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

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$\sin^{- 1} (\frac{2}{3})$

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

$\sin^{- 1} (\frac{1}{4})$

$\sin^{- 1} (\frac{2}{5})$

answer is B.

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Detailed Solution

For first diffraction minimum $a \sin θ' = λ \Rightarrow a = \frac{λ}{\sin θ'}$

For first secondary maximum, we use,

$asinθ = \frac{3 λ}{2}$

$sinθ = \frac{3 λ}{2} \times \frac{1}{a} = \frac{3 λ}{2} \times \frac{sinθ'}{λ}$

$sinθ = \frac{3}{2} \times \sin 30^{o} = \frac{3}{4}$

$θ = \sin^{- 1} (\frac{3}{4})$

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