Understanding the Concept:
In a Young's double slit interference experiment, light waves from two coherent sources superimpose. Let the amplitude of the wave from each individual slit be $E$. When both slits are open, the waves interfere constructively at the central maximum with a combined amplitude of:
\[
E_{\text{max}} = E + E = 2E
\]
Since the intensity of light ($I$) is directly proportional to the square of its wave amplitude ($I \propto E^2$), the maximum central intensity $I_0$ is:
\[
I_0 \propto (2E)^2 = 4E^2 \quad \cdots (1)
\]
Step 1: Analyze the circuit condition when one slit is blocked.
When one of the two slits is covered by an completely opaque shield, interference can no longer take place. Light now emerges from only a single isolated slit. Consequently, the field amplitude at the screen drops down to the individual wave amplitude:
\[
E_{\text{new}} = E
\]
Step 2: Determine the new intensity in terms of $I_0$.
The new intensity ($I'$) produced by this single active wave source is:
\[
I' \propto E^2 \quad \cdots (2)
\]
Comparing equation (2) with equation (1):
\[
I' = \frac{I_0}{4}
\]