Sol-Ch38 - CHAPTER 38 E xercises 1 a sin 1 =/a then y 1 L/a...

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

View Full Document Right Arrow Icon
Exercises 1. (a) sin θ 1 = λ /a, then y 1 ≈ L λ /a = 2.04 cm. width 2y 1 = 4.08 cnl (b) sin θ 2 = 2 λ /a, so y 2 ≈ 2L λ /a = 4.08 cm then y 2 – y 1 = 2.04 cm 2. θ 1 = sin –1 ( λ /a); where λ = 6.8 cm. /width is 2 θ 1 . (a) 116 ° ; (b) 26.2 ° 3. ∆y 2 /∆y 1 = λ 2 / λ 1 , so ∆y 2 = (3)(436/589) = 2.22 cm 4. sin θ ≈ tan θ = y 1 /L = (∆y/2)L = 2 ×   10 –3 , a = λ /sin θ = 0.273 mm 5. ∆y = L(sin θ 2 – sin θ 1 ) = L λ /a. So a = L λ /∆y = 0.045 mm 6. (a) f = v/ λ = 340 ×   4/0.76 = 1790 Hz (b) sin θ 1 = λ /a = 1/4, leads to θ 1 = 14.5˚ 7. d/a = 4, so there are 7 fringes 8. sin θ 1 = 1.22 λ /a = 6.7 ×   10 –4 ; Width is 2 θ 1 . (a) 116 ° ; (b) 26.2 ° 9. θ c = 1.22 λ /a = 9.15 ×   10 –4 ; then s = d θ c = 1.46 cm 10. θ c = 1.22 λ /a = 1.83 ×   10 –2 ; then s = d θ c = 366 m 11. (a) θ c = 1.22 λ /a = 1.34 ×   10 –4 ; d = L θ c = 51.6 km; (b) θ c = 1.49 ×   10 –7 ; d = L θ c = 57.3 m 12. θ c = 1.22 λ /a = 1.14 ×   10 –6 ; s = d θ c = 0.205 m; 13. θ c = 1.22 λ /a = s/25, thus s = 50.8 µm 14. θ c = 1.22 λ /a = 4.47 × 10 –5 ; d = s/ θ c = 44.7 km; 15. (a) θ c = 1.22 λ /a = 1.2 ×   10 –7 ; s = d θ c = 1.2 ×   10 9 m; CHAPTER 38
Image of page 1

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

View Full Document Right Arrow Icon
(b) θ c = 1.22 λ /a = 8.4 ×   10 –4 ; s = d θ c = 8.4 ×   10 –4 ; s = d θ c = 8.4 ×   10 –2 m 16. (a) θ c = 1.22 λ /a = 1.59 ×   10 –4 ; d = s/ θ c = 11.3 km; (b) θ c = 1.22 λ /a = 2.83 ×   10 –7 ; d = 1.8/ θ c = 6.26 ×   10 6 m 17. (a) d = 3.33 ×   10 –6 m; sin θ 1 = λ /d θ A = 7.07 ° ; θ B = 11.35
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern