Slit widths in a young double slit experiment are in the ratio `9:4`. Ratio of intensity at minima to that at maxima is

Options

A. `4:9`
B. `16:81`
C. `1:25`
D. `1:16`

Answer

To solve the problem of finding the ratio of intensity at minima to that at maxima in a Young's double slit experiment with slit widths in the ratio of 9:4, we can follow these steps:

### Step 1: Define the slit widths
Let the width of the first slit be \( A_1 = 9k \) and the width of the second slit be \( A_2 = 4k \), where \( k \) is a constant.

### Step 2: Determine the intensities at the slits
The intensity at each slit is directly proportional to the square of the width of the slit. Therefore, we can express the intensities as:
- \( I_1 \propto A_1^2 = (9k)^2 = 81k^2 \)
- \( I_2 \propto A_2^2 = (4k)^2 = 16k^2 \)

### Step 3: Establish the ratio of intensities
Now, we can find the ratio of the intensities:
\[ \frac{I_1}{I_2} = \frac{81k^2}{16k^2} = \frac{81}{16} \]

### Step 4: Calculate the maximum intensity \( I_{\text{max}} \)
The maximum intensity \( I_{\text{max}} \) in a double slit experiment is given by:
\[ I_{\text{max}} = I_1 + I_2 = 81k^2 + 16k^2 = 97k^2 \]

### Step 5: Calculate the minimum intensity \( I_{\text{min}} \)
The minimum intensity \( I_{\text{min}} \) is given by:
\[ I_{\text{min}} = |I_1 - I_2| = |81k^2 - 16k^2| = 65k^2 \]

### Step 6: Find the ratio of intensities at minima to maxima
Now, we can find the ratio of the minimum intensity to the maximum intensity:
\[ \frac{I_{\text{min}}}{I_{\text{max}}} = \frac{65k^2}{97k^2} = \frac{65}{97} \]

### Step 7: Simplify the ratio
Since \( k^2 \) cancels out, we have:
\[ \frac{I_{\text{min}}}{I_{\text{max}}} = \frac{65}{97} \]

### Final Answer
Thus, the ratio of intensity at minima to that at maxima is \( \frac{65}{97} \).

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