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(B) For a single slit diffraction pattern,the condition for secondary maxima is given by $a \sin 	heta = (n + \frac{1}{2}) \lambda$,where $n = 1, 2,

Vedclass · Accepted Answer

$4560$

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(B) For a single slit diffraction pattern,the condition for secondary maxima is given by $a \sin 	heta = (n + \frac{1}{2}) \lambda$,where $n = 1, 2, 3, \dots$ is the order of the secondary maximum.For the second secondary maximum $(n = 2)$ with wavelength $\lambda_1 = 6384 Å$,the condition is $a \sin 	heta = (2 + \frac{1}{2}) \lambda_1 = \frac{5}{2} \lambda_1$.For the third secondary maximum $(n = 3)$ with wavelength $\lambda_0$,the condition is $a \sin 	heta = (3 + \frac{1}{2}) \lambda_0 = \frac{7}{2} \lambda_0$.Since the positions coincide,we equate the two expressions: $\frac{5}{2} \lambda_1 = \frac{7}{2} \lambda_0$.This simplifies to $5 \lambda_1 = 7 \lambda_0$.Substituting $\lambda_1 = 6384 Å$,we get $\lambda_0 = \frac{5 	imes 6384}{7} = \frac{31920}{7} = 4560 Å$.

$A$ single slit diffraction pattern is formed with light of wavelength $6384 Å$. 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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