A double-slit experiment is set up using red light (A = 706 nm). A first order b

Question

A double-slit experiment is set up using red light (A = 706 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? A = 2) A new experiment is created with the screen at a distance of 1.8 m from the slits (with spacing 0.11 mm). What is the distance between the second order bright fringe of light with A = 694 nm and the third order bright fringe of light with A = 406 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

In double slit

distance of first order bright fringe = t = L1R/d ....(2)

L1 = wavelength = 706 nm

for dark fringe

y = n*L2*R/2d.......(2)

from 1& 2

L1 = n*L2/2

for n = 3

L2 = 2L1/3 = (2/3)*706 = 470.7 nm <<------answer

+++++

2)

the positon of bright fringe from central = x = n*L*R/d

L = wavelength

R = distance between source and screen

d = slit width

n = order

for second order

x2 = 2*L*R/d

for third order = 3*LR/d

distance between second and third order

x3 - x2 = LR/d = (694*10^-9*1.8)/(0.11*10^-3) = 0.0114 m

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