An initially parallel cylindrical beam travels in a medium of refractive index $$\mu(I) = \mu_0 + \mu_2 I$$, where $$\mu_0$$ and $$\mu_2$$ are positive constants and $$I$$ is the intensity of the light beam. The intensity of the beam is decreasing with increasing radius. The speed of light in the medium is

The speed of light in a material medium is related to its refractive index by the basic relation
$$v=\frac{c}{\mu}$$
where $$c$$ is the speed of light in vacuum and $$\mu$$ is the refractive index of the medium.

In this problem the refractive index depends on the local intensity $$I$$ of the cylindrical beam according to
$$\mu(I)=\mu_0+\mu_2 I$$
with $$\mu_0,\mu_2 \gt 0$$.

Because $$\mu_2$$ is positive, $$\mu(I)$$ increases as the intensity $$I$$ increases. Equivalently, the refractive index is highest wherever the intensity is highest.

The statement “the intensity of the beam is decreasing with increasing radius” means that the intensity is greatest at the central axis of the beam and becomes smaller as we move outward from the axis.

Therefore:
• On the beam axis: intensity $$I$$ is maximum ⟹ refractive index $$\mu$$ is maximum.
• Away from the axis: intensity $$I$$ is lower ⟹ refractive index $$\mu$$ is lower.

Since $$v=\dfrac{c}{\mu}$$, a larger $$\mu$$ corresponds to a smaller speed $$v$$. Hence the speed of light attains its minimum value where $$\mu$$ is largest, i.e. on the axis of the beam.

Consequently, the correct statement is:

Option A which is: minimum on the axis of the beam