Solution5

Solution5 - PHYS 434 Assignment 5 solutions Problems...

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PHYS 434: Assignment 5 solutions Problems 5.1-5.3 5.1: Proceed as in example 5.1. Bright fringes will be spaced by Δ y = λ L d = 530 × 10 9 m ( ) 5.00 m 1.00 × 10 4 m = 2.65 cm 5.2: We cannot use the small-angle approximation for an angle this large. Use equation 5.16: sin θ = ± 100 d = ± 100 589 × 10 9 m 1.00 × 10 4 m = ± 0.589 giving = ± sin 1 (0.589) = ± 36.1 Use a tangent to find the separation, Δ y = 2 L tan(36.1 ) = 2(5.00 m )tan(36.1 ) = 7.29 m 5.3: Use equation 5.16 with sin = 1 : m max = d = 1.00 × 10 4 m 589 × 10 9 m = 169.8 Rounding down will fit within the tolerance window. (In fact rounding up or down will still fit within the tolerance window.) Problem 5.4 : Use equation 5.8: I = I 1 + I 2 + 2 1 I 2 cos δ = 3 I 2 + 2 I 2 cos giving a fringe minimum that is 2.94% of the fringe maximum (as compared to the case I 1 = I 2 , where the fringe minimum is 0% of the fringe maximum). Problem 5.5: If the slab is placed over the top slit, the optical path length is increased for the distance r 1 in Figure 5.3. This will shift the pattern up
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This note was uploaded on 04/11/2010 for the course PHYS 434 taught by Professor Kilfoil during the Winter '09 term at McGill.

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Solution5 - PHYS 434 Assignment 5 solutions Problems...

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