In a single slit diffraction pattern,the dist | vedclass

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(C) For a single slit diffraction pattern,the condition for the $n^{th}$ secondary maximum is given by: $a \sin 	heta = (n + \frac{1}{2}) \lambda$,wh

Vedclass · Accepted Answer

$5200$

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(C) For a single slit diffraction pattern,the condition for the $n^{th}$ secondary maximum is given by: $a \sin 	heta = (n + \frac{1}{2}) \lambda$,where $a$ is the slit width,$	heta$ is the angle of diffraction,and $\lambda$ is the wavelength.Since $	heta$ is very small,$\sin 	heta \approx 	an 	heta = \frac{y}{D}$,where $y$ is the distance from the center and $D$ is the distance to the screen.For the second maximum,$n = 2$. Thus,$a \frac{y}{D} = (2 + \frac{1}{2}) \lambda = \frac{5}{2} \lambda$.Given: $a = 0.65 \ mm = 0.65 	imes 10^{-3} \ m$,$D = 1.3 \ m$,and $y = 2.6 \ mm = 2.6 	imes 10^{-3} \ m$.Substituting the values: $(0.65 	imes 10^{-3}) 	imes \frac{2.6 	imes 10^{-3}}{1.3} = \frac{5}{2} \lambda$.$(0.65 	imes 10^{-3}) 	imes (2 	imes 10^{-3}) = 2.5 \lambda$.$1.3 	imes 10^{-6} = 2.5 \lambda$.$\lambda = \frac{1.3 	imes 10^{-6}}{2.5} = 0.52 	imes 10^{-6} \ m = 5200 \ 	imes 10^{-10} \ m = 5200 \ Å$.

In a single slit diffraction pattern,the distance between the plane of the slit and the screen is $1.3 \ m$. The width of the slit is $0.65 \ mm$ and the second maximum is formed at a distance of $2.6 \ mm$ from the center of the screen. The wavelength of light used is: (in $Å$)

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