quiz4 - Daniels, Matthew Quiz 4 Due: Dec 7 2006, midnight...

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: Daniels, Matthew Quiz 4 Due: Dec 7 2006, midnight Inst: Ditmire 1 This print-out should have 23 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points What is the frequency of an electromagnetic wave if it has a wavelength of 1.2 km? The speed of light is 3 10 8 m / s. Correct answer: 250000 Hz. Explanation: Let : = 1 . 2 km and c = 3 . 00 10 8 m / s . The speed is c = f f = c = 3 10 8 m / s 1 . 2 km 1 km 10 3 m = 250000 Hz . keywords: 002 (part 1 of 2) 10 points The second-order bright fringe ( m = 2) is 4 . 45 cm from the center line. 4 . 45cm 1 . 11 m . 0307mm S 1 S 2 viewing screen Determine the wavelength of the light. Correct answer: 615 . 383 nm. Explanation: Let : y = 4 . 45 cm , L = 1 . 11 m , and d = 0 . 0307 mm , r 2 r 1 y L d S 1 S 2 = ta n- 1 y L viewing screen d sin r 2- r 1 P O 6 S 2 QS 1 90 Q r 2 r 1 d S 1 S 2 = ta n- 1 y L d s i n r 2- r 1 6 S 2 Q S 1 9 Q For constructive interference y bright = L d m, (1) with m = 2, y 2 = 0 . 0445 m, L = 1 . 11 m, and d = 3 . 07 10- 5 m = dy 2 mL = (3 . 07 10- 5 m)(0 . 0445 m) (2)(1 . 11 m) = 6 . 15383 10- 7 m = 615 . 383 nm . 003 (part 2 of 2) 10 points Calculate the distance between adjacent bright fringes. Correct answer: 2 . 225 cm. Explanation: From equation (1) and the results of the first part of the problem, we get y m +1- y m = L ( m + 1) d- Lm d = L d Daniels, Matthew Quiz 4 Due: Dec 7 2006, midnight Inst: Ditmire 2 = (6 . 15383 10- 7 m)(1 . 11 m) (3 . 07 10- 5 m) = 0 . 02225 m = 2 . 225 cm . keywords: 004 (part 1 of 1) 10 points The reflecting surfaces of two intersecting flat mirrors are at an angle of 64 , as shown in the figure. A light ray strikes the horizon- tal mirror, reflects off the horizontal mirror, impinges on the raised mirror, reflects off the raised mirror, and proceeds in the right-hand direction. 64 Figure is not drawn to scale. Calculate the angle . Correct answer: 52 . Explanation: Basic Concept: incident = reflected Solution: 1 2 Figure is to scale. The sum of the angles in a triangle is 180 . In the triangle on the left we have angles , 180 - 1 2 , and 180 - 2 2 , so 180 = + 180 - 1 2 + 180 - 2 2 , or 1 + 2 = 2 . (1) In the triangle on the right we have angles 1 , 2 , and . 180 = 1 + 2 + , so 1 + 2 = 180 - . (2) Combining Eq. 1 and 2, we have = 180 - 2 = 180 - 2(64 ) = 52 . As a matter of interest, in the upper-half of the figure the angles (clockwise) in the triangles from left to right are 47 , 47 , and 86 ; 94 , 26 , and 60 ; 120 , 17 , and 43 ; 137 , 17 , and 26 ; and in the lower-half of the figure the angles (counter-clockwise) in the triangles from left...
View Full Document

This note was uploaded on 06/23/2008 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas at Austin.

Page1 / 13

quiz4 - Daniels, Matthew Quiz 4 Due: Dec 7 2006, midnight...

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