Answer to Question #138401 in Physics for victory

Question

A converging lens has a refractive index of 1.6. When the lens is used in air, its focal length is 20 cm. If it is immersed in water, its focal length is ?

Answer 1

The focal length depends on refractive index, and radii of outer surfaces of the lens:

$$\frac{1}{f}=\bigg(\frac{n_\text{lens}}{n_\text{medium}}-1\bigg)\bigg(\frac{1}{R_1}-\frac{1}{R_2}\bigg).$$

Rite this equation for air and water and divide the equations one by another:

$$\frac{1}{f_\text{air}}=\bigg(\frac{n_\text{lens}}{n_\text{air}}-1\bigg)\bigg(\frac{1}{R_1}-\frac{1}{R_2}\bigg),\\\space\\ \frac{1}{f_\text{water}}=\bigg(\frac{n_\text{lens}}{n_\text{water}}-1\bigg)\bigg(\frac{1}{R_1}-\frac{1}{R_2}\bigg).\\\space\\ \frac{f_\text{water}}{f_\text{air}}=\frac{(n_\text{lens}-n_\text{air})n_\text{water}}{(n_\text{lens}-n_\text{water})n_\text{air}},\\\space\\ f_\text{water}=f_\text{air}\frac{(n_\text{lens}-n_\text{air})n_\text{water}}{(n_\text{lens}-n_\text{water})n_\text{air}},\\\space\\ f_\text{water}=20\frac{(1.6-1)1.333}{(1.6-1.333)1}=59.9\text{ cm}.$$