Question

A convex lens is put $10\text{cm}$ from a light source, and it makes a sharp image on a screen, kept $10\text{cm}$ from the lens. Now a glass block $\text{(refractive index 1.5)}$ of $1.5\text{cm}$ 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. $1.1\text{cm}$ away from the lens
• B. $0.55\text{cm}$ towards the lens
• C. $0$
• D. $0.55\text{cm}$ away from the lens

Answer 1

The correct option is D $0.55\text{cm}$ away from the lens


Using lens formula
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

$\Rightarrow \frac{1}{10}-\frac{1}{-10}=\frac{1}{f}\Rightarrow f=5\text{cm}$

$\text{Shift due to slab,}\Delta x=t(1-\frac{1}{\mu })$ in the direction of incident ray

$\Rightarrow \Delta x=1.5(1-\frac{2}{3})=0.5$

$\text{Now}{u}^{\prime }=-9.5\text{cm}$

Again using lens formula
$\frac{1}{v^{\prime }}-\frac{1}{-9.5}=\frac{1}{5}$

$\Rightarrow \frac{1}{v^{\prime }}=\frac{1}{5}-\frac{2}{19}=\frac{9}{95}$

$\Rightarrow v^{\prime }=\frac{95}{9}=10.55\text{cm}$

Thus, the shift in the screen is given by,

$d=10.55-10=0.55\text{cm}$ away from the lens.

Hence, option $\left(D\right)$ is correct.