Question

A beam of light parallel to the principal axis is focused on a screen with the help of lens, forming a circular spot with a diameter of $A$. If we put a plane-parallel glass plate of thickness $h$ and refractive index $n$ between the lens and the screen, the diameter of the bright spot on the screen increases. The diameter of the lens is $d$ and focal length $F$. Find the new diameter in $mm$. (Take $F=10\text{cm},A=1\text{cm},n=1.5,d=2\text{cm},h=3\text{cm}.)$

Answer 1

$\frac{F-y}{F}=\frac{A}{d}$

Normal shift $=h(1-\frac{1}{n})$
$\frac{A}{F-y}=\frac{A^{\prime }}{F-y+h(1-\frac{1}{n})}$
$A^{\prime }=A[1+\frac{h}{F-y}(1-\frac{1}{n})]$
$F-y=\frac{AF}{d}$
$A^{\prime }=A[1+\frac{dh}{AF}(1-\frac{1}{n})]$
since $F-y=\frac{AF}{d}$
$A^{\prime }=A[1+\frac{dh}{AF}(1-\frac{1}{n})]$
$=1[1+\frac{2\times 3}{1\times 10}(1-\frac{1}{1.5})]$
$=1.2\text{cm}=12\text{mm}$.