Question

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

Answer 1

The focal length of the glass lens in air is given by

$\frac{1}{f_a}=({}_a{n}_g-1)(\frac{1}{R_1}-\frac{1}{R_2})$

Here, $f_a=20cm$ and ${}_a{n}_g=\mathrm{1.5.}$

$\therefore \frac{1}{20cm}=(1.5-1)(\frac{1}{R_1}-\frac{1}{R_2})$

(i) The focal length of the glass lens in liquid is given by

$\frac{1}{f_l}=({}_w{n}_g-1)(\frac{1}{R_1}-\frac{1}{R_2})$

where ${}_l{n}_g$ is refractive index of glass with respect to the liquid.

Now, ${}_l{n}_g=\frac{{}_a{n}_g}{{}_a{n}_l}=\frac{1.5}{1.75}=\frac{6}{7}$

$\therefore \frac{1}{f_l}=(\frac{6}{7}-1)(\frac{1}{R_1}-\frac{1}{R_2})$ (ii)

Dividing Eq. $\left(i\right)$ by eq. $\left(ii\right)$, we get

$\frac{f_l}{20cm}=\frac{0.5}{-\left(1/7\right)}=-3.5$

$f_l=-3.5\times 20cm=-70cm$.

Or

Since, the focal length in the liquid comes out to be negative, the natural of the immersed lens is concave.