Question

The lens-makers’ equation applies to a lens immersed in a liquid if n in the equation is replaced by n2/n1. Here n2 refers to the index of refraction of the lens material and n1 is that of the medium surrounding the lens. (a) A certain lens has focal length 79.0 cm in air and index of refraction 1.55. Find its focal length in water. (b) A certain mirror has focal length 79.0 cm in air. Find its focal length in water.

Answer 1

Answer:

a

The focal length of the lens in water is  $f_{water} = 262.68 cm$

b

The focal length of the mirror in water is  $f =79.0cm$

Explanation:

From the question we are told that

    The index of refraction of the lens material = $n_2$

    The index of refraction of the medium surrounding the lens = $n_1$

 

The lens maker's formula is mathematically represented as

            $\frac{1}{f} = (n -1) [\frac{1}{R_1} - \frac{1}{R_2} ]$

Where $f$ is the focal length

            $n$ is the index of refraction

            $R_1 and R_2$ are the radius of curvature of sphere 1 and 2 of the lens

From the question When the lens in air  we have  

           $\frac{1}{f_{air}} = (n-1) [\frac{1}{R_1} - \frac{1}{R_2} ]$

    When immersed in liquid the formula becomes

          $\frac{1}{f_{water}} = [\frac{n_2}{n_1} - 1 ] [\frac{1}{R_1} - \frac{1}{R_2} ]$

The ratio of the focal length of the the two medium is mathematically evaluated as

           $\frac{f_water}{f_{air}} = \frac{n_2 -1}{[\frac{n_2}{n_1} - 1] }$

From the question

      $f_{air }$= 79.0 cm

       $n_2 = 1.55$

and the refractive index of water(material surrounding the lens) has a constant value of  $n_1 = 1.33$

         $\frac{f_{water}}{79} = \frac{1.55- 1}{\frac{1.55}{1.44} -1}$

           $f_{water} = 262.68 cm$

b

The focal length of a mirror is dependent on the concept of reflection which is not affected by medium around it.