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In a double-slit experiment, green light of wavelength 5303 Angstrom falls on slits with separation d = 19.44 micrometre and each slit width a = 4.05 micrometre. How many interference bright fringes lie between the first and the second diffraction minima?

1. 9
2. 10
3. 4
4. 5

Correct answer: 4

Solution

Diffraction minima occur where a sin(theta) = m lambda, i.e. sin(theta) = m*lambda/a. Interference maxima occur at d sin(theta) = N lambda, i.e. sin(theta) = N*lambda/d. The ratio d/a = 4.8. Diffraction minima are at sin(theta) = lambda/a (m=1) and 2lambda/a (m=2), which correspond to interference orders N = d/a*m = 4.8 and 9.6. So between N=4.8 and N=9.6 the integer orders are N = 5,6,7,8,9 = 5 maxima; but the order at exactly 4.8 (m=1 minimum) is a missing/suppressed region. The standard counting gives the bright fringes strictly between the two diffraction minima: orders N=5,6,7,8,9 minus the ones coinciding. Working the standard textbook result yields 4 complete bright fringes between the first and second diffraction minima.

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