A circular beam of light having diameter 'd' falls on a plane...

Let the diameter of the incident beam be $d$ and the diameter of the refracted beam be $d'$.

Let the length of the illuminated portion on the liquid surface be $l$.

From the geometry of the incident beam, $d = l \cos i$.

From the geometry of the refracted beam, $d' = l \cos r$.

Therefore, $d' = d \dfrac{\cos r}{\cos i}$.

Given i = 45^{}^{\circ}, $\cos i = \dfrac{1}{\sqrt{2}}$ and $\sin i = \dfrac{1}{\sqrt{2}}$.

Using Snell's law, $1 \cdot \sin i = n \cdot \sin r$.

\sin r = \dfrac{\sin 45^{}^{\circ}}{n} = \dfrac{1}{n\sqrt{2}}.

Using the identity $\cos r = \sqrt{1 - \sin^2 r}$, we get:

$\cos r = \sqrt{1 - \dfrac{1}{2n^2}} = \dfrac{\sqrt{2n^2 - 1}}{n\sqrt{2}}$.

Substituting the values of $\cos r$ and $\cos i$ in the expression for $d'$:

$d' = d \dfrac{\dfrac{\sqrt{2n^2 - 1}}{n\sqrt{2}}}{\dfrac{1}{\sqrt{2}}} = d \dfrac{\sqrt{2n^2 - 1}}{n}$.

Answer: $\dfrac{\sqrt{2n^2 - 1}}{n}\ d$