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:

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:

cdquestions Admin · Accepted Answer

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: $\sqrt{\frac{2a}{a}}$

Answer

$\sqrt\frac{2a}{a}$

Answer

$\sqrt\frac{b}{2a}$

Answer

$\sqrt\frac{3b}{a}$

Answer

$\sqrt\frac{b}{3a}$

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:

The Correct Option is A

Solution and Explanation

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