Q. White light is used to illuminate the two slits in a Young’s double slit experiment. The separation between slits is b and the screen is at a distance $d( > \ > b)$ from the slits. At a point on the screen directly in front of one of the slits, certain wavelengths are missing, figure. Some of these missing wavelengths are
A) $\lambda = \frac{b^{2}}{d},\frac{2b^{2}}{3d}$
B) $\lambda = \frac{b^{2}}{2d},\frac{3b^{2}}{2d}$
C) $\lambda = \frac{2b^{2}}{3d}$
D) $\lambda = \frac{3b^{2}}{4d}$
Explanation
(a) $S_{2}P = \left( d^{2} + b^{2} \right)^{1/2} = d\left( 1 + \frac{b^{2}}{d^{2}} \right)^{1/2}\ $
$= d\left( 1 + \frac{b^{2}}{d^{2}} \right) = d + \frac{b^{2}}{2d}$
Path difference=$S_{2}P - S_{1}P$
$x = d + \frac{b^{2}}{d^{2}} - d = \frac{b^{2}}{2d}$
For missing wavelengths $(2n - 1)\frac{\lambda}{2} = x = \frac{b^{2}}{2d}$
For $n = 1,\ \lambda = \frac{b^{2}}{d},$
For $n = 2,\ \lambda = \frac{2b^{2}}{3d},$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.