Question

A beam of width t is incident at $45\circ$ on air-water boundary. The width of the beam in water is ____

Answer 1

Given that,

Angle $\theta ={45}^0$

According to figure

In $\Delta ABC$

$AC=\sqrt{2}t.....\left(I\right)$

And in $\Delta ACD$

$CD=AC\cos r$

$d=\sqrt{2}t\cos r.....\left(II\right)$

And we have,

$\frac{\sin i}{\sin r}=\mu$

$\sin r=\frac{1}{\sqrt{2}\mu }$

$\cos r=\frac{\sqrt{2{\mu }^2-1}}{\sqrt{2}\mu }.....\left(III\right)$

Now from equation (II) and (III)

$d=\frac{t\sqrt{2{\mu }^2-1}}{\mu }$

Hence, the width of the beam is $\frac{t\sqrt{2{\mu }^2-1}}{\mu }$