In YDSE, the amplitude of intensity variation fo the two sources is found to be `5%` of the average intensity. The ratio of the intensities of two interfering sources is

Options

A. 2564
B. 1089
C. 1681
D. 869

Explanation

To solve the problem of finding the ratio of the intensities of two interfering sources in the Young's Double-Slit Experiment (YDSE), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Given Information:**
- The amplitude of intensity variation of the two sources is given as 5% of the average intensity.
- Let's denote the average intensity as \( I_{avg} \).

2. **Calculating Maximum and Minimum Intensities:**
- The maximum intensity \( I_{max} \) can be calculated as:
\[ I_{max} = I_{avg} + 5\% \text{ of } I_{avg} = I_{avg} + 0.05 I_{avg} = 1.05 I_{avg} \]
- The minimum intensity \( I_{min} \) can be calculated as:
\[ I_{min} = I_{avg} - 5\% \text{ of } I_{avg} = I_{avg} - 0.05 I_{avg} = 0.95 I_{avg} \]

3. **Using the Relationship Between Intensity and Amplitude:**
- We know that intensity \( I \) is proportional to the square of the amplitude \( a \):
\[ I \propto a^2 \]
- Therefore, we can express the maximum and minimum intensities in terms of the amplitudes \( a_1 \) and \( a_2 \):
\[ I_{max} = (a_1 + a_2)^2 \]
\[ I_{min} = (a_1 - a_2)^2 \]

4. **Setting Up the Ratio of Intensities:**
- The ratio of maximum to minimum intensity can be expressed as:
\[ \frac{I_{max}}{I_{min}} = \frac{(a_1 + a_2)^2}{(a_1 - a_2)^2} \]
- Substituting the values we calculated:
\[ \frac{1.05 I_{avg}}{0.95 I_{avg}} = \frac{(a_1 + a_2)^2}{(a_1 - a_2)^2} \]
- This simplifies to:
\[ \frac{1.05}{0.95} = \frac{(a_1 + a_2)^2}{(a_1 - a_2)^2} \]

5. **Calculating the Ratio:**
- Simplifying \( \frac{1.05}{0.95} \):
\[ \frac{1.05}{0.95} = \frac{21}{19} \]
- Taking the square root of both sides gives:
\[ \frac{a_1 + a_2}{a_1 - a_2} = \sqrt{\frac{21}{19}} \approx 1.05 \]

6. **Expressing Amplitudes in Terms of Each Other:**
- Let’s denote \( a_1 + a_2 = 1.05(a_1 - a_2) \):
\[ a_1 + a_2 = 1.05a_1 - 1.05a_2 \]
- Rearranging gives:
\[ 2.05a_2 = 0.05a_1 \]
- This leads to:
\[ \frac{a_2}{a_1} = \frac{0.05}{2.05} = \frac{1}{41} \]

7. **Finding the Ratio of Intensities:**
- Since intensity is proportional to the square of the amplitude:
\[ \frac{I_1}{I_2} = \left(\frac{a_1}{a_2}\right)^2 = \left(\frac{41}{1}\right)^2 = 1681 \]

### Final Result:
The ratio of the intensities of the two interfering sources is:
\[ \frac{I_1}{I_2} = 1681 : 1 \]