Light of wavelength 6000 , mathring A is inci | vedclass

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(B) For diffraction at a single slit,the condition for the $n^{th}$ minimum is given by $a \sin 	heta = n \lambda$.Given: Wavelength $\lambda = 6000

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$4 	imes 10^{-3} \, m$

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(B) For diffraction at a single slit,the condition for the $n^{th}$ minimum is given by $a \sin 	heta = n \lambda$.Given: Wavelength $\lambda = 6000 \, \mathring A = 6 	imes 10^{-7} \, m$,Slit width $a = 0.30 \, mm = 3 	imes 10^{-4} \, m$,Distance $D = 2 \, m$.For small angles,$\sin 	heta \approx 	an 	heta = \frac{x}{D}$.Substituting this into the condition: $a \left( \frac{x}{D} ight) = n \lambda$.For the first minimum,$n = 1$. Therefore,$x = \frac{n \lambda D}{a}$.$x = \frac{1 	imes (6 	imes 10^{-7} \, m) 	imes (2 \, m)}{3 	imes 10^{-4} \, m}$.$x = \frac{12 	imes 10^{-7}}{3 	imes 10^{-4}} \, m = 4 	imes 10^{-3} \, m$.

Light of wavelength $6000 \, \mathring A$ is incident on a slit of width $0.30 \, mm$. $A$ screen is placed at a distance of $2 \, m$ from the slit. Find the position of the first minimum.

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