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
If $$\dfrac{\Delta n}{n} < 0$$, then $$\dfrac{\Delta f}{f} > 0$$
B
For $$n=1.5, \Delta n=10^{-3}$$ and $$f=20cm$$, the value of $$|\Delta f|$$ will be $$0.02$$cm(round off to $$2$$nd decimal place)
C
$$\left|\dfrac{\Delta f}{f}\right| < \left|\dfrac{\Delta n}{n}\right|$$
D
The relation between $$\dfrac{\Delta f}{f}$$ and $$\dfrac{\Delta n}{n}$$ remains unchanged if both the convex surfaces are replaced by concave surfaces of the same radius of curvature
Solution
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#### Correct option is D. The relation between $$\dfrac{\Delta f}{f}$$ and $$\dfrac{\Delta n}{n}$$ remains unchanged if both the convex surfaces are replaced by concave surfaces of the same radius of curvature$$\dfrac{1}{f_0}=\dfrac{2(n-1)}{R}$$ $$(1)$$$$\dfrac{1}{f_1}=(n-1)\left(\dfrac{1}{R}-\dfrac{1}{\infty}\right)$$$$\dfrac{1}{f_2}=(n+\Delta n-1)\left(\dfrac{1}{R}-\dfrac{1}{\infty}\right)$$$$\dfrac{1}{f_0+\Delta f_0}=\dfrac{(n-1)}{R}+(n+\Delta n-1)\left(\dfrac{1}{R}\right)$$$$\dfrac{1}{f_0+\Delta f_0}=\dfrac{2n+\Delta n-2}{R}$$ .$$(2)$$$$(1)/(2)$$$$\Rightarrow \dfrac{f_0+\Delta f_0}{f_0}=\dfrac{\dfrac{2(n-1)}{R}}{\dfrac{2n+\Delta n-2}{R}}$$$$1+\dfrac{\Delta f_0}{f_0}=\dfrac{2(n-1)}{2n+\Delta n-2}$$$$\dfrac{\Delta f_0}{f_0}=\dfrac{-\Delta n}{(2n+\Delta n-2)}$$$$\dfrac{\Delta f_0}{20}=-\dfrac{10^{-3}}{3+10^{-3}-2}$$$$\Rightarrow \Delta f_0=-2\times 10^{-2}$$$$|\Delta f_0|=0.02$$cm.

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