Homework 3 KEY - Golshani (ag5873) Homework 03 Yao (57045)...

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: Golshani (ag5873) Homework 03 Yao (57045) 1 This print-out should have 14 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points A screen is illuminated by monochromatic light whose wave length is , as shown. The distance from the slits to the screen is L . y L d S 1 S 2 viewing screen At the sixth dark fringe on the screen (po- sition y on the screen), (the corresponding path length difference) and (the phase angle difference) is given by 1. = 5 2 and = 5 2. = 9 2 and = 9 3. = 4 and = 8 4. = 6 and = 12 5. = 11 2 and = 11 correct 6. = 7 and = 14 7. = 2 and = 4 8. = 13 2 and = 13 9. = 3 2 and = 3 10. = 7 2 and = 7 Explanation: For dark fringes = d sin = parenleftbigg m + 1 2 parenrightbigg , where m = 0 , 1 , 2 , 3 , so for the sixth dark fringe ( m = 5), = parenleftbigg m + 1 2 parenrightbigg = 11 2 and = 2 = 11 2 = 11 . 002 10.0 points A double slit arrangement produces an inter- ference fringe for yellow sodium light. To produce narrower-spaced fringes, should red light or blue light be used? 1. Red 2. Blue correct 3. Either Explanation: Blue light has a shorter wavelength than both yellow and red light and will produce narrower-spaced fringes. 003 10.0 points A screen is illuminated by monochromatic light as shown in the figure below. The distance from the slits to the screen is 3 m . 1 . 5cm 3 m . 56mm S 1 S 2 viewing screen What is the wave length if the distance from the central bright region to the sixth dark fringe is 1 . 5 cm . Correct answer: 509 . 085 nm. Explanation: Golshani (ag5873) Homework 03 Yao (57045) 2 Basic Concepts: For bright fringes, we have d sin = m , and for dark fringes, we have d sin = parenleftbigg m + 1 2 parenrightbigg , where m = 0 , 1 , 2 , 3 , . From geometry, we have y = L tan . Let : y = 1 . 5 cm = 0 . 015 m , L = 3 m , and d = 0 . 56 mm = 0 . 00056 m ....
View Full Document

This note was uploaded on 09/27/2010 for the course PHY 58650 taught by Professor Yao during the Fall '10 term at University of Texas.

Page1 / 6

Homework 3 KEY - Golshani (ag5873) Homework 03 Yao (57045)...

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