For the refractive index $n = n_r(\omega) + i n_{im}(\omega)$ of a material, which of the following statements are correct?
Options:
(A) $n_r$ can be obtained from $n_{im}$ and vice versa
(B) $n_{im}$ could be zero
(C) $n$ is an analytic function in the upper half of the complex $\omega$ plane
(D) $n$ is independent of $\omega$ for some materials
Official Answer Key:
A,C
Step-by-Step Solution & Explanation
### **Official Solution & Step-by-Step Derivation (GATE PH 2022 Q.24)**
**1. Kramers-Kronig Relations & Optical Media**:
- The complex refractive index is $n(\omega) = n_r(\omega) + i n_{\text{im}}(\omega)$.
- Causality implies that the complex response function $\chi(\omega) = n^2(\omega) - 1$ (and $n(\omega) - 1$) is analytic in the upper half of the complex $\omega$-plane.
**2. Step-by-Step Property Verification**:
- **(A) [TRUE]**: By Kramers-Kronig relations, the real and imaginary parts of the refractive index are Hilbert transform pairs:
$n_r(\omega) - 1 = \frac{2}{\pi} \mathcal{P} \int_0^\infty \frac{s n_{\text{im}}(s)}{s^2 - \omega^2}\, ds$
Hence $n_r$ can be directly computed from $n_{\text{im}}$ and vice-versa.
- **(B) [FALSE]**: At finite frequencies in physical lossy materials, $n_{\text{im}}$ can never vanish identically across all $\omega$.
- **(C) [TRUE]**: The principle of causality dictates that the response function must have no poles in the upper half of the complex plane $\text{Im}(\omega) > 0$.
- **(D) [FALSE]**: All physical materials exhibit dispersion (variation of $n$ with $\omega$).
**3. Official Verification**:
The verified official IIT Master Answer is **Options (A, C)**.
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