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.
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.
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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 $\tan \theta = \mu_2 / \mu_1$.
Updated On: Feb 2, 2026
- 60$^\circ$
- 30$^\circ$
- 53$^\circ$
- 45$^\circ$
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Question:
A light ray travels from one medium to another. Part of the light is reflected and part is refracted. The angle between the reflected ray and the refracted ray is 90$^\circ$. We need to find the angle of incidence $\theta$. This is a classic Brewster's Law scenario.
Step 2: Key Formula or Approach:
When the reflected and refracted rays are perpendicular to each other, the angle of incidence is called the polarizing angle or Brewster's angle.
From geometry, $i + r + 90^\circ = 180^\circ \Rightarrow r = 90^\circ - i$.
Using Snell's Law: $\mu_{1} \sin i = \mu_{2} \sin r$.
Substituting $r$: $\mu_{1} \sin \theta = \mu_{2} \sin(90^\circ - \theta) = \mu_{2} \cos\theta$.
This gives Brewster's formula: $\tan\theta = \frac{\mu_{2}}{\mu_{1}}$.
Step 3: Detailed Explanation:
Given values:
Refractive index of incident medium, $\mu_{1} = 2$
Refractive index of second medium, $\mu_{2} = 2\sqrt{3}$
Angle of incidence, $i = \theta$
Using the derived relation:
$\tan\theta = \frac{\mu_{2}}{\mu_{1}}$
$\tan\theta = \frac{2\sqrt{3}}{2}$
$\tan\theta = \sqrt{3}$
Since $\tan 60^\circ = \sqrt{3}$, we get:
$\theta = 60^\circ$
Step 4: Final Answer:
The value of $\theta$ is 60$^\circ$.
A light ray travels from one medium to another. Part of the light is reflected and part is refracted. The angle between the reflected ray and the refracted ray is 90$^\circ$. We need to find the angle of incidence $\theta$. This is a classic Brewster's Law scenario.
Step 2: Key Formula or Approach:
When the reflected and refracted rays are perpendicular to each other, the angle of incidence is called the polarizing angle or Brewster's angle.
From geometry, $i + r + 90^\circ = 180^\circ \Rightarrow r = 90^\circ - i$.
Using Snell's Law: $\mu_{1} \sin i = \mu_{2} \sin r$.
Substituting $r$: $\mu_{1} \sin \theta = \mu_{2} \sin(90^\circ - \theta) = \mu_{2} \cos\theta$.
This gives Brewster's formula: $\tan\theta = \frac{\mu_{2}}{\mu_{1}}$.
Step 3: Detailed Explanation:
Given values:
Refractive index of incident medium, $\mu_{1} = 2$
Refractive index of second medium, $\mu_{2} = 2\sqrt{3}$
Angle of incidence, $i = \theta$
Using the derived relation:
$\tan\theta = \frac{\mu_{2}}{\mu_{1}}$
$\tan\theta = \frac{2\sqrt{3}}{2}$
$\tan\theta = \sqrt{3}$
Since $\tan 60^\circ = \sqrt{3}$, we get:
$\theta = 60^\circ$
Step 4: Final Answer:
The value of $\theta$ is 60$^\circ$.
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