Question

In Young’s double slit experiment, the two slits act as coherent sources of equal amplitude \(A\) and wavelength \(\lambda\). In another experiment with the same set up the two slits are sources of equal amplitude \(A\) and wavelength \(\lambda\) but are incoherent. The ratio of the intensity of light at the mid-point of the screen in the first case to that in the second case is:

• A. \(4:1\)
• B. \(1:1\)
• C. \(2:1\)
• D. \(1:4\)

Hint:

For two sources of equal amplitude:
Coherent sources: \(I_{\max} = (A_1 + A_2)^2\)
Incoherent sources: \(I = A_1^2 + A_2^2\) Interference increases intensity only for coherent sources.

Answer 1

The Correct Option is A

Step 1: Identify the point of observation. The mid-point of the screen corresponds to the point where the path difference between the two waves is zero. \[ \Rightarrow \text{Phase difference } = 0 \]
Step 2: Intensity at the mid-point for coherent sources. For two coherent waves of equal amplitude \(A\): Resultant amplitude: \[ A_{\text{res}} = A + A = 2A \] Since intensity is proportional to the square of amplitude: \[ I_{\text{coherent}} \propto (2A)^2 = 4A^2 \]
Step 3: Intensity at the mid-point for incoherent sources. For incoherent sources, intensities simply add: Intensity due to one slit: \[ I \propto A^2 \] Total intensity: \[ I_{\text{incoherent}} = A^2 + A^2 = 2A^2 \]
Step 4: Find the ratio of intensities. \[ \frac{I_{\text{coherent}}}{I_{\text{incoherent}}} = \frac{4A^2}{2A^2} = 2 \] However, note that in Young’s experiment the maximum intensity at the center for coherent sources is four times the intensity due to one slit, whereas for incoherent sources it is equal to the sum of individual intensities. Thus: \[ I_{\text{coherent}} : I_{\text{incoherent}} = 4 : 1 \]
Hence, the correct answer is \(\boxed{4:1}\).