Question

Distance of the second dark fringe from the central maxima on the screen for \( 6000 \, \text{Å} \) wavelength.

Hint:

In the double-slit experiment, dark fringes occur when the path difference is an odd multiple of \( \lambda/2 \).

Answer 1

Step 1: Condition for Dark Fringe in Double Slit Experiment.
The condition for the dark fringe in a double-slit experiment is given by: \[ d \sin \theta = (m + \frac{1}{2}) \lambda \] where: - \( d \) is the distance between the slits, - \( \lambda \) is the wavelength of light, - \( \theta \) is the angle of diffraction, - \( m \) is the order of the dark fringe (starting from \( m = 0 \) for the first dark fringe). For the second dark fringe, \( m = 1 \), and the angle \( \theta \) can be approximated as \( \sin \theta \approx \tan \theta = \frac{y}{L} \), where \( y \) is the distance of the dark fringe from the central maxima and \( L \) is the distance from the slits to the screen.
Step 2: Finding the Distance for the Second Dark Fringe.
Substituting into the condition for the dark fringe: \[ d \times \frac{y}{L} = \left( 1 + \frac{1}{2} \right) \lambda \] \[ y = \frac{(1.5 \times 6000 \times 10^{-10} \, \text{m}) \times 1.0 \, \text{m}}{1.0 \times 10^{-3} \, \text{m}} = 0.9 \, \text{mm} \] Thus, the distance of the second dark fringe from the central maxima is \( 0.9 \, \text{mm} \).