A boy uses monochromatic green light to find the refractive index. When a ray of monochromatic green light enters in a liquid medium from air medium as shown in the figure given below, the angle 1 is 45° and angle 2 is 30°.

(a) Determine the refractive index of the liquid.

(b) Represent in the diagram showing the path of the ray after it strikes the mirror and re-enters air. Mark in the diagram wherever necessary.

(c) Draw the diagram again, if plane mirror becomes normal to the refracted ray inside the liquid. Name the principle used.

(a) Refractive index of the liquid is given by Snell's law and is shown as below,

_airμ_liquid = $\dfrac{\text{sin i}}{\text{sin r}}$

_airμ_liquid = $\dfrac{\text{sin 45}\degree}{\text{sin 30}\degree}$

$= \dfrac{\dfrac{1}{\sqrt{2}}}{\dfrac{1}{2}} \\[1em] = \dfrac{2}{\sqrt{2}} \\[1em] = \sqrt{2} \\[1em] = 1.414 \\[1em]$

Hence, the refractive index of the lens is 1.4

(b) Below labelled diagram shows the path of the ray after it strikes the mirror and re-enters in air:

(c) Below labelled diagram shows the path of the ray when it strikes the plane mirror normally inside the liquid:

The principle of reversibility is used.