Question

A myopic adult has a far point at 0.1 m. His power of accommodation is 4 diopters.

(i) What power lenses are required to see distance objects?

(ii) What is his near point without glasses?

(iii) What is his near point with glasses? (Take the image distance from the lens of the eye to the retina to be 2 cm.)

Answer 1

(i) Let the power at the far point be P_f for the normal relaxed eye.

Then

\(P_f=\frac{1}{f}=\frac{1}{0.1}+\frac{1}{0.02}=60\,D\)

With the corrective lens the object distance at the far point is ∞.

The power required is

\(P_f'=\frac{1}{f'}=\frac{1}{∞}+\frac{1}{0.02}=50\,D\)

The effective power of the relaxed eye with glasses is the sum of the eye and that of the glasses P_g.

\(∴P_f'=P_f+P_g\)

∴ P_g = −10 D.

(ii) His power of accommodation is 4 diopters for the normal eye. Let the power of the normal eye for near vision be P_n.

Then

4 = P_n - P_f or P_n = 64 D.
Let his near point be x_n,

Then

\(\frac{1}{x_n}+\frac{1}{0.02}=64\,or\,\frac{1}{x_n}+50=64\)

\(\frac{1}{x_n}=14,\)

\(∴x_n= \frac{1}{14};0.07m\)

(iii) With glasses

P′_n = P′_f + 4 = 54

54 = \(\frac{1}{x_n'}+\frac{1}{0.02}=\frac{1}{x_n'}+50\)

\(\frac{1}{x_n'}=4,\)

\(∴x_n'=\frac{1}{4}=0.25m.\)