Question

In a Young's double slit experiment, the intensity at the center of the screen is I. If one of the slits is closed, the intensity at the center of the screen now will be:

• A. \(\dfrac{1}{2}\)
• B. 1
• C. \(\dfrac{1}{4}\)
• D. \(\dfrac{1}{3}\)

Hint:

Two waves in phase give \(4I_0\). One wave gives \(I_0 = I/4\).

Answer 1

The Correct Option is C

Step 1: Set Up with Intensity Formula:
For two coherent waves of equal intensity $I_0$ and phase difference $\phi$, the resultant is $I = 4I_0\cos^2(\phi/2)$.

Step 2: Center of the Screen:
At the center $\phi = 0$, so $\cos^2 0 = 1$. This gives $I = 4I_0$, which the question calls $I$. Hence $I_0 = I/4$.

Step 3: Close One Slit:
Interference stops. The screen center is lit by one slit only, and its intensity is $I_0$. Therefore the new intensity is $I/4$.

Step 4: Read Off the Option:
$I/4$ is the third printed option.

Final Answer:
\[\boxed{\frac{I}{4}\ \text{(option 3)}}\]