Question

A convex lens is put $10cm$ from a light source and it makes a sharp image on a screen, kept $10cm$ from the lens. Now a glass block (refractive index 1.5) of $1.5cm$ thickness is placed in contact with the light source. To get the sharp image again, the screen is shifted by a distance $d$. Then $d$ is :

• A. $1.1cm$ away from the lens
• B. $0.55cm$ towards the lens
• C. $0.55cm$ away from the lens
• D. $0$

Answer 1

Answer: (C)

Given, $2f=10cm$

$f=5cm$

Now due to glass plate, shift = $t(1-\frac{1}{\mu })=1.5(1-\frac{2}{3})=0.5cm$

New $u=10-0.5=9.5cm$

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

$v=\frac{47.5}{4.5}$

Shift, d = $v-10=\frac{5}{9}cm$