Concept understanding — Superposition and Interference of Light
Interference is the redistribution of light intensity that results when two light waves overlap (are superposed) at a point, producing bright regions (increased intensity) at some points and dark regions (decreased intensity) at others, in a way that a single wave alone never would. If two waves of the same frequency and amplitudes a1,a2 arrive at a point with a constant phase difference ϕ between them, y1=a1sinωt and y2=a2sin(ωt+ϕ), their resultant is y=Asin(ωt+θ) with resultant amplitude A2=a12+a22+2a1a2cosϕ; since intensity I∝A2, the resultant intensity is I∝I1+I2+2I1I2cosϕ. Constructive interference (maximum brightness) occurs when ϕ=0,±2π,±4π,…, giving Imax∝(a1+a2)2∝I1+I2+2I1I2; destructive interference (minimum brightness, possibly complete darkness) occurs when ϕ=±π,±3π,…, giving Imin∝(a1−a2)2∝I1+I2−2I1I2. For the special case of equal amplitudes (a1=a2=a, I1=I2=I0), this simplifies …
Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.
CBSE 2020Set ANNUAL1 markMCQ
Q.Two light waves from slit S1 and S2 on reaching points P and Q on a screen in Young's double slit experiment have a path difference zero and λ/4 respectively. The ratio of light intensities at P and Q will be :
(a) 4 : 1
(b) 3 : 2
(c) 2:1
(d) 2 : 1
›Reveal solutionSolution
Using the two-source interference formula I=I0+I0+2I0cosδ at phase differences δ=0 and δ=π/2 gives intensities 4I0 and 2I0, a ratio of 2:1.
Working
For two coherent sources of equal intensity I0 each, the resultant intensity at a point with phase difference δ is
I=I0+I0+2I0I0cosδ=2I0(1+cosδ)
At P (path difference = 0): phase difference δ=0, cosδ=1