由图知,i 1 =60° 由折射定律得:n= sin i 1 sin r 1 ,则得sinr 1 = sin i 1 n ① 由几何知识得:i 2 =90°-r 1 ② 要光线经AB和AC两界面折射后最终能从AC面射出,则必须有i 2 <C,C为临界角. 则得 sini 2 <sinC= 1 n ③ 将②代入③得:sin(90°-r 1 )< 1 n 即cosr 1 < 1 n ④ 由①得cosr 1 = 1-si n 2 i 1 = 1-( sin i 1 n ) 2 = 1-( sin60° n ) 2 ⑤ 联立④⑤解得,n<