Question

Suppose that there is a dispersive medium whose refractive index depends on the wavelength as given by \(n(λ) = n_0 +\frac{ a}{λ^2}-\frac{b}{λ^4}\) . The value of 𝜆 at which the group and phase velocities would be the same, is:

• A. \(\sqrt\frac{2a}{a}\)
• B. \(\sqrt\frac{b}{2a}\)
• C. \(\sqrt\frac{3b}{a}\)
• D. \(\sqrt\frac{b}{3a}\)

Answer 1

The Correct Option is A

The refractive index is given by:

\(n(\lambda) = n_0 + \frac{a}{\lambda^2} - \frac{b}{\lambda^4}\)

The phase velocity (\(v_p\)) is given by:

\(v_p = \frac{c}{n(\lambda)}\)

The group velocity (\(v_g\)) is given by:

\(v_g = \frac{d\omega}{dk}\)

Using the relation \(v_g = c / (n - \lambda \frac{dn}{d\lambda})\), we compute \(\frac{dn}{d\lambda}\):

\(\frac{dn}{d\lambda} = -\frac{2a}{\lambda^3} + \frac{4b}{\lambda^5}\)

Equating \(v_g = v_p\), we find:

\(n(\lambda) = n(\lambda) - \lambda \frac{dn}{d\lambda}\)

Simplifying gives:

\(-\lambda \frac{dn}{d\lambda} = 0\)

Substitute \(\frac{dn}{d\lambda}\):

\(-\lambda \left(-\frac{2a}{\lambda^3} + \frac{4b}{\lambda^5}\right) = 0\)

Simplify further:

\(\frac{2a}{\lambda^2} = \frac{2a}{\lambda^2}\)

Therefore, we find:

\(\lambda = \sqrt{\frac{2a}{a}}\)

The correct answer is:

Option 1: \(\sqrt{\frac{2a}{a}}\)