CourseNotes.60

# CourseNotes.60 - refraction are n 1 = 1 . 60, n 2 = 1 . 40,...

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surface it will refract out into the air with an angle given by Snell’s law. If we assume that the index of refraction is 1, then sin θ 5 = n 1 sin θ 1 θ 5 = sin - 1 ( n 1 sin θ 1 ) = 56 . 9 Although the calculation of θ 4 at ﬁrst looks much more complex, it will turn out to be just as easy. Snell’s law for the ﬁrst interface gives n 1 sin θ 1 = n 2 sin θ 2 and for the second interface it gives n 2 sin θ 2 = n 3 sin θ 2 The important thing to notice now is that the θ 2 is the same in bot equations because the two interfaces are parallel to each other. Hence, we can combine these two formulas as well as the one for the ﬁnal interface to ﬁnd n 1 sin θ 1 = n 4 sin θ 4 θ 4 = 1 n 4 sin - 1 ( n 1 sin θ 1 ) = 39 . 21 Figure 40: Figure for Problem 33.53. Problem 33.59 In ﬁgure 41, light initially in material 1 refracts into material 2, crosses that material, and is then incident at the critical angle on the interface between materials 2 and 3. The indexes of
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Unformatted text preview: refraction are n 1 = 1 . 60, n 2 = 1 . 40, and n 3 = 1 . 20. (a) What is angle ? (b) If is increased, is there refraction of light into material 3? Since the light is incident at the critical angle in material 2, the angle with respect to the normal in material 2 must be = arcsin n 3 n 2 We need to relate this to the angle with respect to the normal at the interface between 1 and 2. Since these two interfaces are at right angles to each other, we can use that fact that the 3 angles of a triangle sum to 180 . Hence, the angle with respect to the normal at the other interface is 2 = 90 - c = 90 -arcsin n 3 n 2 60...
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## This note was uploaded on 12/05/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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