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 μ(r)=μ0(rr0) , where μ0=1, r0=3 cm, ro≤r<∞ .The beam’s trajectory lies in the plane that includes C. At distance r1=82cm the beam makes an angle ϕ=30o with r1→ as shown in figure. Find the minimum distance (in cm) the beam reaches from the symmetry centre C.

μ1sin ϕ = μ2sin θ r1sinθ'=r2sinθ sinϕ=r2r1sinθ' μ1sinϕ=μ2r2r1sinθ' r1μ0r1r0sin30°=μ0rmin2r0sin90° (At minimum separation) rmin=r1sin30°=8cm

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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 $μ (r) = μ_{0} (\frac{r}{r_{0}})$ , where $μ_{0} = 1$ , $r_{0} = 3 c m$ , $r_{o} \leq r < \infty$ .The beam’s trajectory lies in the plane that includes C. At distance $r_{1} = 8 \sqrt{2}$ cm the beam makes an angle $ϕ = 30^{o}$ with $\vec{r_{1}}$ as shown in figure. Find the minimum distance (in cm) the beam reaches from the symmetry centre C.

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Detailed Solution

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$μ_{1} sin ϕ = μ_{2} sin θ \frac{r_{1}}{\sin θ^{'}} = \frac{r_{2}}{\sin θ} \sin ϕ = \frac{r_{2}}{r_{1}} \sin θ^{'} μ_{1} \sin ϕ = μ_{2} \frac{r_{2}}{r_{1}} \sin θ^{'} r_{1} μ_{0} \frac{r_{1}}{r_{0}} \sin 30 ° = μ_{0} \frac{r_{\min}^{2}}{r_{0}} \sin 90 °$
(At minimum separation)
$r_{\min} = r_{1} \sqrt{\sin 30 °} = 8 c m$