Question 369: Ray Optics

Question

A thin convex lens is made of two materials with refractive indices n
1
and n
2
, as shown in the 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
![](https://app-content.cdn.examgoal.net/fly/@width/image/1l6zwo9qb/ae7d0ba2-ce5e-4db9-b9d6-a19ea2b32b66/2007a930-1f6e-11ed-8d91-a90ec8c961d9/file-1l6zwo9qc.png?format=png)

Accepted Answer

Given, thin convex lens made of two materials.
Radius of curvature of left spherical surface $=$ radius of curvature of right spherical surface
When $n_1=n_2=n$, focal length $=f$
When $n_1=n$ and $n_2=n+\Delta n$, focal length $=f+\Delta f$
Here, $\Delta n \ll(n-1)$ and $1\left|\frac{\Delta n}{n}ight|
\end{aligned}
$
Hence, option (C) is incorrect.
Now, for $n=1.5, \Delta n=10^{-3}$ and $f=20 \mathrm{~cm}$. Using relation $\frac{\Delta f}{f(f+\Delta f)}=-\frac{\Delta n}{R}$ and Eq. (1), we get
$
\begin{aligned}
& \frac{\Delta f}{f(f+\Delta f)}=-\frac{\Delta n}{2(n-1) f} \Rightarrow \frac{\Delta f}{f+\Delta f}=-\frac{\Delta n}{2 n-2} \\
& \Rightarrow \Delta f(2 n-2)=-\Delta n(f+\Delta f) \\
& \Rightarrow \Delta f(2 n-2)=-\Delta n f-\Delta n \Delta f \\
& \Rightarrow \Delta f(2 n-2+\Delta n)=-\Delta n f \\
& \Rightarrow \Delta f=\frac{-\Delta n 	imes f}{2 n-2+\Delta n} \\
& \Rightarrow|\Delta f|=\frac{\Delta n+f}{2 n-2+\Delta n}
\end{aligned}
$
Substituting all values, we get
$
|\Delta f|=\frac{10^{-3} 	imes 20}{2 	imes 1.5-2+10^{-3}}=0.02 \mathrm{~cm}
$
Hence, option (B) is correct.
Since $\frac{\Delta f}{f}=-\frac{\Delta n}{2(n-1)}$, so if $\frac{\Delta n}{n}0$.
Hence, option $(\mathrm{A})$ is correct.
Since the relation between $\frac{\Delta f}{f}$ and $\frac{\Delta n}{n}$ is independent of radius of curvature so the relation remains unchanged only the sign of focal length changes.
Hence, option (D) is correct.

Answer

If ${{\Delta n} \over n} < 0$ then ${{\Delta f} \over f}$ > 0

Answer

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

Answer

$\left| {{{\Delta f} \over f}} \right| < \left| {{{\Delta n} \over n}} \right|$

Answer

The relation between ${{{\Delta f} \over f}}$ and ${{{\Delta n} \over n}}$ remains unchanged if both the convex surfaces are replaced by concave surfaces of the same radius of curvature.

Ray Optics

Choose the correct answer: