The power of a thin lens is +$\displaystyle 5$ D. When it is immersed in a liquid, it behaves like a concave lens of focal length $\displaystyle 100$ cm. Calculate the refractive index of the liquid. Given refractive index of glass = 1.5.

Finding the refractive index of liquid \[\begin{aligned} & \mathrm{P}=+5 \mathrm{D}, \quad \mathrm{f}_{\mathrm{e}}=-100 \mathrm{~cm} \quad, \quad \mu_{\mathrm{g}}=1.5, \quad \mu_{\mathrm{l}}=? \\ & \mathrm{f}_{a}=\frac{1}{\mathrm{P}}=\frac{1}{5}=0.2 \mathrm{~m}=20 \mathrm{~cm} \\ & \frac{1}{\mathrm{f}_{a}}=\left(\mu_{\mathrm{g} a}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right) \\ & \frac{1}{20}=(1.5-1)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right) \\ & \frac{1}{\mathrm{f}_{\mathrm{l}}}=\left(\frac{\mu_{\mathrm{g}}}{\mu_{\mathrm{l}}}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right) \\ & \frac{1}{(-100)}=\left(\frac{1.5}{\mu_{\mathrm{l}}}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right) \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \end{aligned} \]