Q. A single slit Fraunhofer diffraction pattern is formed with white light. For what wavelength of light the third secondary maximum in the diffraction pattern coincides with the second secondary maximum in the pattern for red light of wavelength$\ 6500\mathring{\mathrm{A}}$?
A) $4400\ \mathring{\mathrm{A}}$
B) $4100\ \mathring{\mathrm{A}}$
C) $4642.8\ \mathring{\mathrm{A}}$
D) $9100\ \mathring{\mathrm{A}}$
Explanation
(c) $x = \frac{(2n + 1)\lambda D}{2a}$
For red light $\ x = \frac{(4 + 1)D}{2a} \times 6500$
For unknown wavelength of light,
$x = \frac{(6 + 1)D}{2a} \times \lambda$
Accordingly
$\therefore 5 \times 6500 = 7 \times \lambda$
$\Rightarrow \lambda = \frac{5}{7} \times 6500 = 4642.8\ \mathring{\mathrm{A}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.