The intensity ratio of the two interfering beams of light is m. What is the value of `I_("max")-I_("min") // I_("max")+I_("min")`?

Options

A. `2sqrt(m)`
B. `(2sqrt(m))/(1+m)`
C. `(2)/(1+m)`
D. `(1+m)/(2sqrt(m))`

Answer

To solve the problem, we need to find the value of \(\frac{I_{\text{max}} - I_{\text{min}}}{I_{\text{max}} + I_{\text{min}}}\) given the intensity ratio of two interfering beams of light as \(m\).

### Step-by-Step Solution:

1. **Define Intensities**:
Let the intensities of the two beams be \(I_1\) and \(I_2\). Given the intensity ratio \(m\), we can express this as:
\[ I_1 = m I_2 \]

2. **Calculate \(I_{\text{max}}\)**:
The maximum intensity \(I_{\text{max}}\) for two interfering beams can be calculated using the formula:
\[ I_{\text{max}} = I_1 + I_2 + 2\sqrt{I_1 I_2} \]
Substituting \(I_1 = m I_2\):
\[ I_{\text{max}} = m I_2 + I_2 + 2\sqrt{m I_2 \cdot I_2} \]
Simplifying this:
\[ I_{\text{max}} = (m + 1) I_2 + 2\sqrt{m} I_2 = (m + 1 + 2\sqrt{m}) I_2 \]

3. **Calculate \(I_{\text{min}}\)**:
The minimum intensity \(I_{\text{min}}\) is given by:
\[ I_{\text{min}} = I_1 + I_2 - 2\sqrt{I_1 I_2} \]
Again substituting \(I_1 = m I_2\):
\[ I_{\text{min}} = m I_2 + I_2 - 2\sqrt{m I_2 \cdot I_2} \]
Simplifying this:
\[ I_{\text{min}} = (m + 1) I_2 - 2\sqrt{m} I_2 = (m + 1 - 2\sqrt{m}) I_2 \]

4. **Calculate \(I_{\text{max}} - I_{\text{min}}\)**:
Now, we find \(I_{\text{max}} - I_{\text{min}}\):
\[ I_{\text{max}} - I_{\text{min}} = [(m + 1 + 2\sqrt{m}) I_2] - [(m + 1 - 2\sqrt{m}) I_2] \]
This simplifies to:
\[ I_{\text{max}} - I_{\text{min}} = (4\sqrt{m}) I_2 \]

5. **Calculate \(I_{\text{max}} + I_{\text{min}}\)**:
Next, we find \(I_{\text{max}} + I_{\text{min}}\):
\[ I_{\text{max}} + I_{\text{min}} = [(m + 1 + 2\sqrt{m}) I_2] + [(m + 1 - 2\sqrt{m}) I_2] \]
This simplifies to:
\[ I_{\text{max}} + I_{\text{min}} = (2(m + 1)) I_2 \]

6. **Calculate the Ratio**:
Now we can find the desired ratio:
\[ \frac{I_{\text{max}} - I_{\text{min}}}{I_{\text{max}} + I_{\text{min}}} = \frac{(4\sqrt{m}) I_2}{(2(m + 1)) I_2} \]
Canceling \(I_2\) from numerator and denominator:
\[ = \frac{4\sqrt{m}}{2(m + 1)} = \frac{2\sqrt{m}}{m + 1} \]

### Final Answer:
\[ \frac{I_{\text{max}} - I_{\text{min}}}{I_{\text{max}} + I_{\text{min}}} = \frac{2\sqrt{m}}{m + 1} \]