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

Answer 1

Answer: (A)

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\Rightarrow \frac{1}{10}-\frac{1}{-10}=\frac{1}{f}\Rightarrow f=5cm$
Shift due to slab = $t\left(1-\frac{1}{\mu }\right)$ n the direction of incident ray
$=1.5\left(1-\frac{2}{3}\right)=0.5$
again, $\frac{1}{v}-\frac{1}{-9.5}=\frac{1}{5}$
$\Rightarrow \frac{1}{u}=\frac{1}{5}-\frac{2}{19}=\frac{9}{95}$
$\Rightarrow v=\frac{95}{9}=10.55cm$