Three sheets of plastic have unknown indices of refraction. Sheet 1 is placed on
ID: 2014211 • Letter: T
Question
Three sheets of plastic have unknown indices of refraction. Sheet 1 is placed on top of sheet 2, and a laser beam is directed onto the sheets from above so that it strikes the interface at an angle of 26.1° with the normal. The refracted beam in sheet 2 makes an angle of 30.6° with the normal. The experiment is repeated with sheet 3 on top of sheet 2, and with the same angle of incidence, the refracted beam makes an angle of 36.7° with the normal. If the experiment is repeated again with sheet 1 on top of sheet 3, what is the expected angle of refraction in sheet 3? Assume the same angle of incidence.Explanation / Answer
Given: Let , the refractive indexes of the three layers are : n1 , n2 , n3 respectevely from snell's law we have , n1 sin1 = n2 sin2 since , first she et is on the second sheet , Angle of incidence = 1 = 26.1o Anlge of refraction = 2 = 30.6o Here , n1 is n1 & n2 is n2 thus, n1 = n2 sin2 / sin1 = n2 sin30.6 / sin26.1 n1 = 1.157 n2 Now , the third sheet on the top of the second , Angle of incidence = 1 = 26.1 o Angle of refraction = 2 = 36.7o Her n1 is n3 and n2 is n2 thus, n3 sin26.1o = n2 sin36.7o n3 = 1.358 n2 Now for the first sheet on top of the third sheet , given Angle of incidence 1 = 26.1o Here , n1 is n1 and n2 is n3 thus, n1 sin1 = n3 sin2 or sin2 = n1 /n3 sin1 = 1.157 n2 / 1.358 n2 ( sin 26.1) = 0.3747 2 = 22.01 o Thus, expected angle of refraction in sheet 3 is : 2 = 22.01o 2 = 22.01 o Thus, expected angle of refraction in sheet 3 is : 2 = 22.01oRelated Questions
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