In Young's double slit experiment the intensity at a point where path difference is $λ/6$ is I. If $I_o$ denotes the maximum intensity, then $I/I_o$ is equal to :

$\frac{3}{4}$

The correct answer is Option (3) → $\frac{3}{4}$

In young's double-slit experiment, the intensity at a point depends on the phase difference caused by the path difference between two waves.

$I=I_0\cos^2\left(\frac{\phi}{2}\right)$  ....(1)

where,

I = Intensity at the point

$I_0$ = Maximum intensity

$\phi$ = Phase difference

The phase difference ($\phi$) is related to its path difference (Δx).

$\phi =\frac{2πΔx}{λ}$

Path difference, $Δx=\frac{λ}{6}$ [given]

$\phi =\frac{12π}{λ}×\frac{λ}{6}=\frac{π}{3}$

substitute $\phi=\frac{π}{3}$ in equation (1)

$I=I_0\cos^2\left(\frac{π}{3}.\frac{1}{2}\right)$

$⇒\frac{I}{I_0}=\left(\frac{\sqrt{3}}{2}\right)^2=\frac{3}{4}$