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In a double-slit experiment, one slit is covered with a thin glass plate of refractive index 1.4 and the other with a thin glass plate of refractive index 1.7. Both plates have equal thickness. The central bright fringe position is now occupied by the 5th bright fringe. If the wavelength is 4800 Angstrom, find the common thickness t of the plates.

1. 8 micrometre
2. 4 micrometre
3. 2 micrometre
4. 16 micrometre

Correct answer: 8 micrometre

Solution

Introducing a plate of index mu and thickness t into one path adds an optical path (mu - 1)t. With both slits covered, the net additional path difference is (mu2 - mu1)t = (1.7 - 1.4)t = 0.3t. The shift equals 5 fringes, so (mu2 - mu1)t = 5*lambda. Thus t = 5*lambda/0.3 = 5*(4800e-10)/0.3 = (2.4e-6)/0.3 = 8e-6 m = 8 micrometre.

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