Question

Hey!!!

Solve the question in attachment. ​

Answer 1

▶ Solution :

Given :

▪ Angle of prism = 150°

▪ Angle of incident = 60°

▪ Refractive index of prism = √3

To Find :

▪ Net deviation of ray

Concept :

✏ This question is completely based on concept of TIR.

✏ We can't calculate this question directly by net deviation formula.

✏ Please, see the attachment for better understanding.

Calculation :

▪ At A point :

$\rightarrow\sf\:(1)\sin60\degree=\sqrt{3}\sin{r_1}\\ \\ \rightarrow\sf\:\sin{r_1}=\dfrac{1}{2}\\ \\ \rightarrow\bf\:r_1=30\degree$

▪ At C point :

$\rightarrow\sf\:\sqrt{3}\sin30\degree=(1)\sin{e}\\ \\ \rightarrow\sf\:\sin{e}=\dfrac{\sqrt{3}}{2}\\ \\ \rightarrow\bf\:e=60\degree$

⏭ Net Deviation :

✴ At A point = 30° CW

✴ At B point = 90° ACW

✴ At C point = 30° CW

⏭ Net deviation = 90°- (60°) = 30°

_________________________________

✏ Net deviation of ray light = 30° (Anti-Clockwise direction)

✒ Nice Question !!

Answer 2

Answer:

Hello mate....

Explanation:

Reflected ray is completely polarised. Hence using Brewster’s law, tan i = η Where η = refractive index & i = 60° ∴ η = tan 60° = √3 also η = (c / v) v = (c / η) = {(3 × 108) / √3} = √3 × 108 m/sR

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