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. As the beam enters the medium, it will

Choose the beam axis as the $$z$$-axis and let $$r$$ denote the radial distance from that axis.

The intensity profile is brightest on the axis and falls outward, so $$I = I(r)$$ with $$\frac{dI}{dr} \lt 0$$.

The refractive index depends on intensity according to
$$\mu(r)=\mu_0+\mu_2 I(r),\qquad \mu_0\gt 0,\; \mu_2\gt 0.$$

Taking the radial derivative,

$$\frac{d\mu}{dr}= \mu_2 \frac{dI}{dr}.$$

Because $$\mu_2\gt 0$$ and $$\frac{dI}{dr}\lt 0$$, we get $$\frac{d\mu}{dr}\lt 0$$. Thus:

• $$\mu$$ is maximum on the axis.
• $$\mu$$ decreases continuously as one moves away from the axis.

In a graded-index medium a light ray is deflected toward the region of higher refractive index (Snell’s law in differential form or Fermat’s principle). Hence rays that start at some $$r\gt 0$$ feel a refractive-index gradient that bends them inward toward the axis where $$\mu$$ is larger.

The initially parallel cylindrical beam therefore narrows in radius—it self-focuses and behaves like it is passing through a converging graded-index lens.

So, after entering the medium, the beam will converge.

Option B which is: converge