Question

A laser beam propagates through a spherically symmetric media. The refraction index varies with distance to the symmetry centre by the law n(r) = $\mathrm { n } _ { 0 } \frac { \mathrm { r } } { \mathrm { r } _ { 0 } }$, where n₀ = 1, r₀ = 30 cm r₀£ r £. The beam trajectory lies in the plane that includes c. At distance r₁ = 80 cm the beam makes an angle f = 30° with as shown. What minimal distance does the beam reach relative to the symmetry center c –

• A. 0.566 m
• B. 0.86 m
• C. 1.6 m
• D. 0.16 m

Answer 1

Answer: (A)

0.566 m

Answer 2

n₁ sin a = n₂ sin b

= $\frac { \mathrm { R } _ { 2 } } { \sin ( \pi - \beta ) }$ =

sin b = $\frac { \mathrm { R } _ { 2 } } { \mathrm { R } _ { 1 } }$sin g

n₁ sin a = $\frac { \mathrm { R } _ { 2 } } { \mathrm { R } _ { 1 } }$ × sin g × n₂

n₁R₁sin a = n₂R₂ sin g =

constant

\ $\frac { \mathrm { n } _ { 0 } \mathrm { r } ^ { 2 } } { \mathrm { r } _ { 0 } }$ sinf = constant

\$\frac { \mathrm { n } _ { 0 } } { \mathrm { R } _ { 0 } } \times \mathrm { R } _ { 1 } ^ { 2 } \sin 30$ =