In Young's double slit experiment, the slits are $2.0 \mathrm{~mm}$ apart and the screen is placed a

Question

In Young's double slit experiment, the slits are $2.0 \mathrm{~mm}$ apart and the screen is placed at a distance of $1.5 \mathrm{~m}$ from the slits. If the distance between two consecutive bright fringes is observed to be $0.45 \mathrm{~mm}$, the wavelength of the monochromatic light used is

Professor PC Thomas Classes · Answer

(a) : The fringe separation (or fringe width) is given by the formula $\beta = \frac{\lambda D}{d}$. Rearranging this to solve for the wavelength, we get $\lambda = \frac{\beta d}{D}$. Substituting the given values: $\beta = 0.45 \mathrm{~mm} = 0.45 	imes 10^{-3} \mathrm{~m}$, $d = 2.0 \mathrm{~mm} = 2.0 	imes 10^{-3} \mathrm{~m}$, and $D = 1.5 \mathrm{~m}$. Therefore, $\lambda = \frac{(0.45 	imes 10^{-3}) 	imes (2.0 	imes 10^{-3})}{1.5} = \frac{0.90 	imes 10^{-6}}{1.5} = 0.6 	imes 10^{-6} \mathrm{~m} = 600 	imes 10^{-9} \mathrm{~m} = 600 \mathrm{~nm}$.