09HW9 - Niu (qn269) Homework #9 antoniewicz (58855) 1 This...

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: Niu (qn269) Homework #9 antoniewicz (58855) 1 This print-out should have 16 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 as shown in the figure below. The distance from the slits to the screen is 7 . 3 m . 1 . 9cm 7 . 3 m . 97mm S 1 S 2 viewing screen What is the wave length if the distance from the central bright region to the fifth dark fringe is 1 . 9 cm . Correct answer: 561 . 033 nm. Explanation: 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 . 9 cm = 0 . 019 m , L = 7 . 3 m , and d = 0 . 97 mm = 0 . 00097 m . r 2 r 1 y L d S 1 S 2 = ta n 1 parenleftBig y L parenrightBig viewing screen d sin r 2- r 1 P O negationslash S 2 Q S 1 90 Q r 2 r 1 d S 1 S 2 = ta n 1 parenleftBig y L parenrightBig d s i n r 2- r 1 negationslash S 2 Q S 1 9 Q Solution: The angle from the slits mid- point to the y position on the screen is = arctan bracketleftBig y L bracketrightBig = arctan bracketleftbigg (0 . 019 m) (7 . 3 m) bracketrightbigg = 0 . 00260273 rad . The wavelength of the light for the fifth dark fringe, m = 4, is = d sin parenleftbigg m + 1 2 parenrightbigg = (0 . 00097 m) sin(0 . 00260273 rad) (4 . 5) = 5 . 61033 10 7 m = 561 . 033 nm . 002 10.0 points A pair of narrow parallel slits separated by a distance of 0.208 mm are illuminated by the green component from a mercury vapor lamp ( = 546 . 6 nm). What is the angle from the central maxi- mum to the second dark fringe on either side of the central maximum? Correct answer: 0 . 225851 . Explanation: Niu (qn269) Homework #9 antoniewicz (58855) 2 Basic Concept: d (sin ) = parenleftbigg m + 1 2 parenrightbigg , m = 0 , 1 , 2 , Given : d = 0 . 208 mm 1 m 1000 mm = 2 . 08 10 4 m = 546 . 6 nm 1 m 10 9 nm = 5 . 466 10 7 m m = 1 , second dark fringe Solution: sin = parenleftbigg m + 1 2 parenrightbigg d = sin 1 parenleftbigg m + 1 2 parenrightbigg d = sin 1 bracketleftbigg 1 . 5(5 . 466 10 7 m) . 000208 m bracketrightbigg = . 225851 . 003 (part 1 of 2) 10.0 points Light of wavelength 747 nm illuminates a sin- gle slit....
View Full Document

Page1 / 7

09HW9 - Niu (qn269) Homework #9 antoniewicz (58855) 1 This...

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