the far point of a myopic person is 80 cm in front of the eye.what is the nature and power of the lens require to correct the problem?
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far point, u =80 cm = 0.8 m
to correct far point should be infinity, v =infinity
applying lens law,
1/v -1/u = 1/f
⇒ 0 - 1/0.8 = 1/f
⇒1/f = (-1/0.8)= - 1.25
now, power of lens = 1/f
= -1.25D
So, to correct that myopic eye concave lens of power (- 1.25 D) is required.
to correct far point should be infinity, v =infinity
applying lens law,
1/v -1/u = 1/f
⇒ 0 - 1/0.8 = 1/f
⇒1/f = (-1/0.8)= - 1.25
now, power of lens = 1/f
= -1.25D
So, to correct that myopic eye concave lens of power (- 1.25 D) is required.
Answered by
7
The person is suffering from myopia so, he need concave lens to correct his vision.
For this,
image distance,v=-80cm
object distance,u=infinitey
now by lens formula
1/v-1/u=1/f
1/-80-1/infinity=1/f
1/-80-0=1/f ( 1/infinity=0)
1/-80=1/f
f=-80cm
=-0.8m
Now power=1/f
=1/-0.8
=-1.25D
therefore the power of concave lens required is -1.25D
For this,
image distance,v=-80cm
object distance,u=infinitey
now by lens formula
1/v-1/u=1/f
1/-80-1/infinity=1/f
1/-80-0=1/f ( 1/infinity=0)
1/-80=1/f
f=-80cm
=-0.8m
Now power=1/f
=1/-0.8
=-1.25D
therefore the power of concave lens required is -1.25D
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