In Young's double slit experiment, the intensity on screen at a...

The phase difference $\phi$ is related to the path difference $\Delta x$ by the formula:

$\phi = \dfrac{2\pi}{\lambda} \Delta x$

For a path difference of $\Delta x = \dfrac{\lambda}{4}$, the phase difference is:

$\phi = \dfrac{2\pi}{\lambda} \times \dfrac{\lambda}{4} = \dfrac{\pi}{2}$

The intensity $I$ at any point on the screen in Young's double slit experiment is given by:

I = I_\max \cos^2 \left(\dfrac{\phi}{2}\right)<br><br>Substituting\phi = $\dfrac{\pi}{2}:<br><br>$ I = I_\max \cos^2\left(\dfrac{\pi}{4}\right) =I_\max \left(\dfrac{1}{\sqrt{2}}\right)^2 = \dfrac{I_\max}{2}<br><br>We are given that the intensity at this point is \dfrac{K}{4}. Therefore:<br><br>\dfrac{I_\max}{2} = \dfrac{K}{4} \Rightarrow I_\max = $\dfrac{K}{2}<br><br>Now,$ for $a point where the path difference is \Delta$ x = \lambda$, the phase difference is:<br><br>$\phi' = $\dfrac{2\pi}{\lambda} \times \lambda =2\pi<br><br>The intensity$ I'$at this point will be:<br><br>$I' = I_\max \cos^2\left(\dfrac{2\pi}{2}\right) =I_\max \cos^2(\pi) =I_\max (1)^2 = I_\max

Substituting the value of I_\max:

$I' =$\dfrac{K}{2}

Answer: $ K/2