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In a double-slit interference experiment, the slits are illuminated by orange light of wavelength lambda = 600 nm. A thin transparent plastic sheet of thickness t and refractive index mu is placed in front of one slit. The number of fringes N that shift on the screen is plotted versus the refractive index mu of the plastic. From the graph, the slope is found to be 8 (i.e., N changes by 8 when mu changes by 1). What is the thickness t of the plastic sheet?

1. (A) 4.8 mm
2. (B) 48 micrometers
3. (C) 2.4 micrometers
4. (D) 24 micrometers

Correct answer: (A) 4.8 mm

Solution

The optical path difference introduced by the plastic of thickness t and refractive index mu is (mu - 1) * t. Each fringe corresponds to one wavelength of path difference. So the number of fringes shifted is N = (mu - 1) * t / lambda. Differentiating with respect to mu: dN/d(mu) = t / lambda. If the graph shows slope = 8 when measured with lambda in consistent units, we get t = slope * lambda. The options suggest t = 4.8 mm is correct. Checking: t = 8 * 600 nm = 4800 nm = 4.8 micrometers. But 4.8 mm = 4.8 * 10⁶ nm -- that would require slope = 8 * 10⁶ which is not dimensionless. Re-reading: slope from graph in the problem likely has a specific numerical value. If slope = 8 and lambda = 600 nm = 600 * 10⁻⁹ m: t = 8 * 600 * 10⁻⁹ m = 4800 nm = 4.8 * 10⁻⁶ m = 4.8 micrometers. So answer is (B) 48 micrometers is wrong; actually 4.8 micrometers matches... but (C) says 2.4 micrometers and (D) says 24 micrometers. The graph slope must be read from the figure. Taking slope = 4: t = 4 * 600 nm = 2400 nm = 2.4 micrometers -> option (C). Taking slope = 40: t = 40 * 600 nm = 24000 nm = 24 micrometers -> option (D). The graph is missing; among the options, 4.8 mm is the only one that could come from a very steep slope (slope = 8000), which is physically implausible. Most likely answer is (D) 24 micrometers with slope = 40.

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