Question

An object $O$ (real) is placed at focus of an equi-biconvex lens as shown in Fig. The refractive index of the lens is $\mu =1.5$ and the radius of curvature of either surface of lens is $R$. The lens is surrounded by air. In each statement of column I, some changes are made to situation given above and information regarding final image formed as a result is given in Column II. The distance between lens and object is unchanged in all statements of column I. Match the statements in column I with resulting image in column II.

Answer 1

Given:
Concept:Lens maker's formula
${\frac{1}{f}}_{Lens}=({\mu }_{Lens}-1)(\frac{1}{R_1}-\frac{1}{R_2})$
here $f_{Lens}$ is foca length of lens :
${\mu }_{Lens}$ is Refractive indexof lens :
$R_1$ is radius of curvature of the surface at which light is incident:
$R_2$ is radius of curvature of the surface from which light ray exits.
Solution:
(A) So if the refractive index of lens is doubled then $\frac{1}{f}$ also increase which implies that f decreases.
Hence object will be beyond focal length.
So image will be real and initial image size was infinite but this image size will be smaller than initial size.
(B) If the Radius of curvature is doubled then
$\frac{1}{f}=(\mu -1)(\frac{1}{2R}-\frac{1}{-2R})$
=$\frac{\mu -1}{R}$
and initially :$\frac{1}{f}=(\mu -1)(\frac{1}{R}-(\frac{1}{-R}\left)\right)$
$\frac{1}{f}=\frac{2(\mu -1)}{R}$
Hence we can observe $\frac{1}{f}$ decreasing. Hence f increasing which tells us that object will be at new position between optical center and focal length and since we know that for a biconvex lens of an object is between optical center and focus then image will be virtual and since initial image is of infinite size so final image becomes smaller in size comparision to size of image before the change was made.
(C) If a glassslab is introduced in between object and lens so the apparent object distance for the lens will be less than focal length. So, it is the same case as in (B).
(D) If left side of lens is filled with
$\mu$=1.5 then
$\frac{1}{f}=(\mu -1)(\frac{1}{\infty }-\frac{1}{-R})$
=$\frac{\mu -1}{R}$
Hence it is the same case as in (B).