Two slits separated by a distance of d = 0.180 mm are located at a distance of D

Question

Two slits separated by a distance of d = 0.180 mm are located at a distance of D = 1.64 m from a screen. The screen is oriented parallel to the plane of the slits. The slits are illuminated by a monochromatic and coherent light source with a wavelength of = 592 nm. A wave from each slit propagates to the screen. The interference pattern shows a peak at the center of the screen (m=0) and then alternating minima and maxima.

A) What is the path length difference between the waves at the second maximum (m=2) on the screen?

B) At what angle from the beam axis will the second (m=2) maximum appear? (You can safely use the small angle approximation.)

Explanation / Answer

From Young Double slit experiment we know -
= xd/nL

= wavelength of light used (m) = 592 nm
x = distance from central fringe (m)
d = distance between the slits (m) = 0.180 * 10^-3 m
n = the order of the fringe
L = length from the screen with slits to the viewing screen (m) = 1.64 m

A)
Formula for Path Difference = m*
For second maximum  Path Difference = 2 *  592 nm = 1184 nm
Path Difference = 1184 nm

B) Using Small angle approximation we infer-

  = d sin(theta) / n
sin(theta) = n /d
sin(theta) = 2 * 592 nm / 0.180 * 10^-3
theta = sin^-1 (0.00657)
theta = 0.377o

angle from the beam axis theta = 0.377o

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