A beam of light of $\lambda=450 \mathrm{~nm}$ falls on a single slit of width $0.5 \mathrm{~mm}$. Th

Question

A beam of light of $\lambda=450 \mathrm{~nm}$ falls on a single slit of width $0.5 \mathrm{~mm}$. The resulting diffraction pattern is observed on a screen $3 \mathrm{~m}$ away. The linear width of the central bright fringe is

Professor PC Thomas Classes · Answer

(a) : The linear width of the central maximum in a single slit diffraction pattern is given by $W = \frac{2\lambda D}{a}$. Substituting the given values: $\lambda = 450 	imes 10^{-9} \mathrm{~m}$, $D = 3 \mathrm{~m}$, and $a = 0.5 	imes 10^{-3} \mathrm{~m}$, we get $W = \frac{2 	imes 450 	imes 10^{-9} 	imes 3}{0.5 	imes 10^{-3}} = \frac{2700 	imes 10^{-9}}{0.5 	imes 10^{-3}} = 5400 	imes 10^{-6} \mathrm{~m} = 5.4 \mathrm{~mm}$.