A biconvex lens of refractive index 1 . 5 has a focal length of...

Given, the refractive index of the material of the lens is ${\mu}_{1} = 1 . 5$, the refractive index of the medium is ${\mu}_{m} = 1 . 6$ and the focal length of the lens in air is ${f}_{a} = 20cm$.

In air:

$\dfrac{1}{{f}_{a}} = \left({\mu}_{1} - 1\right)\left[\dfrac{2}{R}\right]$

In liquid:

$\dfrac{1}{{f}_{m}} = \left(\dfrac{{\mu}_{1}}{{\mu}_{m}} - 1\right)\left[\dfrac{2}{R}\right]$

The focal length as measured in air and that as measured inside the liquid medium are related by, 

$\dfrac{{f}_{m}}{{f}_{a}} = \dfrac{\left({\mu}_{1} - 1\right){\mu}_{m}}{\left({\mu}_{1} - {\mu}_{m}\right)} . . . \left(1\right)$

Equation (1) implies that

$\dfrac{{f}_{m}}{20} = \dfrac{( 1 . 5 - 1 ) 1 . 6}{( 1 . 5 - 1 . 6 )} ⇒ {f}_{m} = - 160cm$