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(D) For the $n^{th}$ secondary maxima in a single-slit diffraction pattern,the path difference $x$ between the rays from the edges of the slit is give

Vedclass · Accepted Answer

$3 \pi$

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(D) For the $n^{th}$ secondary maxima in a single-slit diffraction pattern,the path difference $x$ between the rays from the edges of the slit is given by the condition $x = (2n + 1) \frac{\lambda}{2}$,where $n = 1, 2, 3, \dots$ represents the order of the secondary maxima.For the first secondary maximum $(n = 1)$,the path difference is $x_1 = (2(1) + 1) \frac{\lambda}{2} = \frac{3\lambda}{2}$.The phase difference $\phi$ is related to the path difference $x$ by the formula $\phi = \frac{2\pi}{\lambda} x$.Substituting the value of $x_1$,we get $\phi_1 = \frac{2\pi}{\lambda} \left( \frac{3\lambda}{2} \right) = 3\pi$.

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