Question

In a diffraction experiment, the slit is illuminated by light of wavelength 600 nm. The first minimum of the pattern falls at \( \theta = 30^\circ \). Calculate the width of the slit.

Hint:

For diffraction, the angle for minima is determined by the equation \( a \sin \theta = m \lambda \), where \( a \) is the slit width and \( m \) is the order of the minimum.

Answer 1

The condition for the first minimum in diffraction is given by: \[ a \sin \theta = m \lambda \quad \text{where} \quad m = 1 \quad \text{for the first minimum} \] Here: - \( \lambda = 600 \, \text{nm} = 6 \times 10^{-7} \, \text{m} \), - \( \theta = 30^\circ \). Substitute the values: \[ a \sin 30^\circ = 1 \cdot 6 \times 10^{-7} \] \[ a \cdot \frac{1}{2} = 6 \times 10^{-7} \] \[ a = 1.2 \times 10^{-6} \, \text{m} \] Thus, the width of the slit is \( 1.2 \times 10^{-6} \, \text{m} \).

Answer 2

We are given a diffraction experiment where the slit is illuminated by light of wavelength \( \lambda = 600 \, \text{nm} \) (which is \( 600 \times 10^{-9} \, \text{m} \)), and the first minimum of the diffraction pattern occurs at an angle \( \theta = 30^\circ \). We are tasked with calculating the width of the slit.

1. Understanding the Diffraction Pattern: 

- In a single-slit diffraction pattern, the angular positions of the minima are given by the condition:

\[ a \sin(\theta) = m\lambda \]

where: - \( a \) is the width of the slit, - \( \theta \) is the angle of diffraction, - \( m \) is the order of the minima (for the first minimum, \( m = 1 \)), - \( \lambda \) is the wavelength of the light.

2. Applying the Formula:

For the first minimum, \( m = 1 \), and the given angle is \( \theta = 30^\circ \). The formula becomes:

\[ a \sin(30^\circ) = \lambda \]

3. Solving for the Slit Width \( a \):

We know that \( \sin(30^\circ) = \frac{1}{2} \), so the equation becomes:

\[ a \times \frac{1}{2} = 600 \times 10^{-9} \, \text{m} \]

Multiplying both sides by 2 to solve for \( a \):

\[ a = 2 \times 600 \times 10^{-9} \, \text{m} = 1200 \times 10^{-9} \, \text{m} = 1.2 \times 10^{-6} \, \text{m} \]

4. Final Answer:

The width of the slit is \( \boxed{1.2 \, \mu\text{m}} \).