This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: PHYS1200 PHYSICS II SPRING 2006 HOMEWORK SOLUTIONS CLASS 21. 36.06, 36.17, 36.65; Q38.04, Q38.06 36.06. a) The minima of single slit diffraction are determined by λ θ m a = sin . Using the small angle approximation, we have λ m D y a m = . Here m y is the distance from the center of central maximum to the m th minimum. So we get ( 29 m 10 35 . ) m 40 . )( m 10 550 ( 4 4 so , 5 3 9 1 5 1 5 × × = = = y y D a y y D a λ λ λ a = 2.5 ×103 m = 2.5 mm b) m 10 5 . 2 m 10 550 3 9 1 × × = ≈ a λ θ θ 1 = 2.2 ×104 rad = 0.013° 36.17. a) θ λ R d = = × × 122 122 550 10 50 10 9 3 . . . m m θ R = 1.34 ×104 rad = 7.69 ×103 deg = 27.7 arc seconds ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ b) Let s be the distance between the headlights, and D the distance from the headlights to the eye. Then, 4 10 1.34 m .4 1 so , × = = = R R s D D s θ θ D = 1.0 ×10 4 m = 10 km 36.65. The first minimum will occur at an angle given by...
View
Full
Document
This note was uploaded on 04/07/2008 for the course PHYS 1200 taught by Professor Stoler during the Spring '06 term at Rensselaer Polytechnic Institute.
 Spring '06
 Stoler
 Physics, Work, Diffraction

Click to edit the document details