Refraction Physics Example 4

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Example 4

hard
Light enters a diamond (n=2.42n = 2.42) from air (n=1.00n = 1.00) at an angle of incidence of 45°45°. Find the angle of refraction. Then calculate the critical angle for light leaving diamond into air.

Solution

  1. 1
    Snell's law: n1sin⁡θ1=n2sin⁡θ2n_1 \sin\theta_1 = n_2 \sin\theta_2. 1.00×sin⁡45°=2.42×sin⁡θ21.00 \times \sin 45° = 2.42 \times \sin\theta_2.
  2. 2
    sin⁡θ2=0.70712.42=0.2922\sin\theta_2 = \frac{0.7071}{2.42} = 0.2922. θ2=sin⁡−1(0.2922)≈17.0°\theta_2 = \sin^{-1}(0.2922) \approx 17.0°.
  3. 3
    Critical angle (diamond → air): sin⁡θc=nairndiamond=1.002.42=0.4132\sin\theta_c = \frac{n_{\text{air}}}{n_{\text{diamond}}} = \frac{1.00}{2.42} = 0.4132. θc=sin⁡−1(0.4132)≈24.4°\theta_c = \sin^{-1}(0.4132) \approx 24.4°.

Answer

θ2≈17.0°;θc≈24.4°\theta_2 \approx 17.0°; \quad \theta_c \approx 24.4°
Diamond's high refractive index causes strong bending of light and a small critical angle. This small critical angle means most light hitting the back facets undergoes total internal reflection — this is what gives diamonds their brilliance.

About Refraction

The change in direction of a wave as it passes from one medium into another where it travels at a different speed.

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More Refraction Examples

Example 1 medium

Light passes from air ([formula]) into glass ([formula]) at an angle of incidence of [formula]. What

Example 2 hard

What is the critical angle for light going from glass ([formula]) to air ([formula])?

Example 3 medium

Light travels from water ([formula]) into air ([formula]) at an angle of incidence of [formula]. Wha

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