Two slits separated by a distance of d = 0.130 mm are located at a distance of D

Question

Two slits separated by a distance of d = 0.130 mm are located at a distance of D = 2.09 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 lambda= 618 nm. A wave from each slit propagates to the screen. The interference pattern shows a peak at the center of the screen (m=0) What is the path length difference between the waves at the first maximum (m=1) on the screen? At what angle from the beam axis will the first (m=1) maximum appear? (You can safely use the small angle approximation.)

Explanation / Answer

distance between the two slits = 0.130 mm = 0.130 * 10-3 m

distance between the screen and slit D = 2.09 m

wavelength of light = 618 * 10-9 m

for costructive interfrence path difference = m lambda (where m = 0,1,2,....)

d sin theta = m lambda

for m=1

d sin theta = lambda

= 618 * 10-9 m = 6.18 * 10-4 mm

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The formula for double slit to find the location of maxima is

y = m lambda L / d

y = (1)(618 * 10-9)(2.09) / (0.130 * 10-3)

y = 9.94 * 10-3 m

tan theta = y / L = 9.94 * 10-3 / 2.09

theta = 0.2720

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