Q. The Fraunhofer diffraction pattern of a single slit is formed in the focal plane of a lens of focal length 1 m. The width of slit is 0.3 mm. If third minimum is formed at a distance of 5 mm from central maximum, then wavelength of light will be
A) 5000$\mathring{\mathrm{A}}$
B) 2500$\mathring{\mathrm{A}}$
C) 7500$\mathring{\mathrm{A}}$
D) 8500$\mathring{\mathrm{A}}$
Explanation
(a) $a\ \sin{\theta = n\lambda}$
$\frac{ax}{f} = 3\lambda\ or\ \lambda = \frac{ax}{3f} = \frac{0.3 \times 10^{- 3} \times 5 \times 10^{- 3}}{3 \times 1}$
$= 5 \times 10^{7}\ m = 5000\ \mathring{\mathrm{A}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.