As depicted in the image on the right, material B, characterized by a dielectric constant of 4, is applied as a coating on substrate A, which possesses a dielectric constant of 9. In this scenario

Question

As depicted in the image on the right, material B, ﻿characterized by a dielectric constant of 4, ﻿is applied as a coating on substrate A, ﻿which possesses a dielectric constant of 9. ﻿In this scenario, an electromagnetic wave with a photon energy of 2 ﻿eV is vertically incident from the external environment(air).(a)Determine the wavelength, frequency, and speed of the electromagnetic wave in air, in the B layer, and in the A substrate.   (b)Explain the reason for the decrease in the speed of light in materials with a non-zero dielectric constant.  (c)Solely considering the Fresnel equations, calculate the percentage of incident light (perpendicular to the surface of substrate A/B) ﻿that is reflected, disregarding reflection at the B–air interface (i.e.,﻿only consider reflection at the A–B interface).(d)In addition to the Fresnel equation, the interference phenomenon significantly impacts reflection.Specifically, constructive interference between light reflected from the air–B interface and light reflected from the A–B interface maximizes reflection, while destructive interference minimizes it.To minimize reflection through destructive interference, we aim to coat material B with an

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Let's address each part of your question using the given from the image text:

- **Material B (coating):** Dielectric constant $ \epsilon_B = 4 $, refractive index $ n_B = 2 $.
- **Material A (substrate):** Dielectric constant $ \epsilon_A = 9 $, refractive index $ n_A = 3 $.
- **Air:** $ n_{\text{air}} = 1 $.
- **Photon energy:** $ E = 2\,\text{eV} $.

---

## (a) Wavelength, Frequency, and Speed in Each Medium

### **Step 1: Find Frequency**

Photon energy:  
$$
E = h f
$$
Where  
$ E = 2\,\text{eV} = 2 \times 1.602 \times 10^{-19}\,\text{J} = 3.204 \times 10^{-19}\,\text{J} $  
$ h = 6.626 \times 10^{-34}\,\text{J}\cdot\text{s} $

$$
f = \frac{E}{h} = \frac{3.204 \times 10^{-19}}{6.626 \times 10^{-34}} \approx 4.83 \times 10^{14}\,\text{Hz}
$$

### **Step 2: Wavelength in Air**

Speed of light in air: $ c = 3.00 \times 10^8\,\text{m/s} $  
$$
\lambda_{\text{air}} = \frac{c}{f} = \frac{3.00 \times 10^8}{4.83 \times 10^{14}} \approx 6.21 \times 10^{-7}\,\text{m} = 621\,\text{nm}
$$

### **Step 3: Wavelength and Speed in B and A**

- **General:** $ v = \frac{c}{n} $, $ \lambda = \frac{v}{f} = \frac{c}{n f} = \frac{\lambda_{\text{air}}}{n} $

#### In Material B ($ n_B = 2 $):
- Speed: $ v_B = \frac{c}{2} = 1.50 \times 10^8\,\text{m/s} $
- Wavelength: $ \lambda_B = \frac{\lambda_{\text{air}}}{2} = 310.5\,\text{nm} $

#### In Material A ($ n_A = 3 $):
- Speed: $ v_A = \frac{c}{3} = 1.00 \times 10^8\,\text{m/s} $
- Wavelength: $ \lambda_A = \frac{\lambda_{\text{air}}}{3} = 207\,\text{nm} $

### **Summary Table**

| Medium      | Speed ($ v $)        | Wavelength ($ \lambda $) | Frequency ($ f $)          |
|-------------|------------------------|----------------------------|------------------------------|
| Air         | $ 3.00 \times 10^8 $ m/s | $ 621 $ nm               | $ 4.83 \times 10^{14} $ Hz |
| B ($ n=2 $) | $ 1.50 \times 10^8 $ m/s | $ 310.5 $ nm             | $ 4.83 \times 10^{14} $ Hz |
| A ($ n=3 $) | $ 1.00 \times 10^8 $ m/s | $ 207 $ nm               | $ 4.83 \times 10^{14} $ Hz |

---

## (b) Why Does the Speed of Light Decrease in a Medium?

- **Reason:**  
  The speed of light decreases in a material with nonzero dielectric constant because the electromagnetic wave interacts with the material's electric charges (mainly electrons). The wave induces oscillations in the charges, which in turn emit their own electromagnetic waves, causing a delay and thus a reduction in the net propagation speed.  
- **Mathematically:**  
  $$
  v = \frac{c}{n}
  $$
  where $ n = \sqrt{\epsilon_r} $ is the refractive index, and $ \epsilon_r $ is the relative dielectric constant.

---

## (c) Fresnel Reflection at the A–B Interface

- **Fresnel reflection coefficient for normal incidence:**
  $$
  r = \frac{n_A - n_B}{n_A + n_B}
  $$
  $$
  R = |r|^2
  $$
  Where $ n_A = 3, n_B = 2 $:

$$
  r = \frac{3-2}{3+2} = \frac{1}{5} = .2
  $$
  $$
  R = (.2)^2 = .04
  $$

- **Percentage reflected:**  
  $$
  R = 4\%
  $$

---

## (d) Optimal Thickness of B for Destructive Interference (Minimum Reflection)

To minimize reflection, the optical path difference between the two reflected rays should be half a wavelength in B (destructive interference at the air–B interface):

- **Condition:**  
  For minimum reflection, the optical thickness should be a quarter wavelength in B:
  $$
  2 n_B d = \frac{\lambda_B}{2} \implies d = \frac{\lambda_B}{4 n_B}
  $$
  But $\lambda_B = \frac{\lambda_{\text{air}}}{n_B}$:

$$
  d = \frac{1}{4} \cdot \frac{\lambda_{\text{air}}}{n_B}
  $$
  $$
  d = \frac{1}{4} \cdot \frac{621\,\text{nm}}{2} = \frac{1}{4} \cdot 310.5\,\text{nm} = 77.6\,\text{nm}
  $$

### **Recommended thickness for B:**

$$
\boxed{d = 78\,\text{nm}}
$$

---

### **Summary**

1. **Wavelength, frequency, speed:** See summary table above.
2. **Reason for speed decrease:** Interaction with material's charges delays the wave, lowering speed.
3. **Reflection at A–B:** $ 4\% $ of incident light is reflected at the interface.
4. **Optimal thickness B (for minimum reflection):** $ 78\,\text{nm} $.

---

**Image Alt Text:**  
A schematic showing layers: air on top, a thin B layer ($ n_B = 2 $), then a thick A substrate ($ n_A = 3 $). The question asks about reflection and optimal anti-reflection coating thickness.

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