Consider two coherent, monochromatic (wavelength `lambda`) sources `S_(1) and S_(2)`, separated by a distance d. The ratio of intensities of `S_(1)` and that of `S_(2)` (which is responsible for interference at point P, where detector is located) is 4. The distance of point P from `S_(1)` is (if the resultant intensity at point P is equal to `(9)/(4)` times of intensity of `S_(1)`) `("Given : "angleS_(2)S_(1)P=90^(@), d gt0 and n" is a positive integer")`

Options

A. `(d^(2)-n^(2)lambda^(2))/(2n lambda)`
B. `(d^(2)+n^(2)lambda^(2))/(2n lambda)`
C. `(nlambdad)/(sqrt(d^(2)-n^(2)lambda^(2)))`
D. `(2nlambdad)/(sqrt(d^(2)-n^(2)lambda^(2)))`

Answer

To solve the problem, we need to find the distance of point P from source S1 given the conditions of interference from two coherent sources S1 and S2. Here’s a step-by-step solution:

### Step 1: Understand the Given Information
- We have two coherent sources, S1 and S2, separated by a distance \( d \).
- The ratio of their intensities is given as \( \frac{I_1}{I_2} = 4 \).
- The resultant intensity at point P is \( \frac{9}{4} I_1 \).
- The angle \( \angle S_2 S_1 P = 90^\circ \).

### Step 2: Express Intensities
Let the intensity of source S1 be \( I_1 \) and the intensity of source S2 be \( I_2 \). From the given ratio:
\[ I_2 = \frac{I_1}{4} \]

### Step 3: Write the Expression for Resultant Intensity
The resultant intensity \( I \) at point P due to two coherent sources is given by:
\[ I = I_1 + I_2 + 2 \sqrt{I_1 I_2} \cos \phi \]
Where \( \phi \) is the phase difference between the two waves at point P.

### Step 4: Substitute Known Values
Substituting \( I_2 \) into the intensity equation:
\[ I = I_1 + \frac{I_1}{4} + 2 \sqrt{I_1 \cdot \frac{I_1}{4}} \cos \phi \]
This simplifies to:
\[ I = I_1 + \frac{I_1}{4} + 2 \cdot \frac{I_1}{2} \cos \phi \]
\[ I = I_1 + \frac{I_1}{4} + I_1 \cos \phi \]
\[ I = \frac{5}{4} I_1 + I_1 \cos \phi \]

### Step 5: Set the Resultant Intensity Equal to Given Value
We know \( I = \frac{9}{4} I_1 \):
\[ \frac{5}{4} I_1 + I_1 \cos \phi = \frac{9}{4} I_1 \]
Subtracting \( \frac{5}{4} I_1 \) from both sides:
\[ I_1 \cos \phi = \frac{9}{4} I_1 - \frac{5}{4} I_1 \]
\[ I_1 \cos \phi = \frac{4}{4} I_1 \]
\[ \cos \phi = 1 \]

### Step 6: Determine the Phase Difference
Since \( \cos \phi = 1 \), the phase difference \( \phi \) is:
\[ \phi = 2n\pi \quad (n \text{ is a positive integer}) \]

### Step 7: Relate Phase Difference to Path Difference
The phase difference is related to the path difference \( \Delta \) by:
\[ \phi = \frac{2\pi}{\lambda} \Delta \]
Thus,
\[ 2n\pi = \frac{2\pi}{\lambda} \Delta \]
Cancelling \( 2\pi \) gives:
\[ \Delta = n\lambda \]

### Step 8: Calculate Path Difference
The path difference \( \Delta \) can also be expressed in terms of the distances:
\[ \Delta = \sqrt{x^2 + d^2} - x \]
Setting the two expressions for path difference equal:
\[ \sqrt{x^2 + d^2} - x = n\lambda \]
Rearranging gives:
\[ \sqrt{x^2 + d^2} = n\lambda + x \]

### Step 9: Square Both Sides
Squaring both sides:
\[ x^2 + d^2 = (n\lambda + x)^2 \]
Expanding the right side:
\[ x^2 + d^2 = n^2\lambda^2 + 2n\lambda x + x^2 \]
Cancelling \( x^2 \) from both sides:
\[ d^2 = n^2\lambda^2 + 2n\lambda x \]

### Step 10: Solve for \( x \)
Rearranging gives:
\[ 2n\lambda x = d^2 - n^2\lambda^2 \]
\[ x = \frac{d^2 - n^2\lambda^2}{2n\lambda} \]

### Final Result
Thus, the distance of point P from source S1 is:
\[ x = \frac{d^2 - n^2\lambda^2}{2n\lambda} \]