The correct answer is option 3) i.e. I = 4 I0 CONCEPT: Young's double-slit experiment: The experiment uses a screen with two pinholes that act as two coherent sources of light. [ alt="Physics ST 1 Group X Jitendra Sunny 15.5.2020 5" storage.googleapis.com tb-img production f1_p.y_madhu_13.04.20_d3.png src="//storage.googleapis.com/tb-img/production/20/05/Physics%20ST%201_Group%20X_Jitendra_Sunny_15.5.2020%205.png" style="width: 212px; height: 147px;"> As light passes through the two slits, a series of a pattern of dark and bright fringes are noticed. This is due to interference - the superposition of light waves originating from the two slits. For two coherent sources of light with intensities I1 and I2, the resultant intensity I' obtained at any point on the screen is given by I' = I1 + I2 + 2√(I1I2)cosϕ Where ϕ is the phase difference. CALCULATION: Let us assume a monochromatic light wave of intensity Is at the source, and the same is passing through the slits. ⇒ I1 = I2 = Is Case 1 - When both the slits are open We know, I' = I1 + I2 + 2√(I1I2)cosϕ For maximum resultant intensity cosϕ = 1 ⇒ Imax = I1 + I2 + 2√(I1I2) Given that the intensity of the central fringe = I. The central fringe is a bright spot and will have maximum intensity (Imax) ∴ Imax = I = I1 + I2 + 2√(I1I2) = Is + Is + 2√(IsIs) = (√Is + √Is)2 = (2√Is)2 ⇒ I = 4Is ----(1) Case 2 - When one slit is closed The interference pattern occurs only when the two slits are open. In the absence of one slit, interference will not take place and the screen will be illuminated with light of the same intensity as the source. Given that the intensity at the central point when one slit is closed = I0 Therefore, the intensity at the central point, I0 is the same as the intensity of the source, Is ⇒ I0 = Is ----(2) Substituting (2) in (1) we get I = 4I0