# 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.
in case of  liquid lens the considered refractive index should be refrective index of liquid with respect to surrounding, and surrounding in case of combination with lens will be lens , as light enters from lens to liquid. so it would be 1.333/1.5