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Introductory Physics Homework Help
Young's slit: find wavelenght, double slit separation
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[QUOTE="moenste, post: 5266911, member: 553611"] [h2]Homework Statement [/h2] 1. The distance between the 1st bright fringle and the 21st bright fringe in a Young's double slit arrangement was found to be 2.7 mm. The slit separation was 1 mm and the distance from the slits to the plane of the fringes was 25 cm. What was the wavelength of the light? Answer: 5.4 * 10[SUP]-7[/SUP] m 2. In a Young's double-slit experiment a total of 23 bright fringes occupying a distance of 3.9 mm were visible in the traveling microscope. The microscope was focused on a plane which was 31 cm from the double slit and the wavelength of the light being used was 5.5 * 10 [SUP]-7[/SUP] m. What was the separation of the double slit? Answer: 0.96 mm (not 1.0 mm) [h2]Homework Equations[/h2] y = (λD) / a [h2]The Attempt at a Solution[/h2] 1. a. Everything in m. b. y = 2.7 * 10[SUP]-3[/SUP] / 20 fringes = 1.35 * 10[SUP]-4[/SUP] m. c. λ = (y a) / D = ((1.35 * 10[SUP]-4[/SUP]) *10[SUP]-3[/SUP]) / 0.25 = 5.4 * 10[SUP]-7[/SUP] m. 2. a. Everything in m. b. y = 3.9 * 10[SUP]-3[/SUP] / 22 fringes = 1.77 * 10[SUP]-4[/SUP] m. c. a = (λD) / y = ((5.5 * 10[SUP]-7[/SUP]) * 0.31) / (1.77 * 10[SUP]-4[/SUP]) = 9.62 * 10[SUP]-4[/SUP] m or 0.96 mm. Question: why do I need to decrease the number of fringles by 1 to get the right answer? If I use the given 21 and 23 numbers I get wrong answers. And in a different book which has "Five fringes were found to occupy a distance of 4 mm on the screen" the solution method is: "five fringes occupy 4 mm. So the fringe separation is 4 / 5 = 0.8 mm". Any help please? [/QUOTE]
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Introductory Physics Homework Help
Young's slit: find wavelenght, double slit separation
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