In a Young's double slit experiment, λ = 500 nm , d = 1.0 mm , and...

The intensity in Young's double slit experiment is given by I = I_\max \cos^2 \left( \dfrac{\phi}{2} \right).<br><br>For the intensity to be half of the maximum intensity,I = \dfrac{I_\max}{2}.<br><br>\dfrac{I_\max}{2} =I_\max \cos^2 \left( \dfrac{\phi}{2} \right)$<br><br>$\cos \left( \dfrac{\phi}{2} \right) = \dfrac{1}{\sqrt{2}}$<br><br>For the minimum distance,$\dfrac{\phi}{2} = \dfrac{\pi}{4} \Rightarrow \phi = \dfrac{\pi}{2}$.<br><br>The phase difference$\phi$is related to the path difference$\Delta x$by$\phi = \dfrac{2\pi}{\lambda} \Delta x$.<br><br>$\dfrac{\pi}{2} = \dfrac{2\pi}{\lambda} \Delta x \Rightarrow \Delta x = \dfrac{\lambda}{4}$<br><br>Using the relation$\Delta x = \dfrac{y d}{D}$, we get:<br><br>$\dfrac{y d}{D} = \dfrac{\lambda}{4} \Rightarrow y = \dfrac{\lambda D}{4 d}$<br><br>Substituting the given values:<br><br>$y = \dfrac{500 \times 10^{-9} \times 1.0}{4 \times 1.0 \times 10^{-3}}$<br><br>$y = 125 \times 10^{-6} \text{ m} = 1.25 \times 10^{-4} \text{ m}