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(r/r₀), where μ₀=1, r₀=3 cm, r₀≤r

Question

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(r/r₀), where μ₀=1, r₀=3 cm, r₀≤r<∞. The beam's trajectory lies in the plane that includes C. At distance r₁=8√2 cm the beam makes an angle φ=30° with r₁ as shown in figure. Find the minimum distance (in cm) the beam reaches relative to the symmetry centre C.

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1: In a spherically symmetric refractive-index profile, the conserved optical impact parameter is b=μ(r) r sinφ. Step 2: Here μ(r)=r/r₀, so b=[r² sinφ]/r₀. At r₁=8√2 cm and φ₁=30°, b=(8√2)²(1/2)/r₀=64/r₀. Step 3: At the closest approach the ray is perpendicular to the local radial direction, so sinφₘᵢₙ=1 and rₘᵢₙ²/r₀=b. Step 4: Therefore rₘᵢₙ²=64 cm² and rₘᵢₙ=8 cm.