$(a)\;f \\ (b)\;\frac{f}{2} \\ (c)\;\frac{4}{3} f \\ (d)\;\frac{3f}{4} $

In effect we have a combination of 3 lens , the liquid between the two double convex lens forming a double concave lens.

We need to find the focal length of this liquid lens, through it has the same radius of curvature R.

For double convex lens,

$\large\frac{1}{f} =(\mu -1) \bigg[ \large\frac{1}{R} +\frac{1}{R}\bigg]$

$\therefore \large\frac{1}{f} =(\frac{3}{2}-1) \bigg[ \large\frac{1}{R} +\frac{1}{R}\bigg]$

$\therefore f=R$

For the liquid lens :

$\large\frac{1}{f'} =(\mu _2 -1) \bigg[ -\large\frac{1}{R} -\frac{1}{R}\bigg]$

$\qquad= \large\frac{1}{3} \bigg(\large\frac{-2}{R}\bigg)$

$\large\frac{1}{f'}=\large\frac{-2}{3f}$

Now the focal length of combination

$\large\frac{1}{F} =\frac{1}{f} +\frac{1}{f} +\frac{1}{f}$

$\qquad= \large\frac{2}{f} -\frac{2}{3f}$

$\qquad= \large\frac{4}{3f}$

$\therefore F= \large\frac{3f}{4}$

Hence d is the correct answer.

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