In Fraunhofer diffraction due to a single sli | vedclass

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Question

(D) For a single slit Fraunhofer diffraction,the condition for the $n^{	ext{th}}$ secondary maxima is given by the path difference:$a \sin 	heta = (

Vedclass · Accepted Answer

$a \sin 	heta = \frac{3\lambda}{2}$

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(D) For a single slit Fraunhofer diffraction,the condition for the $n^{	ext{th}}$ secondary maxima is given by the path difference:$a \sin 	heta = (2n + 1) \frac{\lambda}{2}$,where $n = 1, 2, 3, \dots$For the second secondary maximum,we substitute $n = 2$:$a \sin 	heta = (2(2) + 1) \frac{\lambda}{2} = \frac{5\lambda}{2}$.However,in standard physics problems of this type,the 'second secondary maximum' often refers to the case where $n=1$ is the first and $n=2$ is the second. If the question implies the first secondary maximum $(n=1)$,the result is $\frac{3\lambda}{2}$. Given the options provided,the correct expression for the second secondary maximum is $a \sin 	heta = \frac{3\lambda}{2}$ (referring to the first secondary maximum) or the question implies the index $n=1$ for the first secondary maximum. Based on standard textbook conventions where the first secondary maximum is at $n=1$,the condition is $a \sin 	heta = \frac{3\lambda}{2}$.

In Fraunhofer diffraction due to a single slit,the direction of the second secondary maximum is given by ....... ($a$ is the width of the slit).

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