Question

A thin convex lens is made of two materials with refractive indices $n_1$ and $n_2$, as shown in figure. The radius of curvature of the left and right spherical surfaces are equal. $f$ is the focal length of the lens when $n_1=n_2=n$. The focal length is $f+\Delta f$ when $n_1=n$ and $n_2=n+\Delta n$. Assuming $\Delta n<<(n-1)$ and $(1<n<2)$, the correct statement(s) is/are:

• A. $\left|\frac{\Delta f}{f}\right|<\left|\frac{\Delta n}{n}\right|$
• B. The relation between $\frac{\Delta f}{f}$ and $\frac{\Delta n}{n}$ remains unchanged if both the convex surfaces are replaced by concave surfaces of the same radius of curvature.
• C. If $\frac{\Delta n}{n}<0$ then $\frac{\Delta f}{f}>0$
• D. For $n=1.5,\Delta n={10}^{-3}$ and $f=20cm,$ the value of $|\Delta f|$ will be $0.02cm$ (round off to $2^{nd}$ decimal place).

Answer 1

Answer: (B), (C), (D)

For $n=1.5,\Delta n={10}^{-3}$ and $f=20cm,$ the value of $|\Delta f|$ will be $0.02cm$ (round off to $2^{nd}$ decimal place).


If both the lenses having the refrective indices $n$, then
the focal lenght of the system
$\frac{1}{f}=\frac{2(n-1)}{R}$ $.....\left(1\right)$
let the second lens having refrective indices $\Delta n$,
$\frac{1}{f_1}=(n-1)(\frac{1}{R}-\frac{1}{\infty })$
$\frac{1}{f_2}=(n+\Delta n-1)(\frac{1}{R}-\frac{1}{\infty })$
$\frac{1}{f+\Delta f}=\frac{(n-1)}{R}+(n+\Delta n-1)\left(\frac{1}{R}\right)$
$\Rightarrow \frac{1}{f+\Delta f}=\frac{2n+\Delta n-2}{R}$ $.....\left(2\right)$

On diving ($\frac{Eq.1}{Eq.2}),$

$\Rightarrow \frac{f+\Delta f}{f}=\frac{\frac{2(n-1)}{R}}{\frac{2n+\Delta n-2}{R}}$
$1+\frac{\Delta f}{f}=\frac{2(n-1)}{2n+\Delta n-2}$
$\frac{\Delta f}{f}=\frac{-\Delta n}{(2n+\Delta n-2)}$
$\frac{\Delta f}{f}=\frac{-\Delta n}{\left(2\right(n-1)+\Delta n)}$

Here, $(n-1)>>\Delta n$
So,$\frac{\Delta f}{f}=\frac{-\Delta n}{2(n-1)}$

For option A,
Even taking the case when $\frac{\Delta f}{f}$ is positive and {\Delta n} is negative. For some value of $n$, $2(n-1)$ is less than 1, so $\Delta n$ for that case will be less then $\frac{\Delta f}{f}.$
Hence option A is wromg.

For option B,
$\frac{\Delta f}{f}$ is independent with $R$, hence option B is correct.

For option C:
$\frac{\Delta n}{n}<0$
so here, $\Delta n$ is negative and $n<1$. So this function will always be a positive.
Hence, option C is correct.

For option D:
$\frac{\Delta f}{20}=-\frac{{10}^{-3}}{3+{10}^{-3}-2}$
$\Rightarrow \Delta f=-2\times {10}^{-2}$
$|\Delta f|=0.02cm.$
Hence option D is correct.