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 Å`, then wavelength of first maximum will be:

Options

A. `3300 Å`
B. `4400 Å`
C. `5500 Å`
D. `6600 Å`

Explanation

To solve the problem step by step, we will use the principles of single slit diffraction.

### Step 1: Understand the conditions for minima and maxima in single slit diffraction.
In a single slit diffraction pattern, the position of the first minimum is given by the formula:
\[ y_{min} = \frac{n \lambda D}{d} \]
where:
- \( y_{min} \) = position of the nth minimum
- \( n \) = order of the minimum (for the first minimum, \( n = 1 \))
- \( \lambda \) = wavelength of the light
- \( D \) = distance from the slit to the screen
- \( d \) = width of the slit

For the first minimum (\( n = 1 \)):
\[ y_{min} = \frac{\lambda D}{d} \]

The position of the first maximum is given by:
\[ y_{max} = \frac{(2n + 1) \lambda D}{2d} \]
For the first maximum (\( n = 0 \)):
\[ y_{max} = \frac{\lambda D}{2d} \]

### Step 2: Set up the equation for the given scenario.
According to the problem, the first minimum for red light (wavelength \( \lambda_1 = 6600 \, \text{Å} \)) coincides with the first maximum of another wavelength \( \lambda_2 \).

Thus, we can equate the two positions:
\[ \frac{\lambda_1 D}{d} = \frac{(2 \cdot 0 + 1) \lambda_2 D}{2d} \]
This simplifies to:
\[ \frac{\lambda_1 D}{d} = \frac{\lambda_2 D}{2d} \]

### Step 3: Cancel common terms and solve for \( \lambda_2 \).
Since \( D \) and \( d \) are common in both terms, we can cancel them out:
\[ \lambda_1 = \frac{\lambda_2}{2} \]

Now, rearranging gives:
\[ \lambda_2 = 2 \lambda_1 \]

### Step 4: Substitute the value of \( \lambda_1 \).
Substituting \( \lambda_1 = 6600 \, \text{Å} \):
\[ \lambda_2 = 2 \times 6600 \, \text{Å} = 13200 \, \text{Å} \]

### Step 5: Correct the interpretation of maxima.
However, we need to consider that the first maximum corresponds to \( n = 1 \) in the maximum formula:
\[ y_{max} = \frac{3 \lambda_2 D}{2d} \]
Thus, we should have:
\[ \frac{\lambda_1 D}{d} = \frac{3 \lambda_2 D}{2d} \]

Setting these equal gives:
\[ \lambda_1 = \frac{3}{2} \lambda_2 \]

### Step 6: Solve for \( \lambda_2 \) again.
Rearranging gives:
\[ \lambda_2 = \frac{2}{3} \lambda_1 \]
Substituting \( \lambda_1 = 6600 \, \text{Å} \):
\[ \lambda_2 = \frac{2}{3} \times 6600 \, \text{Å} = 4400 \, \text{Å} \]

### Final Answer:
The wavelength of the first maximum will be \( \lambda_2 = 4400 \, \text{Å} \).

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