Geometrical Optics

Concept:The power of a thin lens immersed in a liquid depends on the relative refractive index of the lens material and the surrounding medium.Explanation:The Lens Maker's formula for a thin lens in a medium of refractive index $n_L$ is $$P = \frac{1}{f} = \left(\frac{n_g}{n_L} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right).$$Given $n_g = 1.5$, $R_1 = +20\ \mathrm{cm}$, $R_2 = -20\ \mathrm{cm}$. Converting to meters, the formula simplifies to $$P(n_L) = 10\left(\frac{1.5}{n_L} - 1\right) = \frac{15}{n_L} - 10\ \mathrm{D}.$$At $n_L = 1$ (air), $P = +5\ \mathrm{D}$ so the graph passes through $(1, 5)$.At $n_L = 1.5$ (matching index), $P = 0\ \mathrm{D}$ so the graph crosses the $n_L$-axis at $1.5$.At $n_L = 2$, $P = -2.5\ \mathrm{D}$. For $n_L > 1.5$, $P$ is negative.As $n_L \to \infty$, $P \to -10\ \mathrm{D}$.The relation $P = \frac{15}{n_L} - 10$ is a rectangular hyperbola shifted vertically down by $10$ units.Among the given plots, only Option (A) shows a hyperbola passing through $(1,5)$ and $(1.5,0)$, approaching $-10$ as $n_L$ increases.Answer:Option (A) is the correct plot.