Solution6

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

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
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 .
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 04/11/2010 for the course PHYS 434 taught by Professor Kilfoil during the Winter '09 term at McGill.

Page1 / 2

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online