Solution: Solution: (1) Let the critical angle of glass brick be c, then there is sinC= 1n= 12.

D: c = 45

According to geometry knowledge, the incident angle I 1 = 60 > C is on the AB plane, so the light is totally reflected on AB.

According to the reflection law and geometric relationship, the incident angle I2 of light on BC plane is 30 < C, and it is known that light will exit the glass brick from BC plane.

Let the refraction angle of the light when it first exits the prism be R2.

Then it is: n=sinr2sini2

Get: sinr2 = nsini2 = 2× sin30 = 22,

r2=45。

(2) According to geometry knowledge, AD = AMCOS60 = 2am = A4.

BD=AB-AD=a-a4=34a

So EB =12bdcos30 =12× 34a33 = 34a.

Answer: (1) The refraction angle of light when it first exits the prism is 45;

(2) When the light exits the prism for the first time, the distance between the exit point and point B is 34a. ..