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: Understanding the Question:
At the center of the screen the path difference is zero. Both waves arrive in phase. When one slit is closed, only one wave remains.

Step 2: Key Formula or Approach:
If each slit alone gives intensity \(I_0\), the two waves in phase have amplitude \(2a\), where \(a\) is the amplitude of one wave. Intensity goes as the square of the amplitude.

Step 3: Both Slits Open:
\[ I = k(2a)^2 = 4ka^2 = 4I_0 \] So \(I_0 = I/4\).

Step 4: One Slit Closed:
Only one wave of amplitude \(a\) reaches the center. The intensity is \[ I' = ka^2 = I_0 = \frac{I}{4} \]

Step 5: Checking Each Option:
Option 1 (I/2) is what you would get if intensity were linear in amplitude. That is wrong. Option 2 says no change. Wrong. Option 4 (I/3) has no basis. Option 3 (I/4) matches.

Final Answer:
The intensity at the center becomes one fourth of I, so option 3 is correct. \[ \boxed{\frac{I}{4}} \]