Question:medium

The critical angle of a medium for a specific wavelength, if the medium has relative permittivity 3 and relative permeability \( \frac{4}{3} \) for this wavelength, will be:

Show Hint

The refractive index \( n \) is fundamentally linked to the speed of light in the medium, which is determined by its electromagnetic properties (\( \epsilon \) and \( \mu \)). For most transparent materials, \( \mu_r \approx 1 \), but always check both values if provided!
Updated On: Apr 16, 2026
  • 15°
  • 30°
  • 45°
  • 60°
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The Correct Option is B

Solution and Explanation

Step 1: Understanding the Concept:
The critical angle ($\theta_c$) is the angle of incidence in an optically denser medium for which the angle of refraction in the less dense medium (usually air or vacuum) is $90^\circ$. The refractive index of a medium is related to its relative permittivity ($\epsilon_r$) and relative permeability ($\mu_r$).
Step 2: Key Formula or Approach:
The refractive index $n$ of a medium is given by: \[ n = \sqrt{\epsilon_r \mu_r} \] The critical angle $\theta_c$ is related to the refractive index $n$ by: \[ \sin(\theta_c) = \frac{1}{n} \] Step 3: Detailed Explanation:
Given values:
Relative permittivity, $\epsilon_r = 3$
Relative permeability, $\mu_r = \frac{4}{3}$
First, calculate the refractive index $n$ of the medium: \[ n = \sqrt{3 \times \frac{4}{3}} \] \[ n = \sqrt{4} = 2 \] Now, use the critical angle formula: \[ \sin(\theta_c) = \frac{1}{n} = \frac{1}{2} \] Taking the inverse sine: \[ \theta_c = \arcsin\left(\frac{1}{2}\right) = 30^\circ \] Step 4: Final Answer:
The critical angle of the medium is $30^\circ$. The correct option is (B).
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