Two slits separated by a distance of d = 0.11 mm are located at a distance of D

Question

Two slits separated by a distance of d = 0.11 mm are located at a distance of D = 0.87 m from a screen. The screen is oriented parallel to the plane of the slits. The slits are illuminated by a coherent light source with a wavelength of ? = 533 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 pathlength difference between the waves at the first maximum (m=1) on the screen?
What is the pathlength difference between the waves at the first minimum (m=0) on the screen?
Calculate the distance on the screen between the central maximum (m=0) and the first maximum (m=1) (you can assume sin(?) = tan(?) = ? with ? expressed in radians).

Explanation / Answer

a) Path difference (for m=1)= 1 to combine constructively again

so path difference is 544nm

b) path difference (for m=o), d = ½ to combine (for the first time) destructively.

so path difference is 272nm

c) The mean separation (y) of double-slit fringes is

y = D/d

here D= slit to screen distance,

d = slit separation

y = (544x10-9) x (0.8m) / (0.11x10-3m)

y = 4.352x10-7m / 0.11x10-3

y = 3.96x10-3m or 3.96mm

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