In a single-slit Fraunhofer diffraction experiment, the first secondary minimum is observed at an angle of $3 0^{\circ}$ for light of wavelength $λ$ . If the same slit is now illuminated by light of the same wavelength, the angular position ( $θ$ ) of the first secondary maximum is
( $sin 3 0^{\circ} = \frac{1}{2}$ ).

Question

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For a single-slit Fraunhofer diffraction, the condition for the $n$ -th secondary minimum is given by:

$a sin θ = nλ$

For the first secondary minimum ( $n = 1$ ) at $θ = 3 0^{\circ}$ :

$a sin 3 0^{\circ} = λ$

$a (\frac{1}{2}) = λ \Rightarrow a = 2 λ$

The condition for the $n$ -th secondary maximum is given by:

$a sin θ^{'} = (n + \frac{1}{2}) λ$

For the first secondary maximum ( $n = 1$ ):

$a sin θ^{'} = \frac{3 λ}{2}$

Substituting $a = 2 λ$ into the equation:

$(2 λ) sin θ^{'} = \frac{3 λ}{2}$

$sin θ^{'} = \frac{3}{4}$

$θ^{'} = sin^{- 1} (\frac{3}{4})$

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

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