A plane monochromatic electromagnetic wave travels from one medium (refractive index = 1.1) to another (refractive index = 2.2) with electric field oscillating within the plane of incidence. What will be the reflection and transmission coefficients for the system if incident wave makes an angle 60° with the normal to the interface?
The Correct Answer and Explanation is:
To calculate the reflection and transmission coefficients for the given system, we will use the Fresnel equations, which describe the behavior of light when it hits an interface between two media.
Given:
- Refractive index of medium 1 (incident medium), n1=1.1n_1 = 1.1n1=1.1
- Refractive index of medium 2 (transmitting medium), n2=2.2n_2 = 2.2n2=2.2
- Angle of incidence θ1=60∘\theta_1 = 60^\circθ1=60∘
- The electric field oscillates within the plane of incidence, which corresponds to the p-polarized (parallel) case.
Step 1: Calculate the angle of refraction
We can use Snell’s law to find the angle of refraction θ2\theta_2θ2:n1sin(θ1)=n2sin(θ2)n_1 \sin(\theta_1) = n_2 \sin(\theta_2)n1sin(θ1)=n2sin(θ2)
Substitute the given values:1.1sin(60∘)=2.2sin(θ2)1.1 \sin(60^\circ) = 2.2 \sin(\theta_2)1.1sin(60∘)=2.2sin(θ2)sin(θ2)=1.1sin(60∘)2.2=1.1×0.8662.2≈0.431\sin(\theta_2) = \frac{1.1 \sin(60^\circ)}{2.2} = \frac{1.1 \times 0.866}{2.2} \approx 0.431sin(θ2)=2.21.1sin(60∘)=2.21.1×0.866≈0.431θ2≈sin−1(0.431)≈25.7∘\theta_2 \approx \sin^{-1}(0.431) \approx 25.7^\circθ2≈sin−1(0.431)≈25.7∘
Step 2: Calculate the reflection and transmission coefficients
For p-polarization, the Fresnel equations for reflection RRR and transmission TTT are given by:R=∣n1cos(θ1)−n2cos(θ2)n1cos(θ1)+n2cos(θ2)∣2R = \left| \frac{n_1 \cos(\theta_1) – n_2 \cos(\theta_2)}{n_1 \cos(\theta_1) + n_2 \cos(\theta_2)} \right|^2R=n1cos(θ1)+n2cos(θ2)n1cos(θ1)−n2cos(θ2)2T=∣2n1cos(θ1)n1cos(θ1)+n2cos(θ2)∣2×n2n1T = \left| \frac{2 n_1 \cos(\theta_1)}{n_1 \cos(\theta_1) + n_2 \cos(\theta_2)} \right|^2 \times \frac{n_2}{n_1}T=n1cos(θ1)+n2cos(θ2)2n1cos(θ1)2×n1n2
Now, calculate RRR:R=∣1.1cos(60∘)−2.2cos(25.7∘)1.1cos(60∘)+2.2cos(25.7∘)∣2R = \left| \frac{1.1 \cos(60^\circ) – 2.2 \cos(25.7^\circ)}{1.1 \cos(60^\circ) + 2.2 \cos(25.7^\circ)} \right|^2R=1.1cos(60∘)+2.2cos(25.7∘)1.1cos(60∘)−2.2cos(25.7∘)2cos(60∘)=0.5,cos(25.7∘)≈0.899\cos(60^\circ) = 0.5, \quad \cos(25.7^\circ) \approx 0.899cos(60∘)=0.5,cos(25.7∘)≈0.899
Substitute the values:R=∣1.1×0.5−2.2×0.8991.1×0.5+2.2×0.899∣2R = \left| \frac{1.1 \times 0.5 – 2.2 \times 0.899}{1.1 \times 0.5 + 2.2 \times 0.899} \right|^2R=1.1×0.5+2.2×0.8991.1×0.5−2.2×0.8992R=∣0.55−1.9780.55+1.978∣2=∣−1.4282.528∣2≈0.324R = \left| \frac{0.55 – 1.978}{0.55 + 1.978} \right|^2 = \left| \frac{-1.428}{2.528} \right|^2 \approx 0.324R=0.55+1.9780.55−1.9782=2.528−1.4282≈0.324
Now, calculate TTT:T=∣2×1.1×0.51.1×0.5+2.2×0.899∣2×2.21.1T = \left| \frac{2 \times 1.1 \times 0.5}{1.1 \times 0.5 + 2.2 \times 0.899} \right|^2 \times \frac{2.2}{1.1}T=1.1×0.5+2.2×0.8992×1.1×0.52×1.12.2T=∣1.12.528∣2×2=(0.435)2×2≈0.379T = \left| \frac{1.1}{2.528} \right|^2 \times 2 = \left( 0.435 \right)^2 \times 2 \approx 0.379T=2.5281.12×2=(0.435)2×2≈0.379
Final Results:
- Reflection coefficient R≈0.324R \approx 0.324R≈0.324
- Transmission coefficient T≈0.379T \approx 0.379T≈0.379
Explanation:
The reflection and transmission coefficients give us the fraction of the incident wave that is reflected and transmitted, respectively. The reflection coefficient RRR is the ratio of the reflected intensity to the incident intensity, and similarly, TTT is the ratio of the transmitted intensity to the incident intensity.
These values depend on the refractive indices of the media and the angle of incidence. Since the refractive index of the second medium is higher, the transmitted wave bends towards the normal (as expected from Snell’s law), and part of the wave is reflected back into the first medium.
