11. The Human Eye and Colourful World

CHAPTER  11.
THE HUMAN EYE AND COLOURFUL WORLD
PAGE No: 190

1. What is meant by power of accommodation of eye ?

The ability of the eye lens to adjust its focal length is called the power of accommodation of eye. It is the maximum variation of its power for focusing on near and far objects. For normal eye , the power of accommodation is about four dioptres.

2. A person with a myopic eye cannot see objects beyond 1.2 metes distinctly. What should be        the type of corrective lens used to restore proper vision?

Concave lens ( diverging )of suitable wavelength should be used.

3. What is the far point and near point of the human eye with normal vision? 

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

4. A student has difficulty reading the blackboard while sitting in the last row. What could be        the defect the child is suffering from ?. How can it be corrected?

The child is suffering from myopia. It can be corrected by using a concave lens ( diverging lens ) of suitable power.

EXERCISES
PAGE No: 197

1. The human eye can focus objects at different distances by adjusting the focal length of the           eye lens. This is due to 

Accommodation

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

Retina.

3. The least distance of distant vision for a young adult with normal vision is about

25 cm

4. The change in focal length of an eye lens is caused by the action of the

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 dioptres .What is the focal length of the lens            required for correcting (i) distant vision and (ii) near vision.

(i) The power of the lens used for distant viewing is  -5.5 D
     Power = 1/f

Focal length = 1 / Power  = 1 / -5.5   =   -0.181818m

 ie  Focal length  = -18.1818cm

Since focal length is negative the lens used is concave

The person should use a concave lens of focal length 18.1818cm

(ii)  The near viewing section of the lens is corrected by power = +1.5 D

     Power = 1/f

Focal length = 1 / Power  = 1 / 1.5   =   0.67 m
 ie  Focal length  = 67 cm

Since focal length is positive the lens used is convex

The person should use a convex lens of focal length 167 cm

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

Since this person is having myopia, the lens used is concave.
The far point distance (v)and the distance of the object at infinity (u), both are negative, since they are in front of the lens

u = -80     v = infinity

By lens formula  1/f  = 1/v - 1/u

1/f  = 1/-80  -  1/infinity  =  -1/80  + 0   =  -1/80

Hence f = -80 cm

Power = 1/f  ( f in meter )

P = (1/ -0.8 ) D  =  -1.25D

Negative sign again indicates the lens is concave

The person should use a concave lens of 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     eye  is   25 cm

The near point distance of hypermetropic eye  (v)and the near point distance of a normal eye  (u), both are negative, since they are in front of the lens

u = -25 cm     v =  -100 cm

By lens formula  1/f  = 1/v - 1/u

1/f  = 1/-100cm  -  1/-25cm  =  = 3/100cm

Hence f = 100/3 cm  = 33.3cm

Power = 1/f  ( f in meter )

P = (1/ 0.333 ) D  =  3D

Positive  sign again  indicates the lens is convex

The person should use a convex lens of 3 D

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

The least distance of distinct vision for a normal eye is 25 cm.

When an object is placed less than  25cm , the ciliary muscles cannot bulge the eye lens anymore, and hence cannot be focused  on the retina and a blurred image is seen

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

For a normal eye the image distance is fixed. It is equal to the distance of retina from the eye lens.

When we increase the distance of an object , to make the image distance constant, the adjustment is made on the focal length of the eye lens by the help of ciliary muscles

10. Why do stars twinkle

Due to the atmospheric refraction the star twinkles.

When the star light enters the earths atmosphere it undergoes continuous refraction before it reaches the earth. The atmospheric refraction occurs in a medium of gradually changing refractive index . Since the atmosphere bend the starlight , the position of the star is slightly different from its original position when  viewed from here

The stars appear a little higher when viewed from the horizon and the apparent position of the star keeps on changing because of the change in physical features of the atmosphere like density, temperature etc

As a result the fluctuating image  of the star makes them twinkle to the observer

11. Explain why the planets do not twinkle

Planets do not emit light. They become visible due to the reflection of light falling on them.
The planets are closer to the earth and are considered as an extended source of light.

The fluctuations of the light coming from a large number of point sources due to refraction gets nullified.Hence no twinkling of planets can be seen.

12.  Why does the sun appears reddish early in the morning

During the sun set and sun rise the sun is closer to the horizon. The solar rays have to travel a longer distance as compared to than at noon when the sun is overhead.
Most of the shorter wave length colours like blue gets scattered by the particles such as dust , humidity and molecules of the air
The longer wave length colour like orange and  red are least scattered
Hence the red colour  reaches the observers eye and so the red colour at dawn and dusk

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

At a huge height there is not much atmosphere for the light rays to scatter. Hence the sky appears dark for an astronaut.