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
Answer & explanation
Correct answer: option 1
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:
$\lambda_1 = 1.5 \lambda_2 \Rightarrow \lambda_2 = \frac{\lambda_1}{1.5} = \frac{600}{1.5} = 400\ \text{nm}$
$λ_2 = 400 nm$