Q143 / 160 · AP EAMCET 2020 (18 Sep Shift 2) · Physics

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

  • A
    \(\frac{3 D \lambda}{a}\)
  • Correct answer
    B
    \(\frac{3 D \lambda}{2 a}\)
  • C
    \(\frac{2 D \lambda}{3 a}\)
  • D
    \(\frac{2 D \lambda}{a}\)

Solution

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 \(x=\theta D=\frac{3 \lambda}{2 a} \cdot D=\frac{3 D \lambda}{2 a}\)

Hence, the correct answer is option: 'B'.

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