Question

A thin convex lens which is made of glass (refractive index 1.5) has a focal length of 20 cm. It is now completely immersed in a transparent liquid having refractive index 1.75. Find the new focal length of the lens.

Answer 1

Given,

_an_g = 1.5, f_a = 20 cm, _an_l = 1.75, f_l = ?

Now,

\(\frac{1}{f_a}=(_an_g -1)\,\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\)

\(\frac{1}{20}=(1.5 -1)\,\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\,...(1)\)

When lens is immersed in liquid

\(\frac{1}{f_l}=(_ln_g)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\)

\(\frac{1}{f_l}=\left(\frac{_an_g}{_an_l}-1\right)\,\left(\frac{1}{R_1}-\frac{1}{R_2}\right),\) \(\left[\because\,_ln_g=\frac{_an_g}{_an_l}\right]\)

\(\frac{1}{f_l}=\left(\frac{1.5}{1.75}-1\right)\,\left(\frac{1}{R_1}-\frac{1}{R_2}\right)...(2)\)

Dividing equation (1) by (2), we get

\(\frac{f_l}{20}=\frac{0.5\times175}{-25}\)

\(f_l =\frac{-20\times0.5\times175}{25}=-70\, cm\)