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A light ray traveling in diamond (n=2.42) strikes the diamond-air interface. Wha

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Question: A light ray traveling in diamond (n=2.42) strikes the diamond-air interface. What is the critical angle?

Options:

  1. 24.4°
  2. 36.9°
  3. 42.5°
  4. 49.5°

Correct Answer: 24.4°

Solution: 

Using sin(θc) = n2/n1, where n1 = 2.42 (diamond) and n2 = 1.0 (air), we find sin(θc) = 1.0/2.42, leading to θc ≈ 24.4°.

A light ray traveling in diamond (n=2.42) strikes the diamond-air interface. Wha

Practice Questions

Q1
A light ray traveling in diamond (n=2.42) strikes the diamond-air interface. What is the critical angle?
  1. 24.4°
  2. 36.9°
  3. 42.5°
  4. 49.5°

Questions & Step-by-Step Solutions

A light ray traveling in diamond (n=2.42) strikes the diamond-air interface. What is the critical angle?
Correct Answer: 24.4°
  • Step 1: Identify the refractive index of diamond (n1) which is 2.42.
  • Step 2: Identify the refractive index of air (n2) which is approximately 1.0.
  • Step 3: Use the formula for critical angle: sin(θc) = n2/n1.
  • Step 4: Substitute the values into the formula: sin(θc) = 1.0 / 2.42.
  • Step 5: Calculate the value of sin(θc): sin(θc) ≈ 0.4132.
  • Step 6: Use the inverse sine function to find θc: θc = sin⁻¹(0.4132).
  • Step 7: Calculate θc to find that θc ≈ 24.4°.
  • Refraction and Critical Angle – The question tests the understanding of Snell's law and the concept of critical angle in the context of light traveling between two media with different refractive indices.
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