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

Step-by-step explanation:

Formula which holds true for a leans with radii
`R_(1)` and
`R_(2)` and index refraction n is given as follows.

`(1)/(f) = (n - 1) [(1)/(R_(1)) - (1)/(R_(2))]`

Since, the lens is immersed in liquid with index of refraction
`n_(1)`. Therefore, focal length obeys the following.

`(1)/(f_(1)) = (n - n_(1))/(n_(1)) [(1)/(R_(1)) - (1)/(R_(2))]`

`(1)/(f(n - 1)) = [(1)/(R_(1)) - (1)/(R_(2))]`

and,
`(n_(1))/(f(n - n_(1))) = (1)/(R_(1)) - (1)/(R_(2))`

or,
`f_(1) = (fn_(1)(n - 1))/((n - n_(1)))`

`f_(w) = (10 * 1.33 * (1.56 - 1))/((1.56 - 1.33))`

= 32.4 cm

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

`(1)/(f) = (1)/(s_(o)) + (1)/(s_(i))`

Hence, image distance can be calculated as follows.

`(1)/(s_(i)) = (1)/(f) - (1)/(s_(o)) = (s_(o) - f)/(fs_(o))`

`s_(i) = (fs_(o))/(s_(o) - f)`

`s_(i) = (32.4 * 100)/(100 - 32.4)`

= 47.9 cm

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