In stand Young's double - slit experiment , intensity of light at the centre of screen is found to be `I_1`. There is another point on the screen where difference beteen the waves is `lambda//4` and intensity of light at this point is `I_2`. Calculate `I_1//I_2`.
`{:(0,1,2,3,4,5,6,7,8,9):}`

Explanation

To solve the problem, we need to find the ratio \( \frac{I_1}{I_2} \) where \( I_1 \) is the intensity at the center of the screen and \( I_2 \) is the intensity at a point where the path difference is \( \frac{\lambda}{4} \).

### Step-by-Step Solution:

1. **Understanding the Central Maximum Intensity**:
At the center of the screen (central maximum), the phase difference \( \phi \) is 0. The formula for intensity at any point in Young's double-slit experiment is given by:
\[ I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos(\phi) \]
Since at the center \( \phi = 0 \), we have:
\[ I_1 = I + I + 2\sqrt{I \cdot I} \cos(0) = 2I + 2I = 4I \]
Thus, we can conclude:
\[ I_1 = 4I \]

2. **Calculating Intensity at Point P**:
At point P, the path difference is \( \frac{\lambda}{4} \). We need to convert this path difference into phase difference:
\[ \text{Phase difference} \, \phi = \frac{2\pi}{\lambda} \times \frac{\lambda}{4} = \frac{\pi}{2} \]
Now, substituting \( \phi = \frac{\pi}{2} \) into the intensity formula:
\[ I_2 = I + I + 2\sqrt{I \cdot I} \cos\left(\frac{\pi}{2}\right) \]
Since \( \cos\left(\frac{\pi}{2}\right) = 0 \), we have:
\[ I_2 = 2I \]

3. **Finding the Ratio \( \frac{I_1}{I_2} \)**:
Now we have \( I_1 = 4I \) and \( I_2 = 2I \). Therefore, the ratio is:
\[ \frac{I_1}{I_2} = \frac{4I}{2I} = 2 \]

4. **Final Answer**:
Thus, the ratio \( \frac{I_1}{I_2} \) is:
\[ \frac{I_1}{I_2} = 2 \]