Q129 / 180 · COMEDK 2018 · Physics

In a Fraunhofer diffraction experiment at a single slit using a light of wavelength $400 \mathrm{~nm}$, the first minimum is formed at an angle of $30^{\circ}$. The direction $\theta$ of the first secondary maximum is given by

  • A
    $\sin ^{-1}(2 / 3)$
  • B
    $\sin ^{-1}(3 / 4)$
  • Correct answer
    C
    $\sin ^{-1}(1 / 4)$
  • D
    $\tan ^{-1}(2 / 3)$

Solution

Given, wavelength of used light, $$ \lambda=400 \mathrm{~nm} $$ For minima in Fraunhofer diffraction, $\therefore \quad d \sin 30^{\circ}=1 \times 400$ $\Rightarrow \quad d=800 \mathrm{~nm}$ For maxima in diffraction, $$ d \sin \theta=(2 n+1) \frac{\lambda}{2} $$ For first maxima $n=0$ $$ \begin{array}{ll} \therefore & d \sin \theta=(0+1) \frac{\lambda}{2} \\ \Rightarrow & d \sin \theta=\frac{\lambda}{2} \\ \Rightarrow & 800 \sin \theta=\frac{400}{2} \\ \Rightarrow & \sin \theta=\frac{1}{4} \\ \Rightarrow & \theta=\sin ^{-1}\left(\frac{1}{4}\right) \end{array} $$

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

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