In YDSE, $β$ is the fringe width and $I_0\, \cos^2 (πy/β)$ is the intensity at a distance y from central bright fringe. In this case $I_0$ is
Answer & explanation
Correct answer: option 3
Phase difference = $\phi = \frac{2π}{λ}$ × path difference
$=\frac{2π}{λ}(Δx)$
Let a = amplitude at screen due to each slit
⇒ $I_0$ = Intensity at central bright fringe
$= k (2a)^2 = 4ka^2$, k = constant
For phase difference $\phi$, A = amplitude = $2a\, \cos (\phi/2)$
$∴ I = kA^2 = k(2a)^2\, \cos^2(\phi/2) = 4ka^2\, \cos^2(\phi/2)$
$=I_0\cos^2(\frac{π}{λ}Δx)$
$=I_0\cos^2(\frac{π}{λ}.\frac{yd}{D})$
$=I_0\cos^2(\frac{πy}{β})$