Problem

Problem

A parallel beam of monochromatic light is used in a Young's double slit experiment. The slits are separated by a distance dd and the screen is placed parallel to the plane of the slits. If the incident beam makes an angle θ=sin⁡−1(λ2d)\theta = \sin^{-1}\left(\frac{\lambda}{2d}\right) with the normal to the plane of the slits, there will be.....at the centre P0P_0 of the pattern. (a) 14\frac{1}{4} the maximum intensity (b) half the maximum intensity (c) bright (d) dark
Intuitive explanation

In Young's double slit experiment, the interference pattern depends on the path difference between light from the two slits—but when the incident beam is tilted, it adds an extra path difference before the light even reaches the slits. The key insight is that this incident angle can exactly compensate for the geometric path difference that would normally create a bright fringe at the center, shifting which order of interference appears there. You need to calculate the total path difference (incident tilt plus geometric separation) to determine what actually happens at P0P_0.

Intuitive explanation

In Young's double slit experiment, the interference pattern depends on the path difference between light from the two slits—but when the incident beam is tilted, it adds an extra path difference before the light even reaches the slits. The key insight is that this incident angle can exactly compensate for the geometric path difference that would normally create a bright fringe at the center, shifting which order of interference appears there. You need to calculate the total path difference (incident tilt plus geometric separation) to determine what actually happens at P0P_0.

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