SATHEE: Optics Question 982

Optics Question 982

Question: Light wave enters from medium 1 to medium 2. Its velocity in 2nd medium is double from 1st. For total internal reflection the angle of incidence must be greater than

[CPMT 2002]

Options:

A) $30^{o}$

B) $60^{o}$

C) $45^{o}$

D) $90^{o}$

Show Answer

Answer:

Correct Answer: A

Solution:

$\frac{μ_{2}}{μ_{1}} = \frac{v_{1}}{v_{2}} = \frac{1}{2} \Rightarrow \frac{μ_{1}}{μ_{2}} = 2 (μ_{1} > μ_{2})$

For total internal reflection $_{2} μ_{1} = \frac{1}{\sin C} \Rightarrow \frac{μ_{1}}{μ_{2}}$

$= \frac{1}{\sin C} \Rightarrow 2 = \frac{1}{\sin C} \Rightarrow C = 30^{o}$

So, for total (Internal reflection angle of incidence must be greater than 30°).

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Optics Question 982

Question: Light wave enters from medium 1 to medium 2. Its velocity in 2nd medium is double from 1st. For total internal reflection the angle of incidence must be greater than

Options:

Answer:

Solution:

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