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(D) When $n_1 = n_2 = n$,the lens acts as a single convex lens with focal length $f$ given by the Lens Maker's Formula:$\frac{1}{f} = (n - 1) \left( \

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$1, 4$

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(D) When $n_1 = n_2 = n$,the lens acts as a single convex lens with focal length $f$ given by the Lens Maker's Formula:$\frac{1}{f} = (n - 1) \left( \frac{1}{R} - \frac{1}{-R} ight) = (n - 1) \frac{2}{R} \implies f = \frac{R}{2(n - 1)}$.In the second case,the lens is composed of two plano-convex lenses in contact. The focal lengths are:$\frac{1}{f_1} = \frac{n - 1}{R}$ and $\frac{1}{f_2} = \frac{(n + \Delta n) - 1}{R}$.The equivalent focal length $f' = f + \Delta f$ is given by:$\frac{1}{f + \Delta f} = \frac{1}{f_1} + \frac{1}{f_2} = \frac{n - 1 + n + \Delta n - 1}{R} = \frac{2n + \Delta n - 2}{R} = \frac{2(n - 1) + \Delta n}{R}$.Since $f = \frac{R}{2(n - 1)}$,we have $\frac{1}{f} = \frac{2(n - 1)}{R}$.$\frac{1}{f + \Delta f} = \frac{1}{f} + \frac{\Delta n}{R} = \frac{1}{f} \left( 1 + \frac{\Delta n}{R} \cdot f ight) = \frac{1}{f} \left( 1 + \frac{\Delta n}{R} \cdot \frac{R}{2(n - 1)} ight) = \frac{1}{f} \left( 1 + \frac{\Delta n}{2(n - 1)} ight)$.Using the binomial approximation $(1 + x)^{-1} \approx 1 - x$ for small $x$:$\frac{1}{f + \Delta f} \approx \frac{1}{f} \left( 1 - \frac{\Delta n}{2(n - 1)} ight) \implies f + \Delta f \approx f \left( 1 - \frac{\Delta n}{2(n - 1)} ight) = f - f \frac{\Delta n}{2(n - 1)}$.Thus,$\frac{\Delta f}{f} = -\frac{\Delta n}{2(n - 1)}$.$(1)$ The relation $\frac{\Delta f}{f} = -\frac{\Delta n}{2(n - 1)}$ is independent of $R$,so it holds for concave surfaces as well. Correct.$(2)$ Since $1  1$ for $n  \left| \frac{\Delta n}{n} ight|$. Incorrect.$(3)$ $|\Delta f| = f \cdot \frac{\Delta n}{2(n - 1)} = 20 \cdot \frac{10^{-3}}{2(1.5 - 1)} = 20 \cdot \frac{10^{-3}}{1} = 0.02 	ext{ cm}$. The statement says $0.04 	ext{ cm}$. Incorrect.$(4)$ If $\frac{\Delta n}{n}  0$. Correct.

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