Question

The lens-to-retina distance is 2.04 cm, and the relaxed power of the eye is 51.0 D.

What is the far point in m?

What eyeglass power in D will help the person see distant objects clearly, if the glasses are 1.80 cm from the eyes?

Answer 1

Solution:

Part (a) Solution:

Given:

Power (P) = 51 D

f = 1/P = 1/51 = 0.0196 m = 1.96 cm

And, s' = 2.04 cm

Now: 1/f = 1/s + 1/s'

1/1.96 = (1/s) + (1/2.04)

s = 50.495 cm

Part (b)Solution:

Here, s = infinity

s' = - (50.495 - 1.8) = - 48.695 cm

So, Using formula : 1/f = (1/s) + (1/s')

f = - 48.695 cm = - 0.48695 m

Therefore: P = 1/f = 1 / (- 0.48695 m) = - 2.054 D