Question

Calculate the power of the eye - lens of the normal eye when it is focused at its (a) far point, infinity and (b) near point, 25 cm from the eye. Assume the distance of the retina from the eye - lens to be 2.5 cm.

Answer 1

A) The object at infinity and image at retina. As eye lens is a convex lens

so focal length is $f=2.5cm=.025m$

Now the power $P=\frac{1}{f}=\frac{1}{.025}=40D$

B) in this case object distance is $u=-25$ cm

and image is at retina $v=2.5$ cm

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

$\frac{1}{f}=\frac{1}{2.5}-\frac{1}{-25}$

$f=25/11cm=\frac{25}{1100}m$

Now the power $P=\frac{1}{f}=\frac{1}{25/1100}=44D$