A double-slit experiment is set up using red light (I = 719 nm). A first order b

Question

A double-slit experiment is set up using red light (I = 719 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? I = nm A new experiment is created with the screen at a distance of 1.8 m from the slits (with spacing 0.1 mm). What is the distance between the second order bright fringe of light with I = 695 nm and the third order bright fringe of light with I=413 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| = m

Explanation / Answer

Here ,

a) wavelength of red light , l = 719 nm

for the dark fringe at the same location

as y = m * wavelength * D/d

719 * 1 * D/d = wavelength * 1.5 * D/d

wavelength = 479.3 nm

the wavelength of the visible light is 479.3 nm

b)

Distance , D = 1.8 m

slit spacing , d = 0.1 mm = 1 *10^-4 m

l1 = 695 nm

l2 = 413 nm

distance between the second order and third order bright fringe = -3 * L2 * D/d + 2 * L1 * D/d

distance between the second order and third order bright fringe = -1.8/(1 *10-4) * (3 * 413 - 2 * 695) * 10^-9

distance between the second order and third order bright fringe = 4.53 *10^-8 m

the distance between the second order and third order bright fringe is 4.53 *10^-8 m

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