ch33-p064 - 1.56 sin 24.3 = n air sin air air = 40 . (d) It...

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Unformatted text preview: 1.56 sin 24.3 = n air sin air air = 40 . (d) It strikes the top surface (side 2) at an angle (measured from the normal axis there, which in this case would be a vertical axis) of 90 2 = 66 which is much greater than the critical angle for total internal reflection (sin 1 ( n air /1.56 ) = 39.9). Therefore, no refraction occurs when the light strikes side 2. (e) In this case, we have n air sin 70 = 1.56 sin 2 which yields 2 = 37.04 if we use the common approximation n air = 1.0, and yields 2 = 37.05 if we use the more accurate value for n air found in Table 33-1. This is greater than the 33.7 mentioned above (regarding the upper-right corner), so the ray strikes side 2 instead of side 3. (f) After bouncing from side 2 (at a point fairly close to that corner) to goes to side 3. (g) When it bounced from side 2, its angle of incidence (because the normal axis for side 2 is orthogonal to that for side 1) is 90 2 = 53 which is much greater than the critical...
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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