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A combination of lens is formed by keeping two identical equiconvex lens of focal length f kept in contact and the space between them filled with liquid of refractive index $\mu_2 =\large\frac{4}{3}$. Focal length of the combination is

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

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.