SATHEE: Optics Question 892

Optics Question 892

Question: The angle of polarisation for any medium is $60^{\circ}$ The critical angle for this is

Options:

A) $\sin^{- 1} \frac{1}{\sqrt{3}}$

B) $\cos^{- 1} \sqrt{3}$

C) $\sin^{- 1} \sqrt{3}$

D) $\tan^{- 1} \sqrt{3}$

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Answer:

Correct Answer: A

Solution:

[a] $μ = \tan i_{p} = \tan 60^{o} = \sqrt{3}$

$\sin i_{C} = \frac{1}{μ} = \frac{1}{\sqrt{3}} \Rightarrow i_{C} = \sin^{- 1} \frac{1}{\sqrt{3}}$

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Optics Question 892

Question: The angle of polarisation for any medium is 60∘ The critical angle for this is

Options:

Answer:

Solution:

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Question: The angle of polarisation for any medium is $60^{\circ}$ The critical angle for this is