Physics - Geometrical Optics Question with Solution | TestHub

Let the diameter of the incident beam be $d$ and the diameter of the refracted beam be $d^{\prime }$.

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, $d^{\prime }=l\cos r$.

Therefore, $d^{\prime }=d\frac{\cos r}{\cos i}$.

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

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

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

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

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

Substituting the values of $\cos r$ and $\cos i$ in the expression for $d^{\prime }$:

$d^{\prime }=d\frac{\frac{\sqrt{2n^2-1}}{n\sqrt{2}}}{\frac{1}{\sqrt{2}}}=d\frac{\sqrt{2n^2-1}}{n}$.

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