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 = 2.30 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 = 623 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.What is the path length difference between the waves at the second maximum (m=2) on the screen?

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

Explanation / Answer

Here ,

wavlength = 623 nm

for the second maximum on the screen , m = 2

as path difference = m * wavelength

path difference = 2 * 623

path difference = 1246 nm

path difference = 1.246 *10^-6 m

the path length difference between the waves at the second maximum (m=2) on the screen is 1.246 *10^-6 m

for the angle

d*sin(theta) = m*wavelrngth

0.180 * 10^-3 * sin(theta) = 1.246 *10^-6

theta = 0.396 degree

the angle from the beam axis will the second (m=2) maximum appear is 0.396 degree

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