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 widths of the two slits be \( A_1 \) and \( A_2 \). Given that the ratio of the slit widths is \( A_1 : A_2 = 9 : 4 \), we can express the widths as:
\[ A_1 = 9k \quad \text{and} \quad A_2 = 4k \]
where \( k \) is a constant.

### Step 2: Relate the amplitudes to the slit widths
In a Young's double slit experiment, the amplitude of the wave from each slit is proportional to the width of the slit. Therefore, the amplitudes \( E_1 \) and \( E_2 \) from the two slits can be expressed as:
\[ E_1 \propto A_1 = 9k \quad \text{and} \quad E_2 \propto A_2 = 4k \]
Thus, we can write:
\[ E_1 = 9k \quad \text{and} \quad E_2 = 4k \]

### Step 3: Calculate the intensities
The intensity \( I \) is proportional to the square of the amplitude:
\[ I_1 \propto E_1^2 = (9k)^2 = 81k^2 \]
\[ I_2 \propto E_2^2 = (4k)^2 = 16k^2 \]

### Step 4: Calculate the maximum intensity
The maximum intensity \( I_{\text{max}} \) occurs when the waves from both slits constructively interfere:
\[ I_{\text{max}} = I_1 + I_2 = 81k^2 + 16k^2 = 97k^2 \]

### Step 5: Calculate the minimum intensity
The minimum intensity \( I_{\text{min}} \) occurs when the waves from both slits destructively interfere:
\[ I_{\text{min}} = |I_1 - I_2| = |81k^2 - 16k^2| = |65k^2| = 65k^2 \]

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

### Final Answer
Thus, the ratio of intensity at minima to that at maxima is:
\[ \frac{I_{\text{min}}}{I_{\text{max}}} = \frac{65}{97} \]