(2) The incident angle and refraction angle of light at the interface of point B are I and R, respectively.
sini=AMR=22=45
Refractive index n=sinisinr
Get: sinr= 12.
The propagation time t of light in the ball
T=BCV=2R？ cosθ2cn=63× 10-9(s)
(3) The exit angle β at point C is
sinrsinβ= 1n
Sin β = 22 β = 45。
∠COP=π-θ 1-∠BOC= 15
α+∠COP=β
Get α = 30.
Answer: (1) Complete the light path diagram as shown in the figure;
(2) The propagation time of light from point B to point C is 63×10-9 (s);
(3) 3) The angle α between CD and MN is 30.