In a fraunhofer's diffraction obtained by a single slit aperture, the value of path difference for `n^(th)` order of minima is-

Options

A. `nlamda`
B. `2nlamda`
C. `((2n-1)lamda)/(2)`
D. `(2n-1)lamda`

Answer

To determine the path difference for the nth order minima in Fraunhofer diffraction through a single slit, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have a single slit of width \(d\) and we are observing the diffraction pattern on a screen at a distance. The light passing through the slit will spread out and create a pattern of bright and dark fringes.

2. **Path Difference Concept**:
- For destructive interference (minima) to occur, the path difference between light rays coming from different parts of the slit must be equal to an odd multiple of half the wavelength. This is because we want the waves to cancel each other out.

3. **Condition for Minima**:
- The condition for the nth order minima can be expressed as:
\[ \text{Path Difference} = n \lambda \]
where \(n\) is an integer (1, 2, 3, ...), and \(\lambda\) is the wavelength of the light.

4. **Deriving the Path Difference**:
- For a single slit, the path difference for light coming from the top edge of the slit to the bottom edge when viewed at an angle \(\theta\) is given by:
\[ d \sin \theta \]
where \(d\) is the width of the slit.

5. **Setting Up the Equation**:
- For the nth order minima, we set the path difference equal to \(n \lambda\):
\[ d \sin \theta = n \lambda \]

6. **Final Expression**:
- Rearranging gives us the condition for the nth order minima:
\[ \sin \theta = \frac{n \lambda}{d} \]

### Conclusion:
The path difference for the nth order minima in Fraunhofer diffraction through a single slit is given by:
\[ d \sin \theta = n \lambda \]