Question

If a person eye has near point at 75 cm from his eye..What is the focal lenght and power of eye lens ?? ​

Answer 1

Answer:

Focal Length = + 37.5 cm ; Power = +2.67 D

Explanation:

Since , near point is 75 cm , person must be suffering from hypermetropia . So he should use Convex Lena for this and If we Place the Object At u= -75cm then his eyes want to make the image at v= -25 cm Then By solving it by Lens Formula

$\frac{1}{f} = \frac{1}{v} - \frac{1}{u }$

By placing the values ,

$\frac{1}{f} = - \frac{1}{75} + \frac{1}{25}$

$\: \: \: \: = \frac{ - 1}{75} + \frac{3}{75}$

, which gives

F = +75/2 = 37.5 Cm = +0.375 M

$p = \frac{1}{f}$

$\: \: \: \: \: = \frac{1}{0.375}$

which gives

P = + 2.666666.. D

Therefore , Power = +2.67 D