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 36% reducing its width, then the intensity of light at the same point will be

Options

A. 90
B. 98
C. 67
D. 81

Answer

To solve the problem step by step, we will follow the principles of wave optics and the concept of intensity from coherent sources.

### Step 1: Understand the Maximum Intensity
Given that two coherent sources of equal intensity produce a maximum intensity of 100 units at a point, we can denote the intensity of each source as \( I_1 \) and \( I_2 \). Since they are equal, we have:
\[ I_1 = I_2 = I \]
The maximum intensity \( I_{\text{max}} \) for two coherent sources is given by the formula:
\[ I_{\text{max}} = 4I \]
From the problem, we know:
\[ I_{\text{max}} = 100 \]
Thus, we can write:
\[ 4I = 100 \]

### Step 2: Calculate the Intensity of Each Source
To find the intensity \( I \) of each source, we can rearrange the equation:
\[ I = \frac{100}{4} = 25 \text{ units} \]
So, both sources have an intensity of 25 units.

### Step 3: Reduce the Intensity of One Source
According to the problem, the intensity of one of the sources is reduced by 36%. We will calculate the new intensity \( I_1' \) of the first source:
\[ I_1' = I_1 - (0.36 \times I_1) = 25 - (0.36 \times 25) \]
Calculating this gives:
\[ I_1' = 25 - 9 = 16 \text{ units} \]

### Step 4: Keep the Second Source Intensity Constant
The intensity of the second source \( I_2 \) remains unchanged:
\[ I_2 = 25 \text{ units} \]

### Step 5: Calculate the Resultant Intensity
Now, we need to calculate the resultant intensity \( I_R \) at the point where the two waves interfere. The formula for the resultant intensity is:
\[ I_R = I_1' + I_2 + 2\sqrt{I_1' I_2} \]
Substituting the values we have:
\[ I_R = 16 + 25 + 2\sqrt{16 \times 25} \]
Calculating \( \sqrt{16 \times 25} \):
\[ \sqrt{16 \times 25} = \sqrt{400} = 20 \]
Now substituting back into the equation:
\[ I_R = 16 + 25 + 2 \times 20 = 16 + 25 + 40 = 81 \text{ units} \]

### Final Answer
Thus, the intensity of light at the same point after reducing the intensity of one source is:
\[ \boxed{81 \text{ units}} \]