Ray Optics

Refractive index of the lens, $\mu_l=1.5$ Refractive index of air, $\mu_a=1$ Focal length of the lens in air, $f=20\text{ cm}$ Using the lens maker's formula: $\;\frac{1}{f}= \left(\frac{\mu_1}{\mu_a}-1\right) \left(\frac{1}{R_1}-\frac{1}{R_2}\right)$ Substituting the known values: $\;\frac{1}{20}=(1.5-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$ $\;\frac{1}{10}= \left(\frac{1}{R_1}-\frac{1}{R_2}\right)$ Now, when the lens is immersed in a liquid with refractive index $\mu_{\text{liquid}}$, it behaves as a concave lens: Power of the lens in the liquid, $P=-1\text{ D}$ Calculating focal length in the liquid: $f=\frac{100}{P}=\frac{100}{-1}=-100\text{ cm}$ Applying the lens maker's formula for the lens immersed in the liquid: $\;\frac{-1}{100}= \left(\frac{1.5-\mu_{\text{liquid}}}{\mu_{\text{liquid}}}\right) \left(\frac{1}{R_1}-\frac{1}{R_2}\right)$ Using the earlier result: $\;\frac{-1}{100}= \left(\frac{1.5}{\mu_{\text{liquid}}}-1\right) \times \frac{1}{10}$ Solving for $\mu_{\text{liquid}} :$ $\;\frac{1.5}{\mu_{\text{liquid}}}=1-\frac{1}{10}$ $\;\mu_{\text{liquid}}=\frac{1.5 \times 10}{9}$ $\;\mu_{\text{liquid}}=\frac{5}{3}$