Question

The light waves from two coherent sources have same intensity \(I_1 = I_2 = I_0\). In interference pattern the intensity of light at minima is zero. What will be the intensity of light at maxima ?

• A. \(2 I_0\)
• B. \(5 I_0\)
• C. \(4 I_0\)
• D. \(I_0\)

Hint:

For interference of two coherent sources with equal intensity \(I_0\), the resulting intensity pattern varies from \(I_{min}=0\) to \(I_{max}=4I_0\). This is a fundamental result in wave optics. The total energy is conserved; it is just redistributed in space to form bright and dark fringes. The average intensity over one cycle is \(2I_0\).

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We are considering the interference of two coherent light waves of equal intensity, \(I_0\). We are given that the minimum intensity is zero, and we need to find the maximum possible intensity.
Step 2: Key Formula or Approach:
The resultant intensity \(I_R\) of two interfering waves with intensities \(I_1\) and \(I_2\) and a phase difference \(\phi\) is given by:
\[ I_R = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos\phi \] - For maximum intensity (constructive interference), \(\cos\phi = 1\).
- For minimum intensity (destructive interference), \(\cos\phi = -1\).
The formulas for maximum and minimum intensities are:
\[ I_{\text{max}} = (\sqrt{I_1} + \sqrt{I_2})^2 \] \[ I_{\text{min}} = (\sqrt{I_1} - \sqrt{I_2})^2 \] Step 3: Detailed Explanation:
We are given that \(I_1 = I_2 = I_0\).
First, let's verify the given information about the minimum intensity.
\[ I_{\text{min}} = (\sqrt{I_0} - \sqrt{I_0})^2 = (0)^2 = 0 \] This matches the problem statement that the intensity at minima is zero. This happens when the amplitudes of the waves are equal, which is consistent with their intensities being equal.
Now, let's calculate the maximum intensity using the formula.
\[ I_{\text{max}} = (\sqrt{I_1} + \sqrt{I_2})^2 \] Substitute \(I_1 = I_2 = I_0\):
\[ I_{\text{max}} = (\sqrt{I_0} + \sqrt{I_0})^2 \] \[ I_{\text{max}} = (2\sqrt{I_0})^2 \] \[ I_{\text{max}} = 4(\sqrt{I_0})^2 = 4I_0 \] Step 4: Final Answer:
The intensity of light at maxima will be \(4I_0\).