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 book (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. $0.55\text{cm}$ towards the lens
• B. $0$
• C. $1.1\text{cm}$ away from 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
$u=-10\text{cm},v=10\text{cm},f=?$
Using lens formula,
$\left(\frac{1}{f}\right)=\left(\frac{1}{v}\right)-\left(\frac{1}{u}\right)$
$\frac{1}{f}=[\left(\frac{1}{10}\right)-\left(\frac{1}{-10}\right)]$
$f=5\text{cm}$.
case II:
Due to introduction of slab, shift in the source is $=t[1-\left(\frac{1}{\mu }\right)]=1.5[1-\left(\frac{2}{3}\right)]=0.5\text{cm}$
Now,
$u=-9.5\text{cm}$,
$\left(\frac{1}{f}\right)=\left(\frac{1}{v}\right)-\left(\frac{1}{u}\right)$
$\left(\frac{1}{5}\right)=\left(\frac{1}{v}\right)+\left(\frac{1}{9.5}\right)$
$v=\frac{5\times 9.5}{4.5}=10.55\text{cm}$
Shift in position of image is
$10.55-10=0.55\text{cm}$ away from the lens