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: Position of First Minimum

The condition for the first minimum (zero intensity) in a single-slit diffraction pattern is:

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

Definitions:

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

Provided Information:

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

For the first minimum (\( m = 1 \)):

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

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

Substituting values:

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

For small angles, \( \sin \theta \approx \tan \theta \). The distance \( y \) from the central maximum to the first minimum on the screen is given by \( \tan \theta = \frac{y}{L} \). Therefore:

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

Calculation:

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

The distance from the central maximum to the first point of zero intensity is \( 0.5 \, \text{mm} \).