This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: montes (dam2772) Homework9 shih (56310) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 (part 1 of 2) 10.0 points Two rectangular, optically flat glass plates (of index of refraction 1 . 5) are in contact along one end and are separated along the other end by a sheet of paper that is 55 m thick. The top plate is illuminated by monochromatic light (of wavelength 546 . 1 nm). An interfer- ence pattern results from the reflection of this light by the bottom surface of the top plate and the top surface of the bottom plate. Consider a light beam traveling in the air incident on a glass surface. What is the phase change as the light is reflected from the first (top) glass surface? 1. = 0 2. = 2 3. = 2 4. none of these 5. = correct Explanation: Since glass is optically denser than air, the phase change of light upon reflection from glass is . 002 (part 2 of 2) 10.0 points Calculate the number of dark parallel bands crossing the top plate including the dark band at zero separation along the line of contact. The last dark band is to be counted if and only if its center (where the intensity is zero) is present. Correct answer: 202. Explanation: The largest possible phase difference max arises from using the largest possible path difference d max with the paper of thickness t ; i.e. , max = 2 N d max + = 4 t + , where N = , since the intervening medium is actually air, and the contribution arises from reflection off the bottom slab. We then want the largest possible integer N such that destructive interference occurs; i.e. , max 2 parenleftbigg N- 1 2 parenrightbigg N 2 t + 1 , so N = integer parenleftbigg 2 t + 1 parenrightbigg = bracketleftbigg 2 (5 . 5 10 5 m) 5 . 461 10 7 m + 1 bracketrightbigg = 202 . 003 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 fifth bright fringe on the screen (po- sition y on the screen), (the corresponding path length difference) and (the phase angle difference) is given by 1. = 6 and = 12 2. = 7 and = 14 3. = 2 and = 4 montes (dam2772) Homework9 shih (56310) 2 4. = 5 2 and = 5 5. = 5 and = 10 correct 6. = 9 2 and = 9 7. = 3 2 and = 3 8. = 11 2 and = 11 9. = 13 2 and = 13 10. = 7 2 and = 7 Explanation: For bright fringes = d sin = m , where m = 0 , 1 , 2 , 3 , so for the fifth bright fringe ( m = 5), = m = 5 and = 2 = 5 = 10 ....
View Full Document