(20 13? As shown in the figure, a transparent cuboid with refractive index n=72 is placed in the air, and the rectangular ABCD is its cross section.

Solution: Solution: (1) Let the minimum refraction angle be α, and according to the geometric relationship, it is:

sinα= APDP = 66d(66d)2+D2 = 17

According to the law of refraction, n=sinθsinα, sinθ = nsinα = 72×17 =12.

The minimum value of the angle θ is θ = 30.

(2) Let the critical angle of total reflection be c, then sinc =1n. As shown in the figure, the total reflection of this beam on the AD plane requires that the incident angle β on the AD plane should satisfy β≥C, that is, when the incident angle θ on the AB plane is maximum, β = C. 。

That is, sinθ=nsinα=ncosβ=ncosC=n2? 1

Substituting n=72 for the solution, the maximum value of θ is obtained: θ = 60.

So the range of angle θ is: 30 ≤ θ≤ 60.

Answer: (1) If the beam enters a cuboid and can hit the AD plane, the minimum value of angle θ is 30.

(2) If the light beam is totally reflected on the AD plane, the range of angle θ is: 30 ≤ θ≤ 60.