Question

50 points
The near point of an eye is 1m. what is the defect of the eye? calculate the power of the lens required to correct this defect

Answer 1

Person is suffering from hypermetropia.

We have,

d = 100 cm

So, 1 / f = 1 / 25 + 1 / -d

=> 1 / f = 1 / 25 - 1 / 100

=> 1 / f = 3 / 100

=> f = 100 / 3

=> f = 33.3 cm

So, power of lens = 100 / f(in cm)
= 100 / 33.3
= 1000 / 333
= 3.003 D

Answer 2

Here is your answer :-


According to question we have given,

u = - 25 cm
v = - 1 m
So v in cm = - 100 cm
f = ?

Formula = 1/f = 1/v - 1/u

1/f = -1/100 + 1/25
1/f = 100/3 cm
1/f = 100/3 x 1/100 = 1/3 m

Now,
Power (P) = 1/f
P = 1/3
P = +3D

The person is suffering from hypermetropia.


Hope it helps.