A single slit of width d is placed in the path of beam of wavelength `lamda`. The angular width of the principal maximum obtained is-

Options

A. `(d)/(lamda)`
B. `(lamda)/(d)`
C. `(2lamda)/(d)`
D. `(2d)/(lamda)`

Answer

To find the angular width of the principal maximum obtained in a single slit diffraction pattern, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Setup**:
We have a single slit of width \( d \) illuminated by a beam of light with wavelength \( \lambda \). The light passing through the slit will produce a diffraction pattern on a screen.

2. **Condition for Minima**:
The condition for the minima in a single slit diffraction pattern is given by:
\[ d \sin \theta = n \lambda \]
where \( n \) is the order of the minima (n = 1, 2, 3,...).

3. **Finding the First Minima**:
For the first minima on either side of the central maximum, we set \( n = 1 \):
\[ d \sin \theta_1 = \lambda \]
For small angles (which is a common approximation in diffraction problems), we can use the small angle approximation where \( \sin \theta \approx \theta \) (in radians). Thus, we can rewrite the equation as:
\[ d \theta_1 \approx \lambda \]
Therefore, we can express \( \theta_1 \) as:
\[ \theta_1 = \frac{\lambda}{d} \]

4. **Angular Width of the Principal Maximum**:
The principal maximum is located between the first minima on either side. Therefore, the total angular width of the principal maximum is:
\[ \text{Angular Width} = 2\theta_1 = 2 \left(\frac{\lambda}{d}\right) \]
Hence, the angular width of the principal maximum is:
\[ \text{Angular Width} = \frac{2\lambda}{d} \]

5. **Conclusion**:
The angular width of the principal maximum obtained is \( \frac{2\lambda}{d} \).

### Final Answer:
The angular width of the principal maximum is \( \frac{2\lambda}{d} \).

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