A parallel beam of light travels in a medium | vedclass

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Question

(B) The refractive index is given by $\mu = \mu_0 + \mu_2 I$.The speed of light $v$ in a medium is related to the refractive index by $v = \frac{c}{\m

Vedclass · Accepted Answer

Minimum on the axis of the beam

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(B) The refractive index is given by $\mu = \mu_0 + \mu_2 I$.The speed of light $v$ in a medium is related to the refractive index by $v = \frac{c}{\mu}$,where $c$ is the speed of light in a vacuum.Since the beam is parallel and the radius increases as intensity decreases,the intensity $I$ is highest at the axis of the beam and decreases as we move away from the axis.Because $\mu = \mu_0 + \mu_2 I$,a higher intensity $I$ at the axis results in a higher refractive index $\mu$ at the axis.Since $v = \frac{c}{\mu}$,a higher refractive index $\mu$ at the axis leads to a lower speed $v$ at the axis.Therefore,the speed of light is minimum on the axis of the beam.

$A$ parallel beam of light travels in a medium with a refractive index $\mu(I) = \mu_0 + \mu_2 I$,where $\mu_0$ and $\mu_2$ are positive constants and $I$ is the intensity. As the intensity of the light decreases,the radius of the beam increases. The speed of light in the medium is:

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