In Young's double slit experiment, if the slit widths are in the ratio `1:9`, then the ratio of the intensity at minima to that at maxima will be

Options

A. 1
B. `9//1`
C. `4//1`
D. `3//1`

Answer

To solve the problem of finding the ratio of intensity at minima to that at maxima in Young's double slit experiment when the slit widths are in the ratio 1:9, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Relationship Between Width and Intensity**:
The intensity of light from a slit is directly proportional to the square of the amplitude of the wave produced by that slit. The amplitude is also directly proportional to the width of the slit. Thus, if the widths of the slits are in the ratio \( w_1:w_2 = 1:9 \), we can express the intensities as:
\[ I_1:I_2 = w_1^2:w_2^2 = 1^2:9^2 = 1:81 \]

2. **Finding the Amplitudes**:
The amplitudes corresponding to the intensities can be found using the square root of the intensity ratios:
\[ \frac{a_1}{a_2} = \sqrt{\frac{I_1}{I_2}} = \sqrt{\frac{1}{81}} = \frac{1}{9} \]

3. **Calculating the Ratio of Intensities at Maxima and Minima**:
In Young's double slit experiment, the intensity at maxima (\( I_{max} \)) and minima (\( I_{min} \)) can be calculated using the following formulas:
\[ I_{max} = (a_1 + a_2)^2 \]
\[ I_{min} = (a_1 - a_2)^2 \]

4. **Substituting the Amplitudes**:
Let \( a_1 = 1 \) and \( a_2 = 9 \) (based on the amplitude ratio we found):
\[ I_{max} = (1 + 9)^2 = 10^2 = 100 \]
\[ I_{min} = (1 - 9)^2 = (-8)^2 = 64 \]

5. **Finding the Ratio of Intensities**:
Now, we can find the ratio of intensity at minima to that at maxima:
\[ \frac{I_{min}}{I_{max}} = \frac{64}{100} = \frac{16}{25} \]

6. **Final Result**:
Thus, the ratio of intensity at minima to that at maxima is:
\[ \frac{I_{min}}{I_{max}} = \frac{16}{25} \]