The maximum intensity in a Young's double slit experiment is \(I_0\). Distance between the slits (\(d\)) 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 position of the point on the screen exactly opposite to one of the slits is at a distance \(y = \dfrac{d}{2}\) from the central maximum.

The path difference \(\Delta x\) at this point is given by:
\(\Delta x = \dfrac{yd}{D}\)

Substituting \(y = \dfrac{d}{2}\) and \(D = 10d\):
\(\Delta x = \dfrac{\left(\dfrac{d}{2}\right)d}{10d} = \dfrac{d^2}{20d} = \dfrac{d}{20}\)

Given that the distance between the slits is \(d = 5\lambda\), we have:
\(\Delta x = \dfrac{5\lambda}{20} = \dfrac{\lambda}{4}\)

The corresponding phase difference \(\Delta \phi\) is:
\(\Delta \phi = \dfrac{2\pi}{\lambda} \Delta x = \dfrac{2\pi}{\lambda} \left(\dfrac{\lambda}{4}\right) = \dfrac{\pi}{2}\)

The intensity at this point is given by:
\(I = I_0 \cos^2\left(\dfrac{\Delta \phi}{2}\right)\)

Substituting \(\Delta \phi = \dfrac{\pi}{2}\):
\(I = I_0 \cos^2\left(\dfrac{\pi}{4}\right) = I_0 \left(\dfrac{1}{\sqrt{2}}\right)^2 = \dfrac{I_0}{2}\)

Answer: \(\dfrac{I_0}{2}\)