Two waves of same frequency and same amplitude from two monochromatic sources are allowed to superpose at a certain point. If in one case the phase difference is 0 and in other case it is `pi//2` then the ratio of the intensities in the two cases will be

Answer

To solve the problem of finding the ratio of intensities of two waves with different phase differences, we can follow these steps:

### Step 1: Understand the Problem
We have two waves of the same frequency and amplitude from two monochromatic sources. We need to find the ratio of their intensities when the phase difference is 0 in one case and \( \frac{\pi}{2} \) in the other.

### Step 2: Write the Intensity Formula
The intensity \( I \) of two superimposed waves can be expressed in terms of the maximum intensity \( I_{\text{max}} \) and the phase difference \( \phi \) between the waves:
\[ I = I_{\text{max}} \cos^2\left(\frac{\phi}{2}\right) \]

### Step 3: Calculate Intensity for Phase Difference of 0
For the first case, where the phase difference \( \phi_1 = 0 \):
\[ I_1 = I_{\text{max}} \cos^2\left(\frac{0}{2}\right) = I_{\text{max}} \cos^2(0) = I_{\text{max}} \cdot 1 = I_{\text{max}} \]

### Step 4: Calculate Intensity for Phase Difference of \( \frac{\pi}{2} \)
For the second case, where the phase difference \( \phi_2 = \frac{\pi}{2} \):
\[ I_2 = I_{\text{max}} \cos^2\left(\frac{\frac{\pi}{2}}{2}\right) = I_{\text{max}} \cos^2\left(\frac{\pi}{4}\right) \]
We know that \( \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \), so:
\[ I_2 = I_{\text{max}} \left(\frac{1}{\sqrt{2}}\right)^2 = I_{\text{max}} \cdot \frac{1}{2} \]

### Step 5: Find the Ratio of Intensities
Now, we can find the ratio of the two intensities:
\[ \frac{I_1}{I_2} = \frac{I_{\text{max}}}{\frac{I_{\text{max}}}{2}} = \frac{I_{\text{max}} \cdot 2}{I_{\text{max}}} = 2 \]

Thus, the ratio of the intensities \( I_1 : I_2 \) is:
\[ I_1 : I_2 = 2 : 1 \]

### Final Answer
The ratio of the intensities in the two cases is \( 2 : 1 \).

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