SATHEE: Optics Question 452

Optics Question 452

Question: A prism of refractive index $μ$ and angle A is placed in the minimum deviation position. If the angle of minimum deviation is A, then the value of A in terms of $μ$ is

Options:

A) $\sin^{- 1} (\frac{μ}{2})$

B) $\sin^{- 1} \sqrt{\frac{μ - 1}{2}}$

C) $2 \cos^{- 1} (\frac{μ}{2})$

D) $\cos^{- 1} (\frac{μ}{2})$

Show Answer

Answer:

Correct Answer: C

Solution:

[c] Given $δ_{m} = A,$ as $μ = \frac{\sin (\frac{A + δ_{m}}{2})}{\sin (\frac{A}{2})}$

$\Rightarrow μ = \frac{\sin (\frac{A + A}{2})}{\sin (\frac{A}{2})} = 2 \cos \frac{A}{2}$

$\Rightarrow A = 2 \cos^{- 1} (\frac{μ}{2})$

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Optics Question 452

Question: A prism of refractive index μ and angle A is placed in the minimum deviation position. If the angle of minimum deviation is A, then the value of A in terms of μ is

Options:

Answer:

Solution:

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Question: A prism of refractive index $μ$ and angle A is placed in the minimum deviation position. If the angle of minimum deviation is A, then the value of A in terms of $μ$ is