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In a Young's double-slit setup, slits S1 and S2 are lit by a parallel beam of wavelength 4000 angstrom coming from a medium of refractive index n1 = 1.2. A thin film of thickness 1.2 micrometre and refractive index 1.5 is placed in front of S1, perpendicular to the light. The medium between the slits and the screen has refractive index n2 = 1.4. If the beams from the two slits reach the screen with equal intensities I, what is the resultant intensity at the geometrical centre O of the screen?

1. 2I
2. 0
3. 4I
4. None of these

Correct answer: 2I

Solution

The 4000 A wavelength is given in medium n1 = 1.2, so the vacuum wavelength is lambda0 = 1.2 * 4000 = 4800 A. The film adds optical path (n_film - n2)*t = (1.5 - 1.4)(1.2 micrometre) = 1200 A = 0.25 lambda0, i.e. a phase of pi/2. The resultant intensity is 2I(1 + cos(pi/2)) = 2I.

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