In Fraunhofer diffraction pattern,slit width | vedclass

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(D) Given: Slit width $a = 0.3 \ mm = 0.3 	imes 10^{-3} \ m$,Screen distance $D = 1.5 \ m$,Wavelength $\lambda = 4500 \ Å = 4500 	imes 10^{-10} \ m$

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$4.5$

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(D) Given: Slit width $a = 0.3 \ mm = 0.3 	imes 10^{-3} \ m$,Screen distance $D = 1.5 \ m$,Wavelength $\lambda = 4500 \ Å = 4500 	imes 10^{-10} \ m$.For Fraunhofer diffraction,the position of the $n^{th}$ minimum is given by $a \sin 	heta = n \lambda$. For small $	heta$,$\sin 	heta \approx 	heta = \frac{y}{D}$.Thus,$a \left( \frac{y}{D} ight) = n \lambda \implies y_n = \frac{n \lambda D}{a}$.The distance of the first minimum $(n=1)$ from the central maximum is $y_1 = \frac{\lambda D}{a}$.The distance between the first minimum on either side of the central maximum is $2y_1 = \frac{2 \lambda D}{a}$.Substituting the values: $2y_1 = \frac{2 	imes 4500 	imes 10^{-10} 	imes 1.5}{0.3 	imes 10^{-3}}$.$2y_1 = \frac{13500 	imes 10^{-10}}{0.3 	imes 10^{-3}} = 45000 	imes 10^{-7} \ m = 4.5 	imes 10^{-3} \ m = 4.5 \ mm$.

In Fraunhofer diffraction pattern,slit width is $0.3 \ mm$ and screen is at $1.5 \ m$ away from the lens. If wavelength of light used is $4500 \ Å$,then the distance between the first minimum on either side of the central maximum is [ $\theta$ is small and measured in radian.] (in $mm$)

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