A laser beam propagates through a spherically symmetric medium, as shown in the figure. The refractive index varies with the distance to the symmetry centre `C` by the law `mu(r )=mu_(0)(r )/(r_(0))`, where `mu_(0)=1`, `r_(0)=3cm`, `r_(0) le r lt oo`. The beam's trajectory lies in the plane that includes `C`. At distance `r_(1)=8sqrt(2)cm` the beam makes an angle `phi=30^(@)` with `vecr_(1)` as shown in the figure. Find the minimum distance (in `cm`) the beam reach relative to the symmetry centre `C`.

Explanation

`mu_(1)sinphi=mu_(2)sintheta`
`(r_(1))/(sintheta')=(r_(2))/(sintheta)`
`sinphi=(r_(2))/(r_(1))sintheta'`
`mu_(1)sinphi=mu_(2)(r_(2))/(r_(1))sintheta'`
`r_(1)mu_(1)sinphi=r_(2)mu_(2)sintheta'`
`r_(1)mu_(0)(r_(1))/(r_(0))sin30^(@)=mu_(0)(r_(min)^(2))/(r_(0))sin90^(@)` (At minimum separation)
`r_(min)=r_(1)sqrt(sin30^(@))=8cm`