Question

Incident light of wavelength $ \lambda = 800 \, \text{nm} $ produces a diffraction pattern on a screen 1.5 m away when it passes through a single slit of width 0.5 mm. The distance between the first dark fringes on either side of the central bright fringe is

• A. \( 2.4 \, \text{mm} \)
• B. \( 2.4 \, \text{cm} \)
• C. \( 4.8 \, \text{cm} \)
• D. \( 4.8 \, \text{mm} \)

Hint:

For small angles, \( \sin \theta \approx \tan \theta \), making calculations easier. Always convert the units of wavelength and slit width to the same base (meters) when applying formulas.

Answer 1

The Correct Option is D

The angular position of the first dark fringe in a single-slit diffraction pattern is given by: \[ \sin \theta = \frac{m\lambda}{a} \] Where: 
- \( \lambda \) is the wavelength of the light, 
- \( a \) is the width of the slit, 
- \( m \) is the order of the dark fringe (for the first dark fringe, \( m = 1 \)). 
Given: - \( \lambda = 800 \, \text{nm} = 800 \times 10^{-9} \, \text{m} \), 
- \( a = 0.5 \, \text{mm} = 0.5 \times 10^{-3} \, \text{m} \), 
- The distance to the screen is \( L = 1.5 \, \text{m} \). 
The angle for the first dark fringe can be calculated by: \[ \sin \theta = \frac{800 \times 10^{-9}}{0.5 \times 10^{-3}} = 1.6 \times 10^{-3} \] Now, the linear distance between the first dark fringes is given by: \[ y = 2L \tan \theta \approx 2L \sin \theta \quad \text{(for small angles, } \tan \theta \approx \sin \theta) \] Substituting the values: \[ y = 2 \times 1.5 \times 1.6 \times 10^{-3} = 4.8 \times 10^{-3} \, \text{m} = 4.8 \, \text{mm} \] 
Thus, the distance between the first dark fringes on either side of the central bright fringe is \( 4.8 \, \text{mm} \).