the retina, as shown in the following figure.

The rays starting from the normal ear point N' converge on passing through the convex lens, and appear to come from N, the near point of the hypermetropic eye.

Given:

Object distance, u = -d = -25 cm

Image distance, v = -1 m = -100 cm

Focal length, f = ?

Using the lens formula,

begin mathsize 14px style 1 over straight v minus 1 over straight u equals 1 over straight f
fraction numerator 1 over denominator negative 100 end fraction minus fraction numerator 1 over denominator negative 25 end fraction equals 1 over straight f
1 over straight f equals 1 over 25 minus 1 over 100
1 over straight f equals fraction numerator 4 minus 1 over denominator 100 end fraction
1 over straight f equals 3 over 100
straight f equals 100 over 3 equals 33.33 space cm
therefore straight f equals 0.33 space straight m end style

 

begin mathsize 14px style Power comma space straight P equals fraction numerator 1 over denominator straight f space left parenthesis in space metres right parenthesis end fraction equals fraction numerator 1 over denominator 0.33 space straight m end fraction equals plus space 3.0 space straight D end style

A convex lens of power +3.0 D is required to correct the defect.

Concept Insight: Concave lens is used for correcting short sightedness. Convex lens is used for correcting long sightedness.

Solution 7

A normal eye is unable to clearly see the objects placed closer than 25 cm because the ciliary muscles of eyes are unable to contract beyond a certain limit.

If the object is placed at a distance less than 25 cm from the eye, then the object appears blurred and produces strain in the eyes.

Solution 8

When we increase or decrease the distance of an object from the eye, the image distance in the eye (distance of retina from the eye lens) does not change. The increase in the object distance is compensated by the change in the focal length of the eye lens. The focal length of the eyes changes in such a way that the image is always formed at the retina of the eye.  

Concept Insight: The property of the eye to be able to see the objects placed at different distances clearly is also known as accommodation.

Solution 9

Stars emit their own light and they twinkle due to the atmospheric refraction of light. Stars are very far away from the earth. Hence, they are considered as point sources of light. When the light coming from stars enters the earth's atmosphere, it gets refracted at different levels because of the variation in the air density at different levels of the atmosphere.
When the atmosphere refracts more star-light towards us, the star appears to be bright and when the atmosphere refracts less star-light, then the star appears to be dim. Therefore, it appears as if the stars are twinkling at night.

Concept Insight: Remember that twinkling of stars is just an illusion that appears due to the various atmospheric conditions.

Solution 10

Planets do not twinkle because they appear larger in size than the stars as they are relatively closer to Earth. Planets can be considered as a collection of a large number of point-size sources of light.
The different parts of these planets produce either brighter or dimmer effect in such a way that the average of brighter and dimmer effects is zero. Hence, planets not twinkle.

Concept Insight: One should remember the conditions that cause twinkling effect in stars while discussing about twinkling effect of planets.

Solution 11

During sunrise, the light rays coming from the Sun have to travel a greater distance in the earth's atmosphere before reaching our eyes. In this journey, the shorter wavelengths of lights are scattered out and only longer wavelengths are able to reach our eyes.
Since blue colour has a shorter wavelength and red colour has a longer wavelength, the red colour is scattered the least and is able to reach our eyes after the atmospheric scattering of light. Therefore, the Sun appears reddish early in the morning.

Concept Insight: The fact that longer wavelengths get less scattered & hence are able to travel longer distances is very important.

Solution 12

The sky appears dark instead of blue to an astronaut because there is no atmosphere in the outer space that can scatter the sunlight. As the sunlight is not scattered, no scattered light reachs the eyes of the astronauts and the sky appears black to them.

Concept Insight: The prerequisite of the scattering process is the existence of particles of the size comparable to the wavelength of the light.

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