Two wavelengths of sodium light 590 ,nm and 5 | vedclass

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(D) The condition for the $n^{th}$ maximum in a single slit diffraction pattern is given by $d \sin 	heta = (n + \frac{1}{2}) \lambda$,where $n = 1,

Vedclass · Accepted Answer

$6.75$

Vedclass · Answer

(D) The condition for the $n^{th}$ maximum in a single slit diffraction pattern is given by $d \sin 	heta = (n + \frac{1}{2}) \lambda$,where $n = 1, 2, 3, ...$ For the first maximum $(n=1)$,the position $x$ on the screen is given by $x = \frac{3 \lambda D}{2 d}$. Given: $\lambda_1 = 590 \,nm$,$\lambda_2 = 596 \,nm$,$D = 1.5 \,m$,$d = 2 	imes 10^{-6} \,m$. The separation between the positions of the first maximum is $\Delta x = x_2 - x_1 = \frac{3 D}{2 d} (\lambda_2 - \lambda_1)$. Substituting the values: $\Delta x = \frac{3 	imes 1.5}{2 	imes 2 	imes 10^{-6}} 	imes (596 - 590) 	imes 10^{-9} \,m$. $\Delta x = \frac{4.5}{4 	imes 10^{-6}} 	imes 6 	imes 10^{-9} \,m$. $\Delta x = 1.125 	imes 10^6 	imes 6 	imes 10^{-9} \,m = 6.75 	imes 10^{-3} \,m = 6.75 \,mm$.

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