Exams › JEE Advanced › Physics

In a Young's double-slit experiment, the upper slit is covered by a thin sheet of thickness t and refractive index 2*mu, and the lower slit by a sheet of thickness 2t and refractive index mu. Find the position y of the central maximum on the screen (D = slit-to-screen distance, d = slit separation; take downward as negative).

1. Zero
2. tD/d
3. -tD/d
4. None of these

Correct answer: Zero

Solution

Extra optical path from the upper slit's sheet: (2*mu - 1)*t. Extra optical path from the lower slit's sheet: (mu - 1)*(2t) = (2*mu - 2)*t. The net extra path difference introduced between the two slits is (2*mu - 1)t - (2*mu - 2)t = t (not zero), so in general the central maximum would shift. However, comparing the two extra paths: upper = (2*mu - 1)t and lower = (2*mu - 2)t; the upper path is longer by t, meaning the central maximum shifts. Re-evaluating: the path-difference at a point y is delta = (yd/D) + [path added at lower] - [path added at upper]. Setting delta = 0 for the central maximum: yd/D = (2*mu-1)t - (2*mu-2)t = t, giving y = tD/d only if upper has the larger added path. Because (2*mu-1)t > (2*mu-2)t, the central maximum shifts toward the upper slit by y = tD/d in magnitude. The correct nonzero position depends on the sign convention; with the given options the intended balanced design makes the standard answer y = tD/d. Given the ambiguity in sign convention, the magnitude is tD/d.

Related JEE Advanced Physics questions

• What factor influences the visibility of fringes in an interference pattern?
• When the screen is moved further along the x-axis away from the source, which of the following statements is false?
• In the Young's double slit experiment using a monochromatic light of wavelength λ, the path difference (in terms of an integer n) corresponding to any point having half the peak intensity is
• Using the expression 2d sin θ = λ, one calculates the values of d by measuring the corresponding angles θ in the range 0 to 90°. The wavelength λ is exactly known and the error in θ is constant for all values of θ. As θ increases from 0°:
• Two coherent point sources separated by a distance d = 5 micrometers emit light of wavelength lambda = 2 micrometers in phase. A circular wire of radius R = 20 micrometers is placed symmetrically around the two sources (sources lie along a diameter). Points A and B are at the ends of the diameter along the line joining the sources; points C and D are at the ends of the diameter perpendicular to the line joining the sources. Which of the following is correct?
• Young's double slit experiment is performed in a medium of refractive index mu1 = 4/3. A thin transparent slab of refractive index mu2 = 3/2 and thickness t = 8 micrometers is placed in front of slit S2. The magnitude of the optical path difference (with respect to the liquid medium) at the central point O is:

⚔️ Practice JEE Advanced Physics free + battle 1v1 →