Question

A thin double convex glass lens with an index of 1.56 while surrounded by air has a 10 cm focal length. If it is placed under water, which has an index of 1.33, 100 cm beyond a small fish, what is the focal length of the lens in water?

Answer 1

Explanation:

Formula which holds true for a leans with radii $R_{1}$ and $R_{2}$ and index refraction n is given as follows.

          $\frac{1}{f} = (n - 1) [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$

Since, the lens is immersed in liquid with index of refraction $n_{1}$. Therefore, focal length obeys the following.  

            $\frac{1}{f_{1}} = \frac{n - n_{1}}{n_{1}} [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$  

             $\frac{1}{f(n - 1)} = [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$

and,       $\frac{n_{1}}{f(n - n_{1})} = \frac{1}{R_{1}} - \frac{1}{R_{2}}$

or,          $f_{1} = \frac{fn_{1}(n - 1)}{(n - n_{1})}$

              $f_{w} = \frac{10 \times 1.33 \times (1.56 - 1)}{(1.56 - 1.33)}$

                          = 32.4 cm

Using thin lens equation, we will find the focal length as follows.

             $\frac{1}{f} = \frac{1}{s_{o}} + \frac{1}{s_{i}}$

Hence, image distance can be calculated as follows.

       $\frac{1}{s_{i}} = \frac{1}{f} - \frac{1}{s_{o}} = \frac{s_{o} - f}{fs_{o}}$

              $s_{i} = \frac{fs_{o}}{s_{o} - f}$

             $s_{i} = \frac{32.4 \times 100}{100 - 32.4}$

                       = 47.9 cm

Therefore, we can conclude that the focal length of the lens in water is 47.9 cm.