In a single slit diffraction experiment first minimum for red light (660 nm) coincides with first maximum of some other wavelength λ'. The value of λ' is
Answer & explanation
Correct answer: option 1
In a single slit diffraction experiment, position of minima is given by d sin θ = nλ
So for first minima of red $\sin \theta=1 \times\left(\frac{\lambda_R}{d}\right)$
and as first maxima is midway between first and second minima, for wavelength λ, its position will be
$d \sin \theta'=\frac{\lambda'+2 \lambda'}{2} \Rightarrow \sin \theta'=\frac{3 \lambda'}{2 d}$
According to given condition $\sin \theta=\sin \theta'$
$\Rightarrow \lambda'=\frac{2}{3} \lambda_{R}$ so $\lambda'=\frac{2}{3} \times 6600$ = 440 nm = 4400 Å