Question

A lens is made of three thin different mediums. Radius of curvature and refractive index of each medium is shown in Fig. Surface $AB$ is straight. An object is placed at some distance from the lens by which a real image is formed on the screen placed at a distance of $10cm$ from the lens. Find
a. The distance of the object from the lens.
b. A slab of thickness $1.5cm$ and refractive index $1.5$ is introduced between the image and the lens. Find new position of the object so that image is again formed on the screen.
c. Find the position of the image when same slab is placed on the other side of the lens {as in part (b)} with object in the position of part (a).

Answer 1

$\frac{1}{f_1}=(1.6-1)[\frac{1}{20}-\frac{1}{-30}]=\frac{1}{20}c{m}^{-1}$
$\frac{1}{f_2}=(1.5-1)[\frac{4}{-30}-\frac{1}{15}]=\frac{1}{20}c{m}^{-1}$
$\frac{1}{f_3}=(2.5-1)[\frac{1}{15}-\frac{1}{\infty }]=\frac{1}{10}c{m}^{-1}$
$\Rightarrow \frac{1}{f_{eq}}=\frac{1}{20}-\frac{1}{20}+\frac{1}{10}=\frac{1}{10}$
$\Rightarrow f_v=10cm$
Since the image is formed on the screen which is $10cm$ from the lens, so the object is at infinity
b. Shift in the image away from the slab due to introduction of slab is
$△x=(1-\frac{1}{n})t=(1-\frac{1}{1.5})\times 1.5=0.5$
The distance of the object so that image is again formed on the screen
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
$\Rightarrow \frac{1}{9.5}-\frac{1}{u}=\frac{1}{10}\Rightarrow u=190cm$
c. When the slab is placed on the other side of the lens, there will not be any change in the formation.