A biconvex lens made of material with refractive index ${\text{n}}_2$. The radii of curvatures of its left surface and right surface are ${\text{R}}_{1}and{R}_2$. The media on its left and right have refractive indices ${\text{n}}_1$ and ${\text{n}}_3$ respectively. The first and second focal lengths of the lens are respectively ${\text{f}}_{1}and{f}_2$ . The ratio, $\frac{f_1}{f_2}$ , of the two focal lengths is equal to$$\begin{gathered} \mathrm{\backslash begin}\mathrm{gathered}\left(a\right)n_1/n_3 \\ \left(b\right)\frac{n_1-1}{n_3-1} \\ \left(c\right)\frac{n_{1}1}{n_{3}1} \\ \left(d\right)\frac{n_2-n_3}{n_2-n1}\mathrm{\backslash end}\mathrm{gathered} \end{gathered}$$

We need to keep this formula in mind for solving this problem,

$\frac{n_3}{v}-\frac{n_1}{u}=\frac{n_3-n_2}{R_2}\frac{n_2-n_1}{R_1}\dots \text{(1)}$

where,

$n_1,n_2,n_3$

, are the respective refractive indices of the three media,

$v$

and

$u$

are the image distance and the object distance respectively, and

$R_1$

and

$R_2$

are the radius of curvatures of the two respective faces of the lens.

Let us first draw the diagram of the problem:

If we consider,

$u=\infty$

, and

$v=f_2$

, ( here

$f_2$

is the focal length of the second face of the lens), then using equation (1), we have,

$\frac{n_3}{f_2}-\frac{n_2}{\infty }=\frac{n_3-n_1}{-R_2}\frac{n_2-n_1}{R_1}\dots \left(2\right)$

Here, this equation has a negative

$R_2$

, because,

$R_2$

is in the direction opposite to the propagation of light.

$\therefore f_2=\frac{n_3}{\frac{n_3-n_2}{-R_2}\frac{n_2-n_1}{R_1}}\dots \left(3\right)$

If we consider

$u=\infty$

and

$v=f_1$

,( here,

$f_1$

is the focal length of the first face of the lens), then using equation (1), we have,

$\frac{n_1}{f_1}-\frac{n_3}{\infty }=\frac{n_1-{n_2}_{}}{-R_1}\frac{n_2-n_3}{R_2}\dots \left(4\right)$

Here, this equation has a negative value of

$R_2$

, because

$R_2$

is in the direction opposite to the direction of propagation of light.

$\therefore f_1=\frac{n_1}{\frac{n_2-n_1}{R_1}\frac{n_3-n_2}{(-R_2)}}\dots \left(5\right)$

Now dividing eqn. (5) by eqn. (3), we get,

$\therefore \frac{f_1}{f_2}=\frac{n_1}{n_3}$

The required answer is option (a)

$\frac{n_1}{n_3}$

.