The maximum intensity in a Young's double slit experiment is $$I_0$$, distance between the slit(d) is  $$5\lambda$$, where $$\lambda$$ is the wavelength of light used the intensity of the fringe, exactly opposite to one of the slits on the screen, placed at $$D = 10d$$ is_________.

The point is described as being "exactly opposite to one of the slits." Since the two slits are separated by a distance $$d$$ symmetrically about the central axis, the distance from the central maximum to a point directly opposite one of the slits is: $$y = \frac{d}{2}$$

$$\Delta x = \frac{y \cdot d}{D} = \frac{\left(\frac{d}{2}\right) \cdot d}{10d} = \frac{d^2}{20d} = \frac{d}{20}$$

$$\Delta x = \frac{5\lambda}{20} = \frac{\lambda}{4}$$

$$\phi = \frac{2\pi}{\lambda} \Delta x = \frac{2\pi}{\lambda} \left(\frac{\lambda}{4}\right) = \frac{\pi}{2}$$

$$I = I_0 \cos^2\left(\frac{\pi / 2}{2}\right) = I_0 \cos^2\left(\frac{\pi}{4}\right)$$

$$I = I_0 \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{I_0}{2}$$