Question

the far point of a miopic person is 80cm in front of eye.what is the nature and power of the lens required to correct the problem​

Answer 1

Part A

The nature of the lens required is a concave lens

Part B

The power of the lens required is -1.25 D

Step-by-step explanation:

Part A

Myopia is the eye condition whereby the focus of light entering the eye is in front of the retina such that the vision of a myopic person is blurred

The given far point of the myopic person in question = 80 cm

To correct the problem, the focus of light entering the eye will be moved further to reach the retina, by means of a concave lens which makes distant objects appear to come from (form an image at) 80 cm

Part B

The lens formula is given as follows;

`(1)/(f) = (1)/(d_o) + (1)/(d_i)`

Where;

f = The focus of the lens (virtual focus)

`d_o` = The distance of the object = ∞ (infinity)

`d_i` = The distance of the image = -80 cm (virtual image formed in front of the mirror)

Plugging in the values, gives;

`(1)/(f) = (1)/(\infty) + (1)/(-80 \, cm) = (1)/(-80 \, cm)`

`\therefore (1)/(f) = (1)/(-80 \, cm)`

f = -80 cm = -0.8 m

The power of a lens, 'P' is the reciprocal of the focal length, 'f', given in meters

`P = (1)/(f (in \, meters))`

The SI unit for the power of a lens is the dioptre (D)

Therefore;

`P = (1)/(-0.80 \, m) = -1.25 \, D`

Therefore, the power of the lens required, P = -1.25 D.