Q. In the far field diffraction pattern of a single slit under polychromatic illumination, the first minimum with the wavelength $\lambda_{1}$ is found to be coincident with the third maximum at $\lambda_{2}$. So
A) $3\lambda_{1} = 0.3\lambda_{2}$
B) $3\lambda_{1} = \lambda_{2}$
C) $\lambda_{1} = 3.5\lambda_{2}$
D) $0.3\lambda_{1} = 3\lambda_{2}$
Explanation
(c) Position of first minima = position of third maxima $i.e.,$
$\frac{1 \times \lambda_{1}D}{d} = \frac{(2 \times 3 + 1)}{2}\frac{\lambda_{2}D}{d} \Rightarrow \lambda_{1} = 3.5\lambda_{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.