Question

Light wave are incident from a medium of refractive index 2 making an angle $\theta$ with normal on to a medium of refractive index $2\sqrt{3}$. What should be the value of $\theta$ for which reflected wave and refracted wave will be perpendicular to each other.

• A. 60$^\circ$
• B. 30$^\circ$
• C. 53$^\circ$
• D. 45$^\circ$

Hint:

The condition "reflected and refracted rays are perpendicular" is the fundamental condition for Brewster's law. Even if the topic of polarization is sometimes omitted, the wave optics geometric condition ($\tan \theta = \mu_{rel}$) remains standard in ray optics. Always remember $\tan \theta = \mu_2 / \mu_1$.

Answer 1

The Correct Option is A

Step 1: Use relation between angles at reflection–refraction condition

When a light ray is incident at the interface of two media and the reflected ray is perpendicular to the refracted ray, the following relation holds:

i + r = 90°

where i is the angle of incidence and r is the angle of refraction.

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Step 2: Apply Snell’s law

According to Snell’s law:

μ₁ sin i = μ₂ sin r

Using r = 90° − i:

μ₁ sin i = μ₂ cos i

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Step 3: Rearrange the equation

Divide both sides by cos i:

μ₁ tan i = μ₂

So,

tan i = μ₂ / μ₁

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Step 4: Substitute given values

μ₁ = 2
μ₂ = 2√3

tan i = (2√3) / 2 = √3

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Step 5: Determine the angle of incidence

Since:

tan 60° = √3

Therefore,

i = 60°

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Final Answer:

The angle of incidence is
60°