In a Young's double slit experiment, the slits possess equal width,...

Let the intensity of light from each slit be $I$.

When both slits are open, the intensity at the central fringe is maximum because the path difference is zero.

The resultant intensity is given by $I_{0} = I + I + 2\sqrt{I \cdot I} \cos(0^{{}^{{}^{\circ}}}) = 4I$.

When one of the slits is blocked, there is no interference. The intensity at the central point is solely due to the light from the remaining single slit.

Thus, the new intensity is $I$.

From the relation $I_{0} = 4I$, we get $I = \dfrac{I_{0}}{4}$.