Physics - Geometrical Optics Question with Solution | TestHub

Let the refractive index of the denser medium be ${\mu }_1$ and that of the rarer medium be ${\mu }_2$.

The critical angle $C$ is given by $\sin C=\frac{{\mu }_2}{{\mu }_1}$.

According to the law of reflection, the angle of incidence $i$ is equal to the angle of reflection $r$, so $i=r$.

It is given that the reflected and refracted rays are perpendicular to each other. From the geometry of the interface, the sum of the angle of reflection $r$ and the angle of refraction $r^{\prime }$ is $90^{\circ }$.

$r+r^{\prime }=90^{\circ }$

Since $r=i$, we have $i+r^{\prime }=90^{\circ }$, which gives $r^{\prime }=90^{\circ }-i$.

Applying Snell's law at the interface:

${\mu }_1\sin i={\mu }_2\sin r^{\prime }$

Substituting $r^{\prime }=90^{\circ }-i$:

${\mu }_1\sin i={\mu }_2\sin (90^{\circ }-i)$

${\mu }_1\sin i={\mu }_2\cos i$

$\frac{\sin i}{\cos i}=\frac{{\mu }_2}{{\mu }_1}$

$\tan i=\sin C$

$i={\tan }^{-1}(\sin C)$

Answer: ${\tan }^{-1}(\sin C)$