The two coherent sources of equal intensity produce maximum intensity of 100 units at a point. If the intensity of one of the sources is reduced by 50% by reducing its width then the intensity of light at the same point will be

Options

A. `90`
B. `89`
C. `67`
D. `72.85`

Answer

To solve the problem, we need to find the new intensity at a point where two coherent sources of light are present, given that the intensity of one source is reduced by 50%.

### Step-by-Step Solution:

1. **Understanding the Initial Conditions**:
- Let the intensity of each coherent source be \( I_0 \).
- The maximum intensity \( I_{\text{max}} \) at a point is given as 100 units.

2. **Using the Formula for Maximum Intensity**:
- The formula for maximum intensity when two coherent sources are involved is:
\[ I_{\text{max}} = I_1 + I_2 + 2\sqrt{I_1 I_2} \]
- Since both sources are initially equal, we can set \( I_1 = I_2 = I_0 \):
\[ I_{\text{max}} = I_0 + I_0 + 2\sqrt{I_0 I_0} = 2I_0 + 2I_0 = 4I_0 \]

3. **Solving for \( I_0 \)**:
- Given \( I_{\text{max}} = 100 \):
\[ 4I_0 = 100 \implies I_0 = \frac{100}{4} = 25 \]

4. **Adjusting the Intensity of One Source**:
- The intensity of one source is reduced by 50%, so the new intensity \( I' \) of that source becomes:
\[ I' = 0.5 I_0 = 0.5 \times 25 = 12.5 \]

5. **Calculating the New Intensity**:
- The intensity of the other source remains \( I_0 = 25 \).
- Now, we can calculate the new maximum intensity using the same formula:
\[ I_{\text{new}} = I_0 + I' + 2\sqrt{I_0 I'} \]
- Substituting the values:
\[ I_{\text{new}} = 25 + 12.5 + 2\sqrt{25 \times 12.5} \]

6. **Calculating the Square Root**:
- First, calculate \( 25 \times 12.5 = 312.5 \).
- Now, find the square root:
\[ \sqrt{312.5} \approx 17.68 \]

7. **Final Calculation**:
- Substitute back into the intensity equation:
\[ I_{\text{new}} = 25 + 12.5 + 2 \times 17.68 \]
\[ I_{\text{new}} = 25 + 12.5 + 35.36 = 72.86 \text{ units} \]

8. **Conclusion**:
- The new intensity at the same point is approximately \( 72.85 \) units.

### Final Answer:
The intensity of light at the same point after reducing the intensity of one source by 50% is **72.85 units**.