chapter 35 solutions - Solutions to Exercises 12 th Ed....

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

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

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

Unformatted text preview: Solutions to Exercises 12 th Ed. 35.1. I DENTIFY : Compare the path difference to the wavelength. S ET U P : The separation between sources is 5.00 m, so for points between the sources the largest possible path difference is 5.00 m. E XECUTE : (a) For constructive interference the path difference is m λ , m = 0, ± 1, ± 2, ... Thus only the path difference of zero is possible. This occurs midway between the two sources, 2.50 m from A. (b) For destructive interference the path difference is ( m + 1 2 ) λ , m = 0, ± 1, ± 2, ... A path difference of ± λ / 2 = 3.00 m is possible but a path difference as large as 3 λ / 2 = 9.00 m is not possible. For a point a distance x from A and 5.00- x from B the path difference is x- (5.00 m- x ). x- (5.00 m- x ) = + 3.00 m gives x = 4.00 m. x- (5.00 m- x ) = - 3.00 m gives x = 1.00 m . E VALUATE : The point of constructive interference is midway between the points of destructive interference. 35.8.I DENTIFY : The value of y 20 is much smaller than R and the approximate expression y m = R m λ d is accurate. S ET U P : y 20 = 10.6 × 10- 3 m . E XECUTE : d = 20 R λ y 20 = (20)(1.20 m)(502 × 10- 9 m) 10.6 × 10- 3 m = 1.14 × 10- 3 m = 1.14 mm E VALUATE : tan θ 20 = y 20 R so θ 20 = 0.51o and the approximation sin θ 20 ≈ tan θ 20 is very accurate. 35.9. I DENTIFY and S ET U P : The dark lines correspond to destructive interference and hence are located by Eq.(35.5): d sin θ = m + 1 2 λ so sin θ = m + 1 2 λ d , m = 0, ± 1, ± 2, K Solve for θ that locates the second and third dark lines. Use y = R tan θ to find the distance of each of the dark lines from the center of the screen. E XECUTE : 1st dark line is for m = 2nd dark line is for m = 1 and sin θ 1 = 3 λ 2 d = 3(500 × 10- 9 m) 2(0.450 × 10- 3 m) = 1.667 × 10- 3 and θ 1 = 1.667 × 10- 3 rad 3rd dark line is for m = 2 and sin θ 2 = 5 λ 2 d = 5(500 × 10- 9 m) 2(0.450 × 10- 3 m) = 2.778 × 10- 3 and θ 2 = 2.778 × 10- 3 rad (Note that θ 1 and θ 2 are small so that the approximation θ ≈ sin θ ≈ tan θ is valid.) The distance of each dark line from the center of the central bright band is given by y m = R tan θ , where R = 0.850 m is the distance to the screen. tan θ ≈ θ so y m = R θ m y 1 = R θ 1 = (0.750 m)(1.667 × 10- 3 rad) = 1.25 × 10- 3 m y 2 = R θ 2 = (0.750 m)(2.778 × 10- 3 rad) = 2.08 × 10- 3 m ∆ y = y 2- y 1 = 2.08 × 10- 3 m- 1.25 × 10- 3 m = 0.83 mm E VALUATE : Since θ 1 and θ 2 are very small we could have used Eq.(35.6), generalized to destructive interference: y m = R m + 1 2 λ / d . 35.10. I DENTIFY : Since the dark fringes are eqully spaced, R ? y m , the angles are small and the dark bands are located by y m + 1 2 = R ( m + 1 2 ) λ d ....
View Full Document

This note was uploaded on 04/14/2008 for the course PHYS 2303 taught by Professor Hoffman during the Fall '07 term at University of Texas at Dallas, Richardson.

Page1 / 10

chapter 35 solutions - Solutions to Exercises 12 th Ed....

This preview shows document pages 1 - 3. Sign up to view the full document.

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