# Laser travelling through a radially graded index of refraction

A laser beam propagates through a region whose refractive index varies as $\mu=\mu_0(\frac{r}{r_0})$. At a distance of R the beam makes an angle of 30° with the normal.

Find the minimum seperation between the beam and the center of this region(the point where r=0).

I was able to solve this using the Lagrangian method and Fermat's principle.

$$x=r \cos \theta$$ $$y=r \sin \theta$$

$$ds=\sqrt {1+r^2(\theta')^2} dr,\theta'=\frac {d\theta}{dr}$$

$$v= \frac {c}{\mu}$$ Therefore,

$\int dt$ should be minimum

$$\int \frac {kr}{c} \sqrt {1+r^2(d\theta)^2} dr$$ should be minimum

let,

$$L=\frac {kr}{c} \sqrt {1+r^2(d\theta)^2}$$ Then,by Euler-Langrangian Equation,

$$\frac {\partial L}{\partial \theta}=\frac {d}{dr}(\frac {\partial L}{\partial \theta'})=0$$

Therefore,$\frac {\partial L}{\partial \theta'}=e$,e is constant.

Therefore,$\frac{r^3 \theta'}{\sqrt {1+r^2(\theta')^2}}=f$,f is another constant.

Now,this is valid for all $r,\theta$,

Therefore,initial conditions will also satisfy. initially,ray is horizontal. Therefore,$\frac {dy}{dx}=0$ Therefore,$\theta'=\frac {\tan \theta}{r}$ substituting,for initial conditions,

$$f=r_0^2* \sin \theta$$ Also,for minimum,$\frac {1}{\theta'}=0$ Thus, $$r_{min}^2=r_0^2* \sin \theta$$

The problem is,we haven't been taught this method in class.I learned it by myself however,for homework we have to derive it using elementary methods(Snell's Law and some calculus).I don't know how to solve this using elementary methods.