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 on its other side. Now a glass block (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 must be shifted by a distance $d$. Find the value of $d$.

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

Answer 1

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

Given, $2f=10\text{cm}\Rightarrow f=5\text{cm}$

Now, due to glass slab,

Shift $\left(S\right)=t(1-\frac{1}{\mu })$

Substituting the given data we get,

$S=1.5(1-\frac{2}{3})=0.5\text{cm}$

According to the sign convention,

$u=-(10-0.5)=-9.5\text{cm}$

Now, From lens formula

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

Substituting the values of $v,u$ and $f$

$\Rightarrow v=\frac{\left(5\right)\left(9.5\right)}{4.5}=\frac{47.5}{4.5}=10.55\text{cm}$

Hence the screen should be shifted by $0.55\text{cm}$.



Hence, option (d) is the correct answer.