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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?

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$$\dfrac{1}{f_0}=\dfrac{2(n-1)}{R}$$ $$(1)$$

$$\dfrac{1}{f_1}=(n-1)\left(\dfrac{1}{R}-\dfrac{1}{\infty}\right)$$

$$\dfrac{1}{f_2}=(n+\Delta n-1)\left(\dfrac{1}{R}-\dfrac{1}{\infty}\right)$$

$$\dfrac{1}{f_0+\Delta f_0}=\dfrac{(n-1)}{R}+(n+\Delta n-1)\left(\dfrac{1}{R}\right)$$

$$\dfrac{1}{f_0+\Delta f_0}=\dfrac{2n+\Delta n-2}{R}$$ .$$(2)$$

$$(1)/(2)$$

$$\Rightarrow \dfrac{f_0+\Delta f_0}{f_0}=\dfrac{\dfrac{2(n-1)}{R}}{\dfrac{2n+\Delta n-2}{R}}$$

$$1+\dfrac{\Delta f_0}{f_0}=\dfrac{2(n-1)}{2n+\Delta n-2}$$

$$\dfrac{\Delta f_0}{f_0}=\dfrac{-\Delta n}{(2n+\Delta n-2)}$$

$$\dfrac{\Delta f_0}{20}=-\dfrac{10^{-3}}{3+10^{-3}-2}$$

$$\Rightarrow \Delta f_0=-2\times 10^{-2}$$

$$|\Delta f_0|=0.02$$cm.

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