No fringes are seen in a single-slit diffraction pattern if

Options

A. The screen is far away
B. The wavelength is less than the slit width
C. The wavelength is greater than the slit width
D. The distance to the screen is greater than the slit width

Answer

To determine when no fringes are seen in a single-slit diffraction pattern, we need to analyze the conditions under which diffraction occurs. The key factors include the relationship between the wavelength of light and the width of the slit.

### Step-by-Step Solution:

1. **Understanding Single-Slit Diffraction**:
In single-slit diffraction, light passing through a narrow slit spreads out and creates a pattern of light and dark fringes on a screen. The condition for minima (dark fringes) in the diffraction pattern is given by the formula:
\[ a \sin \theta = m \lambda \]
where \( a \) is the slit width, \( \lambda \) is the wavelength of light, \( \theta \) is the angle of the minima from the central maximum, and \( m \) is an integer (1, 2, 3, ...).

2. **Condition for No Fringes**:
For no fringes to be observed, the condition \( a \sin \theta = m \lambda \) must not be satisfied for any integer \( m \). This situation occurs when the wavelength \( \lambda \) is greater than the slit width \( a \) (i.e., \( \lambda > a \)).

3. **Analyzing the Range of \( \sin \theta \)**:
The sine function has a range of values between -1 and 1. Thus, for \( m \lambda / a \) to be valid, it must satisfy:
\[ -1 \leq \frac{m \lambda}{a} \leq 1 \]
If \( \lambda > a \), then \( \frac{\lambda}{a} > 1 \). This means that for any integer \( m \), \( m \lambda / a \) will exceed the range of \( \sin \theta \), leading to no possible solutions for \( \theta \).

4. **Conclusion**:
Therefore, no fringes are seen in a single-slit diffraction pattern if the wavelength of light is greater than the slit width:
\[ \text{No fringes if } \lambda > a \]