Question

A ray of green light enters a liquid from air, as shown in Fig. 4.21. The angle 1 is ${45}^{\circ }$ and angle 2 is ${30}^{\circ }$.
(a) Find the refractive index of the liquid.
(b) Show in the diagram the path of the ray after it strikes the mirror and re-enters in the air. Mark in the diagram the angles wherever necessary.
(c) Redraw the diagram if the plane mirror becomes normal to the refracted ray inside the liquid. State the principle used.

Answer 1

(a)
Refractive index of liquid,
$\mu =\frac{\sin i}{\sin r}$
$\mu =\frac{\sin \left({45}^{\circ }\right)}{\sin \left({30}^{\circ }\right)}$
$\mu =\frac{1/\sqrt{2}}{1/2}=\sqrt{2}$

(b)
Use laws of reflection on the plane mirror and basic geometry, we find that the incident angle when light is leaving water is ${30}^{\circ }$.
After this, using principle of reversibility, we conclude that the angle of refraction would be ${45}^{\circ }$.

(c)
The light is incident normally on the plane mirror, so it would reflect back on the same line. Again, using reversibility of light, it would retrace its path.