Three identical slits are illuminated by monochromatic light. The top and center slits are separated by a?
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distance 2d; and the center and bottom slits are separated by distance d. Derive an expression for light intensity on a distant screen as a function of phase difference between center and bottom slits; plot graph of this function; and roughly estimate the angular position of the first interference minimum. Give your answer in degrees rounding it to the whole integer.
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Answer:
Since the width of each slit is not given, I will assume that it is small compared to the wavelength. You also need the phase differenve between center and top slits. The field at a point at distance r1 from the top slit, r2 from the center slit and r3 from the bottom slit is (in phasor notation): E= e^{i (φ1 + k r1)/r1+e^{i (φ2 + k r2)/r2+e^{i (φ3 + k r3)/r3, where k=2π/λ. The intensity is I = EE*, or, taking r1 ≈r2≈r2≈z I = (3+2 Cos[k (r1-r2)+φ1-φ2]+2 Cos[k (r1-r3)+φ1-φ3]+ 2 Cos[k (r2-r3)+φ2-φ3])/z^2 where r1 = sqrt[z^2 + (x - 2 d)^2]≈ z-2 d x/z r2 = sqrt[z^2 + x^2] ≈ z r3 = sqrt[z^2 + (x + d)^2]≈z+d x/z where z is the distance to the screen. The above approximate expressions are in the Fraunhofer limit, yielding I= 3+2 Cos[(2 d k x)/z-φ1+φ2]+2 Cos[(3 d k x)/z-φ1+φ3] +2 Cos[(d k x)/3+2 z-φ2+φ3] Given φ1-φ2 and φ1-φ3 we can determine the central maximum numnerically, but without that information it is not possible.
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