First slide

The lens

Question

Two identical $(μ_{g} = 3 / 2)$ equiconvex lenses of focal length f are kept in contact. The space between the two lenses is filled with water $(μ_{ω} = 4 / 3)$ . The focal length of the combination is

Moderate

A

f
B

f/2
C

4f/3
D

3f/4

Solution

Let R be the radius of curvature of each surface. Then
$\begin{array}{l} \frac{1}{f} = (1.5 - 1) (\frac{1}{R} + \frac{1}{R}) \\ \Rightarrow f = R \end{array}$
For the water lens,
$\begin{array}{l} \frac{1}{f'} = (\frac{4}{3} - 1) (- \frac{1}{R} - \frac{1}{R}) = \frac{1}{3} (- \frac{2}{R}) \\ \Rightarrow \frac{1}{f'} = - \frac{2}{3 f} \end{array}$
Now, using $\frac{1}{F} = \frac{1}{f_{1}} + \frac{1}{f_{2}} + \frac{1}{f_{3}}$ , we have
$\begin{array}{l} \frac{1}{F} = \frac{1}{f} + \frac{1}{f} + \frac{1}{f^{'}} = \frac{2}{f} - \frac{2}{3 f} = \frac{4}{3 f} \\ \Rightarrow F = \frac{3 f}{4} \end{array}$

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