In Young's double slit experiment the ratio of intensities at the position of maxima and minima is 9/1. the ratio of amplitudes of light waves is

Options

A. `2/1`
B. `3/1`
C. `4/1`
D. `1/4`

Explanation

To solve the problem, we need to find the ratio of amplitudes of light waves when the ratio of intensities at the position of maxima and minima is given as 9:1 in Young's double slit experiment.

### Step-by-Step Solution:

1. **Understand the relationship between intensity and amplitude**:
The intensity (I) of a wave is proportional to the square of its amplitude (A). This can be expressed as:
\[ I \propto A^2 \]

2. **Set up the equations for maxima and minima**:
In Young's double slit experiment:
- The intensity at maxima (\(I_{max}\)) is given by:
\[ I_{max} = (A_1 + A_2)^2 \]
- The intensity at minima (\(I_{min}\)) is given by:
\[ I_{min} = (A_1 - A_2)^2 \]

3. **Write the ratio of intensities**:
We are given that:
\[ \frac{I_{max}}{I_{min}} = \frac{9}{1} \]

4. **Substitute the expressions for intensities**:
Substituting the expressions for \(I_{max}\) and \(I_{min}\) into the ratio gives:
\[ \frac{(A_1 + A_2)^2}{(A_1 - A_2)^2} = \frac{9}{1} \]

5. **Cross-multiply to eliminate the fraction**:
This leads to:
\[ (A_1 + A_2)^2 = 9(A_1 - A_2)^2 \]

6. **Take the square root of both sides**:
Taking the square root gives:
\[ \frac{A_1 + A_2}{A_1 - A_2} = 3 \]

7. **Cross-multiply again**:
Cross-multiplying gives:
\[ A_1 + A_2 = 3(A_1 - A_2) \]

8. **Expand and rearrange**:
Expanding the right side:
\[ A_1 + A_2 = 3A_1 - 3A_2 \]
Rearranging terms gives:
\[ A_1 + A_2 + 3A_2 = 3A_1 \]
\[ 4A_2 = 2A_1 \]

9. **Solve for the ratio of amplitudes**:
Dividing both sides by \(2A_1\) gives:
\[ \frac{A_1}{A_2} = \frac{4}{2} = 2 \]
Thus, the ratio of amplitudes is:
\[ \frac{A_1}{A_2} = 2:1 \]

### Final Answer:
The ratio of amplitudes of the light waves is \(2:1\).