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)9

VIPSolutions · Accepted Answer

# Problem Setup- **Incident Wave**: energy $E = 2 $ eV, from **air** (dielectric constant $ \epsilon = 1 $, $ n_{\text{air}} \approx 1 $)
- **Coating (B)**: Dielectric constant $ \epsilon_B = 4 $ ($ n_B = 2 $)
- **Substrate (A)**: Dielectric constant $ \epsilon_A = 9 $ ($ n_A = 3 $)

## Diagram Summary

```
Air (n ≈ 1)
   |
   v
-------------------
| Coating B (n=2) |
-------------------
| Substrate A (n=3)|
-------------------
```

# (a) Reflection and Transmission at Interfaces

Let’s analyze the **reflection and transmission** of an electromagnetic wave at **normal incidence** through these layers.

## 1. **Reflection at Air–B Interface**
The **reflection coefficient** at normal incidence is:
$$
r_{12} = \frac{n_1 - n_2}{n_1 + n_2}
$$
where:
- $ n_1 = n_{\text{air}} = 1 $
- $ n_2 = n_B = 2 $

So,
$$
r_{AB} = \frac{1 - 2}{1 + 2} = \frac{-1}{3} = -.333
$$

And the **transmission coefficient**:
$$
t_{12} = \frac{2 n_1}{n_1 + n_2}
$$
$$
t_{AB} = \frac{2 \times 1}{1 + 2} = \frac{2}{3} \approx .667
$$

## 2. **Reflection at B–A Interface**
Now, for the wave that passes through B and reaches the B–A interface:

$$
r_{BA} = \frac{n_B - n_A}{n_B + n_A} = \frac{2 - 3}{2 + 3} = \frac{-1}{5} = -.2
$$

$$
t_{BA} = \frac{2 n_B}{n_B + n_A} = \frac{2 \times 2}{2 + 3} = \frac{4}{5} = .8
$$

## 3. **Physical Interpretation**
- **At each interface, some light is reflected and some is transmitted.**
- **Multiple reflections** can occur inside layer B (if the thickness is non-negligible).
- If the thickness is **not specified**, typically only the **first interface reflection** is considered.

# Summary Table

| Interface      | Reflection ($ r $) | Transmission ($ t $) |
|:--------------:|:-------------------:|:----------------------:|
| Air–B          | $ -.333 $        | $ .667 $            |
| B–A            | $ -.2 $          | $ .8 $              |

# Next Steps

- **If the thickness of B is given** (or specified as a quarter-wavelength, etc.), you can analyze **interference effects**.
- For **energy reflectance** (fraction of intensity reflected), use $ R = |r|^2 $.
- For **photon energy**, the refractive index is not directly dependent on photon energy unless dispersion is considered (here, we use the given $ n $).

---

**Let me know if you need to consider the effect of thickness or want to calculate the total reflectance including multiple reflections!**

✨ VIPSolutions

Question:

Final Answer:

📢 Instructions:

Study Help, Fast Answers

✨ VIPSolutions

Question:

Final Answer:

📢 Instructions:

Related Answers

Study Help, Fast Answers