A double slit diffraction pattern is formed on a distant screen using mercury gr

Question

A double slit diffraction pattern is formed on a distant screen using mercury green light of wavelength 546.1 nm. Each slit has a width of 0.1 mm. The pattern shows that the 4th order of the interference maximum is missing. a) What is the separation of the slits? b) What is the intensity of the 2nd order interference maximum relative to the zero order maximum? c) Show that the number of bright fringes seen under the central diffraction peak is always given by 2(a/b)-1 , where a is the separation of the slits and b is the slit width.

Explanation / Answer

a)

Conditions for diffraction minima and interference maxima are asin = m and bsin = p respectively, where a is the slit width and b is the slit separation. Now for a certain value of when the above two conditions are satisfied simultaneously the corresponding interference maximum becomes missing. So the condition for missing order is b a = pm.Asp = 4, m = 1, so b = 4a = 0.4 mm.

We know that for double slit diffraction intensity distribution is

I = [Imax sin^2()/2]*cos^2

= asin / ,

, = bsin /

for second  2nd order interference maximum = 2, so = /2

I2 / Imax = 4 / pi^2 = 0.405

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