Given refractive index dependence

The medium is nonlinear, with $\mu(I)=\mu_0+\mu_2 I,$ where $\mu_2>0$.

Also, the beam intensity decreases with increasing radius, so:

• intensity is maximum at the axis,
• intensity is smaller at the periphery.

Effect on refractive index across the beam

Since $\mu_2>0$, larger intensity means larger refractive index.

Therefore,

• near the axis: $I$ is large $\Rightarrow \mu$ is large,
• near the edges: $I$ is small $\Rightarrow \mu$ is smaller.

So the beam sees a refractive index profile with $\mu_{\text{axis}} > \mu_{\text{edge}}.$

What does this do to the wavefront?

The phase velocity is $v=\frac{c}{\mu}.$ Hence where $\mu$ is larger, light travels more slowly.

So:

• central part of the beam moves slower,
• outer part moves faster.

Therefore, after entering the medium, the central portion of the wavefront lags behind the outer portions.

Shape of the wavefront

A wavefront whose central part lags and edges advance becomes concave (as seen in the direction of propagation).

Equivalently, the medium behaves like a self-focusing lens for the beam, which corresponds to a converging wavefront, i.e. a concave wavefront.

Checking options

• A: convex $\rightarrow$ incorrect
• B: concave $\rightarrow$ correct
• C: convex near the axis and concave near the periphery $\rightarrow$ incorrect
• D: planar $\rightarrow$ incorrect

Derived answer

The wavefront becomes concave.

Comparison with stored answer

Stored correct answer: D (planar)

This does not agree with the physics of intensity-dependent refractive index. Since $\mu$ is higher at the center, the center travels slower, so the wavefront cannot remain planar. It must bend into a concave shape.