A young boy can adjust the power of his eye lens between 50 D and 60 D. His far point is infinity. A. What is the distance of his retina from the eye-lens? b. What is the near point?

Explanation

a. When the eye is fuly relaxed, its focal length is largest and the power of the eye lens is minimum. This power is 50 D according to the given data. The focal lengts is `1/50m=2cm.` As the far point is at infinity, the parallel rays coming from infinity are focussed on the retiN/A in the fuly relaxed condition. HEnce, the distance of the retinN/A from the lens equals the focal length which is 2 cm.
b. when the eye is focussed at the near point, the power is maximum which is 60 D. the focal lenth in this case is `f=1/60m=5/3cm. The image is formed on the retiN/A and thus v=2 cm. We have,
`1/v-1/u=1/f`
or `1/u=1/v-1/f=1/(2cm)-3/(5cm)`
or `u=-10cm`
The near point is at 10 cm