In a single slit diffraction experiment, first minima for $λ_1 = 600\, nm$ coincides with first maxima for wavelength $λ_2$. The value of $λ_2$ is

400 nm

The correct answer is Option (1) → 400 nm

For a single slit diffraction, the first minima occurs at:

$a \sin \theta = \lambda_1$

First maxima occurs approximately at:

$a \sin \theta = 0.6 \lambda_2$ (approximation for first secondary maxima)

Given first minima of λ1 coincides with first maxima of λ2:

$\lambda_1 = 1.5 \lambda_2 \Rightarrow \lambda_2 = \frac{\lambda_1}{1.5} = \frac{600}{1.5} = 400\ \text{nm}$

$λ_2 = 400 nm$