In a single slit diffraction experiment first minimum for red light `(660nm)` coincides with first maximum of some other wavelength `lambda`'. The value of `lambda`' is

Options

A. (a) `4400Å`
B. (b) `6600Å`
C. (c) `2000Å`
D. (d) `3500Å`

Explanation

To solve the problem, we need to find the wavelength \( \lambda' \) such that the first minimum of red light (660 nm) coincides with the first maximum of this other wavelength \( \lambda' \) in a single slit diffraction pattern.

### Step-by-Step Solution:

1. **Understanding Single Slit Diffraction:**
In a single slit diffraction pattern, the positions of the minima and maxima are determined by the formula:
- For minima: \( d \sin \theta = n \lambda \) (where \( n = 1, 2, 3, \ldots \))
- For maxima: \( d \sin \theta = (m + \frac{1}{2}) \lambda \) (where \( m = 0, 1, 2, \ldots \))

2. **Identifying the Conditions:**
We are given that the first minimum for red light (660 nm) coincides with the first maximum of the other wavelength \( \lambda' \). Therefore:
- For red light (first minimum), \( n = 1 \):
\[ d \sin \theta = 1 \cdot 660 \text{ nm} \]

3. **Setting Up the Equation for the First Maximum of \( \lambda' \):**
For the first maximum of \( \lambda' \) (where \( m = 0 \)):
\[ d \sin \theta = (0 + \frac{1}{2}) \lambda' = \frac{1}{2} \lambda' \]

4. **Equating the Two Conditions:**
Since both conditions occur at the same angle \( \theta \), we can equate the two expressions:
\[ 660 \text{ nm} = \frac{1}{2} \lambda' \]

5. **Solving for \( \lambda' \):**
Rearranging the equation gives:
\[ \lambda' = 2 \times 660 \text{ nm} = 1320 \text{ nm} \]

### Final Answer:
The value of \( \lambda' \) is \( 1320 \text{ nm} \).