In Young's double slit experiment, monochromatic light of wavelength 5000 Å is used. The slits are 1.0 mm apart and screen is placed at 1.0 m away from slits. The distance from the centre of the screen where intensity becomes half of the maximum intensity for the first time is ______ $$\times 10^{-6}$$ m.

Find the distance from center where intensity becomes half of maximum for the first time in Young's double slit experiment.

In Young's experiment, the intensity at a point on the screen is:

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

where $$\phi$$ is the phase difference.

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

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

$$ \cos\left(\frac{\phi}{2}\right) = \frac{1}{\sqrt{2}} $$

$$ \frac{\phi}{2} = \frac{\pi}{4} \implies \phi = \frac{\pi}{2} $$

$$ \phi = \frac{2\pi}{\lambda} \times \Delta x $$

where $$\Delta x$$ is the path difference. So:

$$ \frac{\pi}{2} = \frac{2\pi}{\lambda} \times \Delta x \implies \Delta x = \frac{\lambda}{4} $$

Path difference $$\Delta x = \frac{yd}{D}$$ where $$y$$ is distance from center, $$d$$ is slit separation, $$D$$ is screen distance.

$$ \frac{\lambda}{4} = \frac{yd}{D} \implies y = \frac{\lambda D}{4d} $$

$$\lambda = 5000$$ A $$= 5 \times 10^{-7}$$ m, $$d = 1.0$$ mm $$= 10^{-3}$$ m, $$D = 1.0$$ m.

$$ y = \frac{5 \times 10^{-7} \times 1.0}{4 \times 10^{-3}} = \frac{5 \times 10^{-7}}{4 \times 10^{-3}} = 1.25 \times 10^{-4} \text{ m} = 125 \times 10^{-6} \text{ m} $$

The answer is 125 $$\times 10^{-6}$$ m.