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Incident light (500 nm, 1 W, speed = c)

Here, c is the speed of light; λ wavelength, f frequency.

Complete the table below:

| Layer | n | C | λ (nm) | Frequency (Hz) |
|---------|-----|-----|--------|-----------------|
| Air | 1 | C₁ | 91 | f₁ |
| Layer 1 | 1.5 | | | |
| Layer 2 | 2.0 | C₂ | 22 | f₂ |
| Air | | | | |

What is the minimum θ required in order to create total internal reflection between Layer 1 and Layer 2? When θ is 0, what is the light power reflected by the front surface of Layer 1?

1 Answer

4 votes

Final answer:

The minimum angle for total internal reflection between Layer 1 and Layer 2 can be found using the critical angle formula derived from Snell's Law. When the angle of incidence, θ, is 0, the light power reflected by Layer 1 is computed by applying the Fresnel reflection coefficient.

Step-by-step explanation:

To determine the minimum angle, θ, required for total internal reflection to occur at the interface between Layer 1 and Layer 2, we must apply Snell's Law, which can be expressed as n1 sin(θ1) = n2 sin(θ2), where n is the index of refraction. However, for total internal reflection to happen, the incident angle must be greater than the critical angle. The critical angle is calculated using sin(θc) = n2 / n1, where n1 > n2 and θc is the critical angle in the medium with a higher index of refraction (n1 in this case).

When θ = 0, there is no angle of incidence and thus no refraction. The light power reflected by the front surface of Layer 1 can be determined by the Fresnel reflection coefficient for the perpendicular (normal) incidence, R = ((n1 - n0) / (n1 + n0))2, where n0 is the index of refraction of air (assumed to be 1) and n1 is the index of refraction of Layer 1. The reflected power would then be R times the incident power, which is 1 W in this problem.

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