Question

1. Sketch a graph of intensity versus position for a set of slits whose width is 0.0500 mm and whose separation is 0.250 mm. Assume that 2 = 633 nm and that your screen is 65.0 cm from the slits. Pay particular attention to the number of interference maxima visible in the central diffraction maximum and the width of the central diffraction maximum. Notice that you can obtain information about the "diffractor" by looking at the diffraction-interference pattern.

Answer 1

Ans.) seit width = b = 0.05mm = 0.05 x 10²m seit separation d = 0.25mm = 0.25% 10²m A = 633nm = 6338 109 m Distance of the screen from the seit =D DE 65cm 0.65m now first diffraction minima occurs when b sino = d Since o = 633x1059 0.05X163 sino =0.01266 De width of the central diffra- inteifeience maxima ocecs at etion maximum dsino = må m20,1,2,3 d since This gives the number of interference maxima up to ² m A

angie m. 0.25 X103 X0.0126.6 633 X 109 m 55. the interference maxima coresponding to m2 5 will be missing as for since 20.01266 fisst diffraction minima očeurs So the number of interference maxima up to angle o - the number of interference maxima visible in the centsat diffraction maximum = 9 one Control maxima included 4

also from the figure above we see that D' 2D tance for small co tance ~ since co is small follows from o) → D'=aD since 72x0.65 X 0.017266 0.016458 m 1.6458 cm So the graph of intensity ressus position looks like. intensity