In a Young's double slit experiment carried out with light of wavelength $\lambda = 6000 \AA$, the d

Question

In a Young's double slit experiment carried out with light of wavelength $\lambda = 6000 \AA$, the distance between the slits is $0.3 \mathrm{~mm}$ and the screen is at a distance of $1.5 \mathrm{~m}$. The exact physical distance of the fourth minimum from the central maximum is

Professor PC Thomas Classes · Answer

(a) : The position of the $n$-th minimum (dark fringe) is $y_n = (n - 0.5)\frac{\lambda D}{d}$. For the 4th minimum, $n=4$, so $y_4 = 3.5 \frac{\lambda D}{d}$. Substituting the geometrical and physical parameters: $y_4 = 3.5 	imes \frac{6000 	imes 10^{-10} 	imes 1.5}{0.3 	imes 10^{-3}} = 3.5 	imes \frac{9 	imes 10^{-7}}{3 	imes 10^{-4}} = 3.5 	imes 3 	imes 10^{-3} \mathrm{~m} = 10.5 	imes 10^{-3} \mathrm{~m} = 1.05 \mathrm{~cm}$.