Two wavelengths 590 nm and 596 nm of sodium | vedclass

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(A) The condition for the $n^{th}$ secondary maximum in a single slit diffraction pattern is given by $\sin 	heta = (2n + 1) \frac{\lambda}{2a}$.For

Vedclass · Accepted Answer

$7.5 	imes 10^{-6} \ m$

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(A) The condition for the $n^{th}$ secondary maximum in a single slit diffraction pattern is given by $\sin 	heta = (2n + 1) \frac{\lambda}{2a}$.For the first secondary maximum $(n = 1)$,$\sin 	heta = \frac{3\lambda}{2a}$.Since the angle $	heta$ is very small,$\sin 	heta \approx 	an 	heta = \frac{x}{D}$,where $x$ is the distance from the central maximum and $D$ is the distance to the screen.Thus,$x = \frac{3\lambda D}{2a}$.The separation between the positions of the first secondary maxima for two wavelengths $\lambda_1$ and $\lambda_2$ is $\Delta x = \frac{3D}{2a} (\lambda_2 - \lambda_1)$.Given: $\lambda_1 = 590 	imes 10^{-9} \ m$,$\lambda_2 = 596 	imes 10^{-9} \ m$,$D = 2 \ m$,$a = 2.4 	imes 10^{-3} \ m$.Substituting the values:$\Delta x = \frac{3 	imes 2 	imes (596 - 590) 	imes 10^{-9}}{2 	imes 2.4 	imes 10^{-3}}$$\Delta x = \frac{6 	imes 6 	imes 10^{-9}}{4.8 	imes 10^{-3}} = \frac{36 	imes 10^{-9}}{4.8 	imes 10^{-3}} = 7.5 	imes 10^{-6} \ m$.

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