Question

A convex lens focuses a distant object $40 \,cm$ from it on a screen placed $10 \,cm$ away from it. A glass plate $(\mu=1.5)$ and of thickness $3 \,cm$ is inserted between the lens and the screen. Where should the object be placed so that its image is again focused on the screen ? (in $cm$ )

Answer 1

$\frac{1}{ v }-\frac{1}{ u }=\frac{1}{ f } $
$\Rightarrow \frac{1}{+10}-\frac{1}{-40}=\frac{1}{ f } $
$\Rightarrow \frac{1}{ f }=\frac{1}{10}+\frac{1}{40}$ .... (1)
Object shift due to slab $= t \left(1-\frac{1}{\mu}\right)$
$=3\left(1-\frac{1}{1.5}\right) $
$=1 \,cm$
If a glass slab is inserted between lens and screen then $I_{f}$ lens forms the image at $(10-1) cm$ then the slab will shift the image on screen.

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
$\Rightarrow \frac{1}{+9}-\frac{1}{-x}=\frac{1}{f}$
$\frac{1}{9}+\frac{1}{x}=\frac{1}{10}+\frac{1}{40}$
$x=72 \,cm$