Question:

If $θ_p$ is the polarizing angle for a glass plate of refractive index μ and critical angle $θ_c$, then

Updated On: Apr 10, 2025

$θ_p=θ_c $
$ θ_pθ_c=1 $
$ tanθ_p . sinθ_c =1$
$ tanθ_p = sinθ_c $
$ tanθ_p . sinθ_c =μ $

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The Correct Option is C

Approach Solution - 1

The polarizing angle ($ \theta_p $) and the critical angle ($ \theta_c $) are related through the refractive index of the medium ($ \mu $). The polarizing angle is the angle at which light is perfectly polarized when it is incident on a transparent material.

To derive the relationship, we use Brewster's law for the polarizing angle and the Snell's law for the critical angle.

According to Brewster's law, the polarizing angle $ \theta_p $ is related to the refractive index $ \mu $ by the equation:

\[ \tan(\theta_p) = \mu \]

And according to Snell's law, the critical angle $ \theta_c $ is related to the refractive index $ \mu $ by the equation:

\[ \sin(\theta_c) = \frac{1}{\mu} \]

Now, multiplying $ \tan(\theta_p) $ and $ \sin(\theta_c) $, we get:

\[ \tan(\theta_p) \cdot \sin(\theta_c) = \mu \cdot \frac{1}{\mu} = 1 \]

This shows that:

\[ \tan(\theta_p) \cdot \sin(\theta_c) = 1 \]

Correct Answer:

Correct Answer: (C) $ \tan(\theta_p) \cdot \sin(\theta_c) = 1 $

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Approach Solution -2

We are given:

Refractive index of the glass = $ \mu $
$ \theta_p $ = polarizing angle (Brewster's angle)
$ \theta_c $ = critical angle

Step 1: Use Brewster’s Law: \[ \tan \theta_p = \mu \]

Step 2: Use the definition of critical angle: \[ \sin \theta_c = \frac{1}{\mu} \]

Step 3: Multiply the two expressions: \[ \tan \theta_p \cdot \sin \theta_c = \mu \cdot \frac{1}{\mu} = 1 \]

Final Answer: $ \boxed{\tan \theta_p \cdot \sin \theta_c = 1} $