The dispersion for a medium of wavelength lam | vedclass

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Question

(A) According to Cauchy's equation,the refractive index $\mu$ is given by $\mu = a + \frac{b}{\lambda^2}$,where $a$ and $b$ are constants.Dispersion $

Vedclass · Accepted Answer

$D/8$

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(A) According to Cauchy's equation,the refractive index $\mu$ is given by $\mu = a + \frac{b}{\lambda^2}$,where $a$ and $b$ are constants.Dispersion $D$ is defined as the magnitude of the rate of change of refractive index with respect to wavelength: $D = |\frac{d\mu}{d\lambda}|$.Calculating the derivative: $\frac{d\mu}{d\lambda} = \frac{d}{d\lambda}(a + b\lambda^{-2}) = -2b\lambda^{-3}$.Thus,$D = | -\frac{2b}{\lambda^3} | = \frac{2b}{\lambda^3}$.This shows that $D \propto \frac{1}{\lambda^3}$.For a new wavelength $\lambda' = 2\lambda$,the new dispersion $D'$ is given by $\frac{D'}{D} = (\frac{\lambda}{\lambda'})^3 = (\frac{\lambda}{2\lambda})^3 = \frac{1}{8}$.Therefore,$D' = \frac{D}{8}$.

The dispersion for a medium of wavelength $\lambda$ is $D$. Then,the dispersion for the wavelength $2\lambda$ will be:

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