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?

Answer 1

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.