Physics - Geometrical Optics Question with Solution | TestHub

Given that the lens is made of two plano-convex lenses. Let it be $L_1$ or $L_2$ .

Let focal length of $L_1=f_1,$
An focal length of $L_2=f_2$
$\therefore$ Using lens maker's formula,
$\frac{1}{f_1}=\left(n_1-1\right)\left(\frac{1}{R}-\frac{1}{\infty }\right),$ and $\frac{1}{f_2}=\left(n_2-1\right)\left(\frac{1}{\infty }-\frac{1}{\left(-R\right)}\right)$
$\Rightarrow \frac{1}{f_1}=\frac{n_1-1}{R}$
$\Rightarrow \frac{1}{f_2}=\frac{n_2-1}{R}$
Now, for the first situation, given that,
$n_1=n_2=n$ and focal length of combination is $‘f’$
$\Rightarrow \frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2}=\frac{2\left(n-1\right)}{R}$ ....(1)
In the second situation, $n_1=n,$ $n_2=n+\Delta n$ and focal length= $f+\Delta f$
$\Rightarrow \frac{1}{f+\Delta f}=\frac{n-1}{R}+\frac{n+\Delta n-1}{R}$
$\Rightarrow \frac{1}{f+\Delta f}=\frac{2n+\Delta n-2}{R}$ .....(2)
Nos dividing equation (1)/(2) we get
$\frac{f+\Delta f}{f}=\frac{2\left(n-1\right)}{2n+\Delta n-2}\Rightarrow 1+\frac{\Delta f}{f}=\frac{\left(2n-2+\Delta n\right)-\Delta n}{2n+\Delta n-2}$
$\Rightarrow 1+\frac{\Delta f}{f}=1-\frac{\Delta n}{2n+\Delta n-2}\Rightarrow \frac{\Delta f}{f}=\frac{-\Delta n}{2n+\Delta n-2}$
Now, as $\prime \Delta n^{\prime }$ is very small $\Rightarrow \frac{\Delta f}{f}=\frac{-\Delta n}{2\left(n-1\right)}\dots ..\left(3\right)$
Now, option $\left(\mathrm{A}\right)$
$n=1.5,\Delta n={10}^{-3}\mathrm{a}\mathrm{n}\mathrm{d}f=20\mathrm{c}\mathrm{m}$
$\Rightarrow \left|\Delta f\right|=\frac{f\cdot \left(\Delta n\right)}{2(n-1)}=\frac{20\times {10}^{-3}}{2\times (1.5-1)}=0.02cm\Rightarrow \left(\mathrm{A}\right)$ is correct.
Option $\left(\mathrm{B}\right)$
The relation between $\frac{\Delta f}{f}$ or $\frac{\Delta n}{n}$ is dependent of $\mathrm{‘}\mathrm{R}\mathrm{’}$
$\Rightarrow$ Remain unchanged for convex or concave surfaces
$\Rightarrow \left(\mathrm{B}\right)$ is correct.
Option $\left(\mathrm{C}\right)$
As $n<2\Rightarrow 2\left(n-1\right)<n$
$\Rightarrow \frac{\Delta f}{f}=\frac{1}{2}\left|\frac{\Delta n}{n-1}\right|>\frac{\Delta n}{n}$
$\Rightarrow \frac{\Delta f}{f}>\frac{\Delta n}{n}\Rightarrow \left(\mathrm{C}\right)$ is incorrect.
Option $\left(\mathrm{D}\right)$
If $\frac{\Delta n}{n}<0,$ then $\frac{\Delta f}{f}>0,$ using equation $\left(3\right)$
$\Rightarrow \left(\mathrm{D}\right)$ is correct.