Phys 375 Sol HW 2

# Phys 375 Sol HW 2 - Phys 375 HW 2 Fall 2009 Due 21 22...

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Unformatted text preview: Phys 375 HW 2 Fall 2009 Due 21 / 22 September, 2009 1. Pedrotti 3 , 3 rd edition, problem 2-7 (see Fig. 2-33). Solution: See FIGURE 2-33in the text P 3 From the geometry it is clear that h D C 4 / tan = θ , where h is the height of the slab and D is the diameter of the circle of light. From Snell’s law we know that the critical angle occurs when the angle of refraction is 2 π θ = R . Then applying Snell’s Law, 2 2 1 1 sin sin θ θ n n = we have: ( 29 55 . 1 4 / 4 / sin 2 / sin 2 2 = + = = h D D n n C air glass θ π Where I used 1 = air n . 2. Pedrotti 3 , 3 rd edition, problem 3-6. Solution: See FIGURE 3-9 of the text P 3 For the deviation angle, δ , we have from the figure and equation 3-9 and 3-11 : A- + = = ′- ′- + = 2 1 2 1 2 1 θ θ θ θ θ θ δ Combining this with equations 3-(7-10), we get: ( 29 A n A n- - + =-- 1 1 1 1 sin sin sin sin θ θ δ Here is a plot of the deviation angle vs. incident angle for n=1.52 and A=60 o ....
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## This note was uploaded on 12/29/2011 for the course PHYSICS 375 taught by Professor Eno during the Spring '11 term at Maryland.

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Phys 375 Sol HW 2 - Phys 375 HW 2 Fall 2009 Due 21 22...

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