A young boy has accommodation of his eye such that the power of his eye lens can change between 50D and 60D.His far point is at infinity. The distance of his retina from the eye-lens and his near point are respectively in cm

When light rays comes from Infinity, they converge at focus, i.e we confirm that distance between retina and lens =f (relaxed eye maximum focal length minimum power. 

$\mathrm{f}=\frac{100}{50}=2 \mathrm{cm}$

$\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}}\Rightarrow \mathrm{u}=\text{ Infinity }\Rightarrow \mathrm{v}=\mathrm{f}=2\mathrm{cm}$

for near point minimum focal length of maximum power

$\mathrm{f}=\frac{100}{60}=\frac{5}{3}\mathrm{cm}$

image will form on retina so v=2cm

$\text{ For near point }\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}}\Rightarrow \mathrm{u}=-10$