Light of wavelength 6000 , mathring{A} is inc | vedclass

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(A) The condition for the $n^{th}$ minima in a single slit diffraction pattern is given by $a \sin 	heta = n \lambda$. For small angles,$\sin 	heta

Vedclass · Accepted Answer

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

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(A) The condition for the $n^{th}$ minima in a single slit diffraction pattern is given by $a \sin 	heta = n \lambda$. For small angles,$\sin 	heta \approx 	an 	heta = \frac{x}{D}$.Thus,the position of the $n^{th}$ minima is $x = \frac{n \lambda D}{a}$.Given:Wavelength $\lambda = 6000 \, \mathring{A} = 6000 	imes 10^{-10} \, m = 6 	imes 10^{-7} \, m$.Slit width $a = 0.30 \, mm = 0.30 	imes 10^{-3} \, m = 3 	imes 10^{-4} \, m$.Distance $D = 2 \, m$.For the first minima,$n = 1$.Substituting the values:$x = \frac{1 	imes 6 	imes 10^{-7} 	imes 2}{3 	imes 10^{-4}}$$x = \frac{12 	imes 10^{-7}}{3 	imes 10^{-4}}$$x = 4 	imes 10^{-3} \, m$.

Light of wavelength $6000 \, \mathring{A}$ is incident on a slit of width $0.30 \, mm$. The screen is placed $2 \, m$ from the slit. Find the position of the first minima.

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