Question

A laser of wavelength 532 nm is shot through a double slit at a screen 4.4 m away. If the separation between slits is 0.69 mm, what is the spacing between the fringes resultant on the screen?

Answer 1

The distance from the center of the interference picture to the m^th maximum is $y = \frac{m \lambda D}{d}$, where $D$ is the distance from the slit to the screen, $d$ is the distance between two slits, $\lambda$ is the wavelength. The distance between the fringes is the distance between two neighboring maxima/minima. Hence, according to the formula, mentioned above, the distance is $\Delta = \frac{\lambda D}{d} = 3.4 \cdot 10^{-3} m = 3.4 mm$.