Phys_322_HW8

Phys_322_HW8 - Phys 322 HW8 Solution 1. In phase: 1 2 , 2 2...

Info iconThis preview shows pages 1–2. 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: Phys 322 HW8 Solution 1. In phase: 1 2 , 2 2 2 2 2 E0 E01 E02 2 E01 E02 cos 2 2 E01 E02 2 E01 E02 E01 E02 2 Out of phase: 2 1 , 2 2 2 2 2 E0 E01 E02 2 E01 E02 cos 2 2 E01 E02 2 E01 E02 E01 E02 2 2. sin cos , so E 2 4 sin t 4 cos t . 2 2 2 2 2 E3 E1 E2 , so E03 E01 E02 2 E01 E02 cos 2 2 9 16 25 , so E03 5 . tan E01 sin 1 E02 sin 2 4 , therefore 53 . E3 lags E1. E01 cos 1 E02 cos 2 3 g 2 3. Long wavelength limit, v 4. Assume v a , vg v g dv 1 vk , vg v k k dk 2k gv k2 dv a a 2 2v d 5. f sin . It’s an even function, and the wavelength is π, k 2 . Therefore, f A0 Am cos 2m ; 2 m1 2 Am 2 0 sin cos 2md 2 cos2m 1 2m 1 0 0 sin2m 1d 2 2 cos2m 1 2m 1 0 sin2m 1d 0 2 1 1 4 1 2m 1 2m 1 4m 2 1 1 Then, f 2 4 cos 2m 2 1 m 1 4m 6. xc ct c 3 10 8 m / s 10 8 s 3m . 0 ~ 2 / xc 500 10 9 m 0 2 3m 8.3 10 5 nm . v v 0 0 8.3 10 5 nm 500 nm 1.6 10 7 . 7. v v 1 10 10 m 600 10 9 m 1.67 10 4 . So v c 3 10 8 m 600 10 9 m 5 1014 Hz . v 1.67 10 4 5 1014 Hz 8.35 1014 Hz , so t 1.2 10 11 s . lc ct c 3 10 8 m / s 1.2 10 11 s 3.6 10 3 m 2 ...
View Full Document

Page1 / 2

Phys_322_HW8 - Phys 322 HW8 Solution 1. In phase: 1 2 , 2 2...

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