A parallel beam of light of wavelength 500; n | vedclass

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(B) Given:Wavelength of light beam,$\lambda = 500\; nm = 500 	imes 10^{-9}\; m$Distance of the screen from the slit,$D = 1\; m$Distance of the first

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$0.2$

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(B) Given:Wavelength of light beam,$\lambda = 500\; nm = 500 	imes 10^{-9}\; m$Distance of the screen from the slit,$D = 1\; m$Distance of the first minimum from the centre of the screen,$x = 2.5\; mm = 2.5 	imes 10^{-3}\; m$For the first minimum in single slit diffraction,the condition is given by $a \sin 	heta = n \lambda$,where $a$ is the slit width and $n = 1$.For small angles,$\sin 	heta \approx 	an 	heta = \frac{x}{D}$.Thus,$a \left( \frac{x}{D} ight) = n \lambda$.Rearranging for slit width $a$:$a = \frac{n \lambda D}{x} = \frac{1 	imes 500 	imes 10^{-9} 	imes 1}{2.5 	imes 10^{-3}}$$a = \frac{500 	imes 10^{-9}}{2.5 	imes 10^{-3}} = 200 	imes 10^{-6}\; m = 0.2 	imes 10^{-3}\; m = 0.2\; mm$.Therefore,the width of the slit is $0.2\; mm$.

$A$ parallel beam of light of wavelength $500\; nm$ falls on a narrow slit and the resulting diffraction pattern is observed on a screen $1\;m$ away. It is observed that the first minimum is at a distance of $2.5\;mm$ from the centre of the screen. Find the width of the slit. (in $; mm$)

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