Q. A convex lens is put $10 \,cm$ from a light source and it makes a sharp image on a screen, kept $10\, cm$ from the lens. Now a glass block (refractive index $1.5$) of $1.5\, cm$ 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 :
Solution:
$\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} = 10.55\, cm $
