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Solution6

# Solution6 - PHYS 434 Assignment 6 solutions Problem 6.1 Use...

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PHYS 434: Assignment 6 solutions Problem 6.1: Use (1 x 2 ) 1/ 2 1 1 2 x 2 1 8 x 4 With the extra term, Equation 6.7 becomes δ = r r h r 2 r r 1 1 2 h r 2 1 8 h r 4 = h 2 2 r + h 4 8 r 3 Problem 6.5: Repeat Example 6.2 for a rectangular aperture of width a = 2.00 μ m and height b = 6.00 m . Proceed with Fraunhofer calculations as in the Example. The aperture is small, so the small angle approximation no longer gives accurate results. The first diffraction nulls along the horizontal direction are located at angles determined by sin ϕ = λ a = 632.8 × 10 9 m 2.00 × 10 6 m = 0.316 = 18.4 . Since this angle is measured from the centre of the pattern, the horizontal width of the central maximum is Δ x = 2 Z (tan ) = 0.667 m . In the vertical direction, sin θ = b = 632.8 × 10 9 m 6.00 × 10 6 m = 0.105 giving Δ y = 2 Z (tan ) = 0.212 m .

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Solution6 - PHYS 434 Assignment 6 solutions Problem 6.1 Use...

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