In a Fraunhofer diffraction experiment at a single slit using light of wavelength 400 nm, the first minimum is formed at an angle of $30^{0}$ with the direction of incident light. Then the angular position $θ$ (with the direction of incident light) of the first secondary maximum is

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

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

$60^{0}$

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

answer is C.

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Detailed Solution

For $1^{s t}$ minimum, $b \sin θ = λ \Rightarrow b \sin 30 = λ ................ (1)$

For $1^{s t}$ secondary maximum $b \sin θ = \frac{3 λ}{2} ............... (2)$

$\frac{(1)}{(2)} \Rightarrow \frac{\sin 30}{\sin θ} = \frac{λ}{\frac{3 λ}{2}} \Rightarrow \sin θ = \frac{3}{4} \Rightarrow θ = \sin^{- 1} (\frac{3}{4})$

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