A ray of green light enters a liquid from air, as shown in Fig. The angle 1 is 45° and angle 2 is 30°. Find the refractive index of the liquid. Show in the diagram the path of the ray after it strikes the mirror and re-enters in air. Mark in the diagram the angles where ever necessary. Redraw the diagram if plane mirror becomes normal to the refracted ray inside the liquid. State the principal used.

Refractive index of the liquid = `a ^µ i = "sin i" /"sin r"` `a^ µ i= (sin45°)/(sin 30°)=(1//sqrt2)/(1//2)=sqrt2=1.414` Path is shown&ang;i = &ang;45°, sin 45° = 0.7&ang;r = 30° and sin 30° = 0.5 The path of the beam after it hits the mirror and re-enters the air is depicted in the labelled diagram below: The reversibility of the light path principle is the one that is applied.

A ray of green light enters a liquid from air, as shown in Fig. The angle 1 is 45° and angle 2 is 30°. a. Find the refractive index of the liquid. b. Show in the diagram the path of the ray after it - Physics

Question

A ray of green light enters a liquid from air, as shown in Fig. The angle 1 is 45° and angle 2 is 30°.

Find the refractive index of the liquid.
Show in the diagram the path of the ray after it strikes the mirror and re-enters in air. Mark in the diagram the angles where ever necessary.
Redraw the diagram if plane mirror becomes normal to the refracted ray inside the liquid. State the principal used.

Short Note

Solution

Refractive index of the liquid =
`a ^µ i = "sin i" /"sin r"`
`a^ µ i= (sin45°)/(sin 30°)=(1//sqrt2)/(1//2)=sqrt2=1.414`
Path is shown
∠i = ∠45°, sin 45° = 0.7
∠r = 30° and sin 30° = 0.5

The path of the beam after it hits the mirror and re-enters the air is depicted in the labelled diagram below:
The reversibility of the light path principle is the one that is applied.