$A$ myopic adult has a far point at $0.1\, m$. His power of accommodation is $4\, D$.
$(i)$ What power lenses are required to see distant 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 $2\, cm$.)

(A) $(i)$ For a normal relaxed eye,the power required to focus at the far point $(0.1\, m)$ is given by the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$. With $v = 0.02\, m$ and $u = -0.1\, m$,the power $P_f = \frac{1}{0.02} - \frac{1}{0.1} = 50 - (-10) = 60\, D$.
To see distant objects $(u = \infty)$,the required power is $P_f' = \frac{1}{0.02} - \frac{1}{\infty} = 50\, D$.
The power of the corrective lens $P_g$ is $P_f' - P_f = 50 - 60 = -10\, D$.
$(ii)$ The power of accommodation is $4\, D$. The power at the near point $P_n = P_f + 4 = 60 + 4 = 64\, D$.
Using $\frac{1}{f} = P_n = \frac{1}{v} - \frac{1}{u_n}$,where $v = 0.02\, m$:
$64 = \frac{1}{0.02} - \frac{1}{u_n} \implies 64 = 50 - \frac{1}{u_n} \implies \frac{1}{u_n} = 50 - 64 = -14$.
$u_n = -\frac{1}{14} \approx -0.0714\, m$ or $7.14\, cm$.
$(iii)$ With glasses of $P_g = -10\, D$,the total power at the near point is $P_{total} = P_n + P_g = 64 - 10 = 54\, D$.
Using $\frac{1}{v} - \frac{1}{u_n'} = 54$,with $v = 0.02\, m$:
$50 - \frac{1}{u_n'} = 54 \implies \frac{1}{u_n'} = 50 - 54 = -4$.
$u_n' = -\frac{1}{4} = -0.25\, m$ or $25\, cm$.