In a Young's double-slit experiment, using monochromatic light of wavelength λ, the intensity of light at a point on the screen is $I_0$, where path difference between two interfering waves is λ. The path difference between the interfering waves at a point where the intensity is $\frac{I_0}{2}$, will be:

$λ/4$

The correct answer is Option (1) → $λ/4$

The intensity at a point is given by,

$I=I_0\cos^2\left(\frac{δ}{2}\right)$

The intensity at the point $I_0/2$ is,

$\frac{I_0}{2}=I_0\cos^2\left(\frac{δ}{2}\right)$

$⇒\frac{1}{\sqrt{2}}=\cos\left(\frac{δ}{2}\right)⇒δ=\frac{π}{2}$

The path difference $δ$ is related to the path difference d is,

$δ=\frac{2πd}{λ}$

$⇒\frac{π}{2}=\frac{2πd}{λ}$

$⇒d=\frac{λ}{4}$