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Phys 2214 hw8 Solutions

# Phys 2214 hw8 Solutions - PHYS 2214 Homework#8 Solutions 1...

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Unformatted text preview: PHYS 2214 Homework #8 Solutions 1. Deriving Snell’s Law. (a) From the figure in the book, t = t 1 + t 2 = d 1 v 1 + d 2 v 2 = q h 2 1 + x 2 v 1 + q h 2 2 + ( l- x ) 2 v 2 . (b) The independent variable here is x , since the others are just pa- rameters. So, we find the minimum time by setting the derivative of t with respect to x equal to zero: dt/dx = 0. This yields x v 1 q h 2 1 + x 2 = ( l- x ) v 2 q h 2 2 + ( l- x ) 2 . Using the trig relations sin( θ 1 ) = x/d 1 and sin( θ 2 ) = ( l- x ) /d 2 gives sin( θ 1 ) v 1 = sin( θ 2 ) v 2 which, using n = c/v , becomes Snell’s Law: n 1 sin( θ 1 ) = n 2 sin( θ 2 ) . 2. Two source interference. (a) The phase difference between the two sources at the observer due to their path difference is Δ φ = k Δ x = 2 π Δ x/λ . Constructive interference will take place when Δ φ = 2 πm , or λ = Δ x/m , where m is an integer. Δ x = 2 . 04 × 10- 6 m , so the wavelengths within the visible range, 400 nm to 700 nm, that will give constructive interference are for m = 3 , 4 , and 5, or λ = { 680 , 510 , 408 } nm. 1 (b) The results would be the same as in part (a), since all that mat- ters is the path length difference of the two light sources to the observer. (c) Destructive interference will take place when Δ φ = 2 π Δ x/λ = (2 m + 1) π , or λ = 2Δ x/ (2 m + 1), where m is an integer. The wavelengths within the visible range that will give destructive in- terference are for m =3 and 4, or λ = { 583 , 453 } nm....
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Phys 2214 hw8 Solutions - PHYS 2214 Homework#8 Solutions 1...

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