Question

A young boy can adjust the power of his eye lens between 50 D and 60 D. The distance of his retina from the eye lens is (his far point is infinity)

• A. 1cm
• B. 2cm
• C. 3cm
• D. 4cm

Answer 1

The correct option is B 2cm

When the eye is fully relaxed, its focal length is largest and the power of the eye lens is minimum. This power is P = 50 D, as given in the question. It is given that the far point is at infinity (∞). Thus, the parallel rays coming from infinity are focussed on the retina.
Thus, the distance of the retina of the eye from the eye lens is equal to the focal length of eye lens when the eye is fully relaxed.
∴ Required distance $=\frac{1}{P}=\frac{1}{50}=0.02m$ $(\because Focallength(f)=\frac{1}{Poweroflens\left(P\right)})$
= 2 cm
Hence, the correct answer is option (b).