Red light of wavelength 5400 mathring{A} fro | vedclass

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(A) Given: Wavelength $(\lambda) = 5400 \ \mathring{A} = 5.4 	imes 10^{-7} \ m$. Slit width $(a) = 0.80 \ mm = 8 	imes 10^{-4} \ m$. Distance of scr

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$1.89$

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(A) Given: Wavelength $(\lambda) = 5400 \ \mathring{A} = 5.4 	imes 10^{-7} \ m$. Slit width $(a) = 0.80 \ mm = 8 	imes 10^{-4} \ m$. Distance of screen $(D) = 1.4 \ m$. The distance between the first two dark bands on each side of the central bright band is equivalent to the width of the central maximum. The formula for the width of the central maximum is given by $w = \frac{2 \lambda D}{a}$. Substituting the values: $w = \frac{2 	imes 5.4 	imes 10^{-7} 	imes 1.4}{8 	imes 10^{-4}}$. $w = \frac{15.12 	imes 10^{-7}}{8 	imes 10^{-4}} = 1.89 	imes 10^{-3} \ m$. Converting to $mm$,we get $w = 1.89 \ mm$.

Red light of wavelength $5400 \ \mathring{A}$ from a distant source falls on a slit $0.80 \ mm$ wide. Calculate the distance between the first two dark bands on each side of the central bright band in the diffraction pattern observed on a screen placed $1.4 \ m$ from the slit. (in $mm$)

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