Question

A convex lens has 20 cm focal length in air. What will be its focal length when the lens is immersed in a liquid of refractive indec 1.63. Given

Answer 1

See the question is correct are rong if it is correct apply n21=n1/n2

Answer 2

Let us consider the radius of curvature of the convex lens are denoted as $R_{1}\ and\ R_{2}\ respectively$ .

Let the focal length of the convex lens in air and liquid are denoted as $f_{a}\ and\ f_{l}\ respectively$

The refractive index of the liquid $\mu_{l} =1.63$

The lens was initially in air.

As per lens maker's formula,

 $\frac{1}{ f_{a}}=[\mu_{g}- 1][ \frac{1}{R_{1}}- \frac{1}{ R_{2}}]$

Let it be the equation 1. 

Now the lens is put inside the liquid.

The refractive index of the lens in air $[\mu_{g}]=1.5$

The refractive index of the lens in liquid is = $\mu_{g}^l$

                                                                      = $\frac{\mu_{g}} { \mu_{l}}$

                                                                      = $\frac{1.5}{1.63}$

                                                                      = 0.92

Putting lens maker's formula again, we get-

$\frac{1}{ f_{l}}=[ \mu_{g}^l-1][ \frac{1}{ R_{1}}- \frac{1}{R_{2}}]$

Let it be the equation 2

Dividing 1 by 2, we get-

                                   $\frac{f_{l}} { f_{a}}= \frac{\mu_{g}-1} { \mu_{g}^l-1}$

                                   $f_{l} =f_{a} *\frac{(1.5-1)}{(0.92-1)}$

                                           $=20*(-0.54378)\ cm$

                                           $=\ -10.87\ cm$          [ans]

The negative sign indicates that the lens will behave just like a concave lens now.