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A light ray traveling in diamond (n=2.42) strikes the diamond-air interface. Wha
A light ray traveling in diamond (n=2.42) strikes the diamond-air interface. What is the critical angle?
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Practice Questions
1 question
Q1
A light ray traveling in diamond (n=2.42) strikes the diamond-air interface. What is the critical angle?
24.4°
36.9°
42.5°
49.5°
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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°.
Questions & Step-by-step Solutions
1 item
Q
Q: A light ray traveling in diamond (n=2.42) strikes the diamond-air interface. What is the critical angle?
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°.
Steps: 7
Show Steps
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°.
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