Sodium light of wavelengths 650, nm and 655, | vedclass

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(C) For a single slit diffraction,the condition for the $n^{th}$ maxima is given by $a \sin 	heta = (n + \frac{1}{2}) \lambda$.For the first maxima,$

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$3$

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(C) For a single slit diffraction,the condition for the $n^{th}$ maxima is given by $a \sin 	heta = (n + \frac{1}{2}) \lambda$.For the first maxima,$n = 1$,so $a \sin 	heta = \frac{3 \lambda}{2}$.Since $	heta$ is very small,$\sin 	heta \approx 	heta = \frac{y}{L}$,where $y$ is the distance from the central maxima and $L$ is the distance to the screen.Thus,$y = \frac{3 \lambda L}{2 a}$.The separation between the first maxima for two wavelengths $\lambda_1 = 650\, nm$ and $\lambda_2 = 655\, nm$ is $\Delta y = y_2 - y_1 = \frac{3 L}{2 a} (\lambda_2 - \lambda_1)$.Given $L = 2.0\, m$,$a = 0.5\, mm = 0.5 	imes 10^{-3}\, m$,$\lambda_1 = 650 	imes 10^{-9}\, m$,and $\lambda_2 = 655 	imes 10^{-9}\, m$.$\Delta y = \frac{3 	imes 2.0}{2 	imes 0.5 	imes 10^{-3}} 	imes (655 - 650) 	imes 10^{-9}$.$\Delta y = \frac{6}{10^{-3}} 	imes 5 	imes 10^{-9} = 6 	imes 5 	imes 10^{-6} = 30 	imes 10^{-6} = 3 	imes 10^{-5}\, m$.Comparing with $x 	imes 10^{-5}\, m$,we get $x = 3$.

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