In a YDSE ratio of slit width is 1:9. If `I_(MAX)/I_(MIN) = X/4` then find x?

Answer

To solve the problem, we need to find the value of \( x \) given the ratio of slit widths in a Young's Double Slit Experiment (YDSE) and the relationship between the maximum and minimum intensities.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We are given the ratio of slit widths \( a_1 : a_2 = 1 : 9 \).
- We need to find \( x \) such that \( \frac{I_{max}}{I_{min}} = \frac{x}{4} \).

2. **Setting Up the Ratios**:
- Let the slit widths be \( a_1 = k \) and \( a_2 = 9k \) for some constant \( k \).
- The intensity \( I \) in YDSE is proportional to the square of the amplitude, which is proportional to the slit width. Thus, we can express the amplitudes as:
\[ A_1 \propto a_1 = k \quad \text{and} \quad A_2 \propto a_2 = 9k \]

3. **Calculating the Intensities**:
- The maximum intensity \( I_{max} \) occurs when the waves from both slits are in phase:
\[ I_{max} = (A_1 + A_2)^2 = (k + 9k)^2 = (10k)^2 = 100k^2 \]
- The minimum intensity \( I_{min} \) occurs when the waves are out of phase:
\[ I_{min} = (A_1 - A_2)^2 = (k - 9k)^2 = (-8k)^2 = 64k^2 \]

4. **Finding the Ratio of Intensities**:
- Now, we can find the ratio \( \frac{I_{max}}{I_{min}} \):
\[ \frac{I_{max}}{I_{min}} = \frac{100k^2}{64k^2} = \frac{100}{64} = \frac{25}{16} \]

5. **Equating to Find \( x \)**:
- We know from the problem statement that:
\[ \frac{I_{max}}{I_{min}} = \frac{x}{4} \]
- Setting the two expressions equal gives:
\[ \frac{25}{16} = \frac{x}{4} \]

6. **Solving for \( x \)**:
- Cross-multiplying to solve for \( x \):
\[ 25 \cdot 4 = 16x \implies 100 = 16x \implies x = \frac{100}{16} = 6.25 \]

### Final Answer:
Thus, the value of \( x \) is \( 6.25 \).