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Bài 2: a) n1. Sini = n2. Sinr --> Sin r = 1/2 --> r = 30 A = r + r’--> r’ = 60 - 30 = 30 n2. Sinr’ = n1. Sini’ --> i’ = 45 ==> D = i + i’ – A = 30 b) Ta có : Sin[(Dmin + A) / 2] = nSin A/2 <=> Sin[(Dmin + 60) / 2] = 1,414. Sin 30 <=> Dmin = 30 Vì D = Dmin = 30 nên khi ta tăng...