Question

A monochromatic light of wavelength 800 nm is incident normally on a single slit of width 0.020 mm to produce a diffraction pattern on a screen placed 1 m away. Estimate the number of fringes obtained in Young's double slit experiment with slit separation 0.20 mm, which can be accommodated within the range of total angular spread of the central maximum due to single slit.

• A. 25
• B. 30
• C. 20
• D. 15

Hint:

To calculate the number of fringes in a diffraction pattern, divide the total angular spread of the central maximum by the angular width of a fringe.

Answer 1

The Correct Option is C

For a single slit diffraction pattern, the angular width \( \theta \) of the central maximum is given by: \[ \theta = \frac{\lambda}{a} \] Where: - \( \lambda = 800 \, \text{nm} = 8 \times 10^{-7} \, \text{m} \) (wavelength of the light), - \( a = 0.020 \, \text{mm} = 2 \times 10^{-5} \, \text{m} \) (width of the slit). Now calculate the angular spread of the central maximum: \[ \theta = \frac{8 \times 10^{-7}}{2 \times 10^{-5}} = 4 \times 10^{-2} \, \text{radians} \] The total angular spread of the central maximum due to the single slit is approximately \( 2\theta \), i.e., \[ \text{Total angular spread} = 2 \times 4 \times 10^{-2} = 8 \times 10^{-2} \, \text{radians} \] Next, the angular width of a fringe in the double-slit diffraction pattern is given by: \[ \Delta \theta = \frac{\lambda}{d} \] Where: - \( d = 0.20 \, \text{mm} = 2 \times 10^{-4} \, \text{m} \) (slit separation). Now calculate the angular width of a fringe: \[ \Delta \theta = \frac{8 \times 10^{-7}}{2 \times 10^{-4}} = 4 \times 10^{-3} \, \text{radians} \] The total number of fringes that can be accommodated within the total angular spread is: \[ N = \frac{\text{Total angular spread}}{\Delta \theta} = \frac{8 \times 10^{-2}}{4 \times 10^{-3}} = 20 \]
Thus, the number of fringes that can be accommodated is 20.