Concept:An anti-reflective coating of thickness \( d = \lambda / (4 n_{\text{AR}}) \) produces destructive interference between reflections from the air-coating interface and the coating-silicon interface when its refractive index is the geometric mean of air and silicon.
Formula:$$n_{\text{AR}} = \sqrt{n_{\text{air}} \cdot n_{\text{Si}}} = \sqrt{1.0 \times 3.8} \approx 1.95 \quad (\text{e.g., Silicon Nitride, } \text{Si}_3\text{N}_4)$$
Solution:- For complete destructive cancellation of reflected waves, the reflection amplitudes at both interfaces must be equal: \( \frac{n_{\text{AR}} - n_{\text{air}}}{n_{\text{AR}} + n_{\text{air}}} = \frac{n_{\text{Si}} - n_{\text{AR}}}{n_{\text{Si}} + n_{\text{AR}}} \).
- Solving this yields \( n_{\text{AR}} = \sqrt{n_{\text{air}} \cdot n_{\text{Si}}} \).
- Using \( n_{\text{Si}} \approx 3.8 \) and \( n_{\text{air}} = 1.0 \), the ideal coating index is \( n_{\text{AR}} \approx 1.95 \) (commonly realized using \( \text{Si}_3\text{N}_4 \)).
Why other options are incorrect:- Option A: Adding refractive indices does not satisfy the boundary matching condition.
- Option B: \( n_{\text{Si}} / n_{\text{air}} = 3.8 \), which provides no intermediate index matching.
- Option D: \( n_{\text{Si}}^2 \approx 14.4 \), which would worsen reflection rather than reducing it.
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