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A Fraunhofer diffraction pattern due to a narrow slit is obtained on a screen placed at a distance \(D\) from the slit whose slit width is \(a\). The distance of first secondary maximum from the central maximum is
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The correct answer is:
\(\frac{3 D \lambda}{2 a}\)
In Fraunhofer diffraction pattern, the direction of secondary maximum is given as
\(\begin{aligned}
\theta & =(2 n+1) \frac{\lambda}{2 a}=(2 \times 1+1) \frac{\lambda}{2 a} \\
\Rightarrow \theta & =\frac{3 \lambda}{2 a}
\end{aligned}\)
\(\therefore\) Distance of first secondary maximum from the central maximum is given by
\(x=\theta D=\frac{3 \lambda}{2 a} \cdot D=\frac{3 D \lambda}{2 a}\)
\(\begin{aligned}
\theta & =(2 n+1) \frac{\lambda}{2 a}=(2 \times 1+1) \frac{\lambda}{2 a} \\
\Rightarrow \theta & =\frac{3 \lambda}{2 a}
\end{aligned}\)
\(\therefore\) Distance of first secondary maximum from the central maximum is given by
\(x=\theta D=\frac{3 \lambda}{2 a} \cdot D=\frac{3 D \lambda}{2 a}\)
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