kseeb sslc science solutions for ch 6 the human eye and the colourful world

KSEEB SSLC Science Solutions
Chapter 6 The Human Eye And The Colourful World

LearnFatafat offers free Solutions for SSLC Science Chapter 6 The Human Eye And The Colourful World. Chapter covers the topics like human eye, its defects and corrections, refraction of light through glass prism, atmospheric refraction, scattering of light and more. Check video lessons, notes and MCQ quizzes for SSLC Science Chapter 6 The Human Eye And The Colourful World click here to buy.

Solutions for SSLC Science Chapter 6 The Human Eye And The Colourful World

1. The human eye can focus 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.

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

Answer: (d) retina.

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

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

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

we know that P = 1 / f

∴ f = 1/ P

∴ f = 1/ (-5.5) = – 0.182 m

(ii)Power of lens for correcting new vision = 1.5 D

We know that P = 1 / f

∴ f = 1 / P

∴ f = 1 / (1.5) = 0.66 m

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: To correct defect myopia concave lenses are used.

Given: Distance of object (u) = infinity

Distance of image (v) = – 80 cm

Let f be the focal length.

Let’s use the lens formula to find the focal length.

(1/v) – (1/u) = 1/f

-1/80 – 1 / ∞ = 1/f

∴ 1/f = – 1/ 80

∴ f = – 80 cm = 0.8m

Now power of lens P = 1 /f

∴ P = 1 / – 0.8 = – 1.25 D

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: Hypermetropia is an eye defect in which person can see distant objects clearly but finds difficulty to see near objects. Following ray diagram shows correction in hypermetropic eye.

NCERT Solutions Class 10 Science Chapter 11 Q7
Given: Distance of object (u) = -25 cm

Distance of image (v) = -1 m = -100 cm

Let f be the focal length.

Using lens formula,

(1/v) – (1/u) = 1/f

∴ (1/-100) – (1/-25) = 1/f

∴ 1/f = 1/25 – 1/100

∴ 1/f = (4-1)/100

∴ f = 100/3 = 33.3 cm = 0.33 m

Power P = 1 / f = 1/0.33 = 3.03D

Convex lens with power +3.03 D is needed to correct the defect.

8. Why is a normal eye not able to see clearly the objects placed closer than 25 cm?

Answer: A normal eye is not able to see clearly the objects placed closer than 25 cm because ciliary muscles do not contract after a certain limit.

9. What happens to the image distance in the eye when we increase the distance of an object from the eye?

Answer: Whether the distance of object is increased or decreased, image is always formed at retina. For this eye lens becomes thin. Thus, focal length increases as the object moves away from the the eye.

10. Why do stars twinkle?

Answer: Twinkling of stars is due to refraction of starlight when passing through layers of earth’s atmosphere. Star is at very long distance from the earth hence can be considered as point source of light. Earth’s atmosphere is made of different layers with varying refractive indexes and also earth’s atmosphere is not stable. When light enters in the earth’s atmosphere, it undergoes refraction through different layers in the atmosphere. Thus, starlight entering the eye flickers and hence we observe twinkling of stars.

11. Explain why the planets do not twinkle.

Answer: Planets do not twinkle because planets are much closer to earth and hence can not be considered as point source. Large number of rays coming from planets undergo variations but these are not observable, also, these nullify each other’s effect.

12. Why does the Sun appear reddish early in the morning?

Answer: Early in the morning during sunrise, sunrays travel larger distance in the atmosphere of the earth, before reaching our eyes. While traveling through earth’s atmosphere, short wavelengths are scattered out and larger wavelengths reaches our eyes. Red light has larger wavelength, thus, even after the scattering of light red light reaches our eyes. Therefore, sun appears reddish early in the morning.

13. Why does the sky appear dark instead of blue to an astronaut?

Answer: In space there is not atmosphere. Thus in the absence of atmosphere scattering of sunlight do not takes place. Therefore, no scattered light reaches to the eyes of the astronaut, thus sky appears dark in the space.

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Chapter 6 – The Human Eye And The Colourful World

6.01 The Human Eye

6.02 Defects of Vision and their Correction

6.03 Refraction of Light Through a Glass Prism

6.04 Atmospheric Refraction

6.05 Scattering of Light