In a single slit diffraction experiment, first minimum for red light…

Here, $λ_r = 660 nm ; λ' = ?$

For diffraction minima,

$a \sin θ = nλ, \sin θ =\frac{nλ}{a}$

For first minima of red,

$\sin θ=\frac{(1)λ_r}{a}$

For diffraction maxima

$a \sin θ = (2n + 1)\frac{λ}{2}$

∴ For first maxima of $λ'$

$a \sin θ' =\frac{3λ'}{2};\sin θ' =\frac{3λ'}{2a}$

As the two coincide, therefore,

$sin θ' = sin θ$

$\frac{3λ'}{2a}=\frac{λ_r}{a}$

or $λ' = \frac{2}{3}λ_r$

or  $λ' = \frac{2}{3}(660) 440nm$