In Young's double slit experiment, monochromatic light of...

Let intensity of light on screen due to each slit is ${I}_{0}$.

So internity at centre of screen is $4{I}_{0}$(as $cosϕ = 1$). 

Intensity at distance $y$ from centre, 

$I = {I}_{0} + {I}_{0} + 2\sqrt{{I}_{0}{I}_{0}}cosϕ$

As we know, {I}_\max = 4{I}_{0}. Therefore,

⇒ \dfrac{{I}_\max}{2} = 2{I}_{0} = 2{I}_{0} + 2{I}_{0}cosϕ

$⇒ cosϕ = 0$

$⇒ ϕ = \dfrac{\pi}{2}$

Hence, $k\Deltax = \dfrac{\pi}{2}$

$⇒ \dfrac{2\pi}{\lambda}\left(dsin\theta\right) = \dfrac{\pi}{2}$

$⇒ \dfrac{2}{\lambda}d \times \dfrac{y}{D} = \dfrac{1}{2}$

$⇒ y = \dfrac{\lambdaD}{4d} = \dfrac{5 \times {10}^{- 7} \times 1}{4 \times {10}^{- 3}}$

$= 125 \times {10}^{- 6}$

$= 125$