Correct Answer: I₀
In Young's double slit experiment, the intensity at any point on the screen depends on the path difference between the waves coming from the two slits.
At a point on the screen directly in front of one slit, the perpendicular distance from one slit is zero, while the distance from the other slit is equal to the slit separation d.
Given:
d = 2 mm = 2 × 10−3 m D = 10 m λ = 6000 Å = 6 × 10−7 m
Path difference at that point:
Δ = d² / (2D)
Δ = (2 × 10−3)² / (2 × 10)
Δ = 4 × 10−6 / 20
Δ = 2 × 10−7 m
Now compare path difference with wavelength:
Δ = (1/3) λ
This phase difference does not correspond to complete constructive or destructive interference. Hence, the interference term averages out.
Therefore, the resultant intensity at that point is simply equal to the intensity due to one slit alone.
Thus, the intensity of light on the screen in front of one of the slits is I₀.
Updated for JEE Main 2026: This PYQ is important for JEE Mains, JEE Advanced and other competitive exams. Practice more questions from this chapter.