1) A double-slit experiment is set up using red light (? = 738.0 nm). A first or
ID: 2201184 • Letter: 1
Question
1) A double-slit experiment is set up using red light (? = 738.0 nm). A first order bright fringe is seen at a given location on a screen. What wavelength of visible light (between 380 nm and 750 nm) would produce a dark fringe at the identical location on the screen?
? =
2) A new experiment is created with the screen at a distance of 1.9 m from the slits (with spacing 0.09 mm). What is the distance between the second order bright fringe of light with ? = 686.0 nm and the third order bright fringe of light with ? = 406.0 nm? (Give the absolute value of the smallest possible distance between these two fringes: the distance between bright fringes on the same side of the central bright fringe.)
|x| =
Explanation / Answer
2). ? = x d / nL ==> x = ?nL / d ......where x = distance from central bright fringe ; d = distance between the slits = 0.09 x 10^-3 m ; n = the order of the fringe ; L = length from the screen with slits to the viewing screen = 1.9 m ......so x1 = 686*10^-9 * 2*1.9 / 0.09*10^-3 = 28.964 mm ............and x2 = 406*10^-9 * 3*1.9 / 0.09*10^-3 = 25.713 mm ...so |x| = |x2-x1| = 3.25 mm ............................ 1) for dark fringe we have x = ?(n+1/2)L / d ............. given x is same for 1st dark and 1st bright fringe so......?nL / d = ?'(n+1/2)L / d ==> 738*1 = ?'(3/2) ==> ?' = 492 nm.......
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