Question

Light of wavelength 750 nm is incident normally on a slit of width 1.5 mm. Diffraction pattern is obtained on a screen 1.0 m away from the slit. Find the distance of the nearest point from the central maxima at which the intensity is zero.

Hint:

For diffraction through a single slit, the minima occur where the condition \( a \sin \theta = m \lambda \) is satisfied, where \( m = 1, 2, 3, \dots \). The angle \( \theta \) is small for practical cases, so we can use \( \sin \theta \approx \tan \theta \) to calculate the position of the minima on the screen.

Answer 1

Single-Slit Diffraction Pattern: Distance to the First Minimum 

For a single-slit diffraction pattern, the condition for the first minimum (zero intensity) is given by:

\[ a \sin \theta = m \lambda \quad \text{(for } m = 1, 2, 3, \dots \text{)} \]

Where:

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

Given Data:

• \( \lambda = 750 \, \text{nm} = 750 \times 10^{-9} \, \text{m} \),
• \( a = 1.5 \, \text{mm} = 1.5 \times 10^{-3} \, \text{m} \),
• \( L = 1.0 \, \text{m} \) (distance from the slit to the screen).

For the first minimum, \( m = 1 \), the equation becomes:

\[ a \sin \theta = \lambda \]

\[ \sin \theta = \frac{\lambda}{a} \]

Substitute the known values:

\[ \sin \theta = \frac{750 \times 10^{-9}}{1.5 \times 10^{-3}} = 5 \times 10^{-4} \]

Now, for small angles, \( \sin \theta \approx \tan \theta \), so:

\[ \tan \theta = \frac{y}{L} \]

Where \( y \) is the distance from the central maximum to the first minimum on the screen. Thus:

\[ y = L \cdot \tan \theta = L \cdot \sin \theta \]

Substituting the values:

\[ y = 1.0 \times 5 \times 10^{-4} = 5 \times 10^{-4} \, \text{m} = 0.5 \, \text{mm} \]

Thus, the distance of the nearest point from the central maximum at which the intensity is zero is \( 0.5 \, \text{mm} \).