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(B) For the $n^{	ext{th}}$ secondary maximum in a single slit diffraction pattern,the position is given by $x_n = \frac{(2n+1) \lambda D}{2a}$.For th

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$4425$

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(B) For the $n^{	ext{th}}$ secondary maximum in a single slit diffraction pattern,the position is given by $x_n = \frac{(2n+1) \lambda D}{2a}$.For the second secondary maximum $(n=2)$ with wavelength $\lambda = 6195 Å$,the position is $x_2 = \frac{(2 	imes 2 + 1) \lambda D}{2a} = \frac{5 \lambda D}{2a}$.For the third secondary maximum $(n=3)$ with wavelength $\lambda_0$,the position is $x_3 = \frac{(2 	imes 3 + 1) \lambda_0 D}{2a} = \frac{7 \lambda_0 D}{2a}$.Since the positions coincide,$x_2 = x_3$,which implies $\frac{5 \lambda D}{2a} = \frac{7 \lambda_0 D}{2a}$.Simplifying this,we get $5 \lambda = 7 \lambda_0$.Therefore,$\lambda_0 = \frac{5 \lambda}{7} = \frac{5 	imes 6195 Å}{7} = 5 	imes 885 Å = 4425 Å$.

$A$ single slit diffraction pattern is formed with light of wavelength $6195 Å$. The second secondary maximum for this wavelength coincides with the third secondary maximum in the pattern for light of wavelength $\lambda_0$. The value of $\lambda_0$ is (in $Å$)

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