Fraunhofer diffraction experiment at a single slit using light of wavelength 400 nm, the first minimum is formed at an angle of 30°. Then the direction $θ$ of the secondary maximum is given by :

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$\tan^{- 1} (4 / 3)$

$\sin^{- 1} (3 / 4)$

60°

$\tan^{- 1} (3 / 4)$

answer is B.

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Detailed Solution

Minimum condition of the diffraction
$\begin{array}{rcl} dsin θ & = & nλ \end{array}$

$\begin{array}{rcl} dsin 30^{\circ} & = & 1 \times λ . . . . . . . (1) \end{array}$

Maximum condition of the diffraction

$dsin θ = (2 n - 1) \frac{λ}{2}$

$\Rightarrow dsin θ = (2 \times 2 - 1) \frac{λ}{2} . . . . (2)$

$\frac{(2)}{(1)} = \frac{\sin θ}{1 / 2} = \frac{3 / 2}{1}$

$\Rightarrow \sin θ = \frac{3}{4} \Rightarrow θ = \sin^{- 1} \frac{3}{4}$

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