In `YDSE` ratio of width of slit is `4:1`, then ratio of maximum to minimum intensity

Options

A. 9
B. 27
C. 3
D. 81

Answer

To solve the problem of finding the ratio of maximum to minimum intensity in the Young's Double Slit Experiment (YDSE) when the ratio of the widths of the slits is 4:1, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Intensity Relation**:
The intensity of light from each slit is directly proportional to the width of the slit. Let the width of the first slit be \(4x\) and the width of the second slit be \(x\). Therefore, the intensities can be expressed as:
\[ I_1 \propto 4x \quad \text{and} \quad I_2 \propto x \]
Thus, we can write:
\[ I_1 = 4I \quad \text{and} \quad I_2 = I \]
where \(I\) is a constant.

2. **Using the Formula for Maximum and Minimum Intensity**:
The maximum intensity \(I_{max}\) and minimum intensity \(I_{min}\) in YDSE can be calculated using the following formulas:
\[ I_{max} = \left( \sqrt{I_1} + \sqrt{I_2} \right)^2 \]
\[ I_{min} = \left( \sqrt{I_1} - \sqrt{I_2} \right)^2 \]

3. **Calculating Maximum Intensity**:
Substitute \(I_1\) and \(I_2\) into the formula for \(I_{max}\):
\[ I_{max} = \left( \sqrt{4I} + \sqrt{I} \right)^2 = \left( 2\sqrt{I} + \sqrt{I} \right)^2 = \left( 3\sqrt{I} \right)^2 = 9I \]

4. **Calculating Minimum Intensity**:
Now substitute \(I_1\) and \(I_2\) into the formula for \(I_{min}\):
\[ I_{min} = \left( \sqrt{4I} - \sqrt{I} \right)^2 = \left( 2\sqrt{I} - \sqrt{I} \right)^2 = \left( \sqrt{I} \right)^2 = I \]

5. **Finding the Ratio of Maximum to Minimum Intensity**:
Now, we can find the ratio:
\[ \frac{I_{max}}{I_{min}} = \frac{9I}{I} = 9 \]

6. **Final Result**:
The ratio of maximum to minimum intensity is:
\[ \text{Ratio} = 9:1 \]