In an experiment of single slit diffraction pattern first minimum for red light coincides with first maximum of some other wavelength. If wavelength of red light is `6600 A^(0)`, then wavelength of first maximum will be :

Options

A. `55000A^(0)`
B. `4400 A^(0)`
C. `3300 A^(0)`
D. `6600 A^(0)`

Answer

To solve the problem, we need to find the wavelength of the first maximum that coincides with the first minimum of red light in a single slit diffraction pattern. Let's go through the solution step by step.

### Step 1: Understand the condition for minima and maxima in single slit diffraction
In a single slit diffraction pattern:
- The condition for the first minimum is given by:
\[ a \sin \theta = \lambda \]
where \( a \) is the width of the slit, \( \lambda \) is the wavelength of the light, and \( \theta \) is the angle of diffraction.

- The condition for the first maximum (after the first minimum) is given by:
\[ a \sin \theta = \left(n + \frac{1}{2}\right) \lambda' \]
where \( n \) is the order of the maximum and \( \lambda' \) is the wavelength of the light corresponding to the maximum.

### Step 2: Set up the equation for the given problem
According to the problem, the first minimum for red light coincides with the first maximum of some other wavelength. Therefore, we can equate the two conditions:
\[ \lambda_r = \left(n + \frac{1}{2}\right) \lambda' \]
For the first maximum, we set \( n = 1 \):
\[ \lambda_r = \left(1 + \frac{1}{2}\right) \lambda' = \frac{3}{2} \lambda' \]

### Step 3: Substitute the known value of the wavelength of red light
We know that the wavelength of red light \( \lambda_r \) is \( 6600 \) angstroms. Substituting this value into the equation gives:
\[ 6600 = \frac{3}{2} \lambda' \]

### Step 4: Solve for the wavelength of the first maximum
To find \( \lambda' \), we rearrange the equation:
\[ \lambda' = \frac{2}{3} \times 6600 \]
Calculating this gives:
\[ \lambda' = \frac{13200}{3} = 4400 \text{ angstroms} \]

### Final Answer
Thus, the wavelength of the first maximum is:
\[ \lambda' = 4400 \text{ angstroms} \]