Question

Two identical glass $\left({\mu }_g=\frac{3}{2}\right)$ equiconvex lenses of focal length f are kept in contact. The space between the two lenses is filled with water (${\mu }_w=\frac{4}{3}$). The focal length of the combination is

Answer 1

Step 1: Given data.

Glass is considered to be a convex lens of the refractive index glass.

This suggests that the radius of curvature of all the lenses is the same.

The convex lens of glass formula is: $\frac{\mathbf{1}}{\mathbf{f}}\mathbf{=}\left({\mu }_g-1\right)\left(\frac{2}{R}\right)$

Where,
${\mu }_g=\frac{3}{2}$

${\mu }_g$= Refractive index of glass.

$R$= Radius of curvature.

$f$= focal length.

Replacing the values in the formula.

$$\begin{gathered} \frac{1}{f}=\left({\mu }_g-1\right)\left(\frac{2}{R}\right) \\ \frac{1}{f}=\left(\frac{3}{2}-1\right)\left(\frac{2}{R}\right) \\ \frac{1}{f}=\frac{1}{R} \\ f=R \end{gathered}$$

Step 2: Calculate for the concave lens.

Water in between the lens can be considered to be a concave lens of the refractive index of water.

The focal length of the concave lens is$\frac{\mathbf{1}}{{\mathbf{f}}^{\mathbf{\text{'}}}}\mathbf{=}\left({\mu }_w-1\right)\left(\frac{-2}{R}\right)$

Where,

${\mu }_w=\frac{4}{3}$

${\mu }_w$= Refractive index of water.

$R$= Radius of curvature.

Replace the value in the formula.

$\begin{array}{rcl}\frac{1}{f^{\text{'}}}& =& \left({\mu }_w-1\right)\left(\frac{-2}{R}\right)\\ \frac{1}{f^{\text{'}}}& =& \left(\frac{4}{3}-1\right)\left(\frac{-2}{R}\right)\\ \frac{1}{f^{\text{'}}}& =& \left(\frac{-2}{3R}\right)\\ & & \frac{-2}{3f}\end{array}$

Step 3: Calculate the equivalent focal length.

$\begin{array}{rcl}\frac{1}{f_{eq}}& =& \frac{1}{f}+\frac{1}{f}-\frac{2}{3f}\\ & \Rightarrow & \frac{1}{f_{eq}}=\frac{4}{3f}\\ & \Rightarrow & f_{eq}=\frac{3f}{4}\end{array}$

Therefore the focal length of the combination is $\frac{\mathbf{3}\mathbf{f}}{\mathbf{4}}$