Question

In Young's experiment, the wavelength of monochromatic light used is $6000A^{\circ }$. The optical path difference between the rays from the two coherent sources at point P on the screen is 0.0075 mm and at a point Q on the screen is 0.0075 mm and at a point Q on the screen is 0.0015 mm. How many bright and dark bands are observed between the two points P and Q?

Answer 1

Given:
$\lambda =6000A^{\circ }$ $=6\times {10}^{-7}m$
Path difference, $S_{2}P-S_{1}P$ = 0.0075 mm $=7.5\times {10}^{-6}m$
Path difference, $S_{2}Q-S_{1}Q$ = 0.0015 mm $=1.5\times {10}^{-6}m$
For point P,
$S_{2}P-S_{1}P=(2n-1)\frac{\lambda }{2}$
$\therefore \frac{S_{2}P-S_{1}P}{\lambda }=\frac{(2n-1)}{2}$
$\frac{7.5\times {10}^{-6}}{6\times {10}^{-7}}=\frac{2n-1}{2}$
$\frac{75}{6}=\frac{2n-1}{2}$
$\frac{25}{2}=\frac{2n-1}{2}$
$\frac{26-1}{2}=\frac{2n-1}{2}$
$\therefore \frac{2\times 13-1}{2}=\frac{2n-1}{2}$
Above equation gives,
n = 13
$\therefore {13}^{th}$ band will occur at P.
$\therefore$ No. of bright bands between P and central bright band O = 12
Similarly, for point Q,
$S_{2}Q-S_{1}Q=(2n-1)\frac{\lambda }{2}$
$\therefore \frac{1.5\times {10}^{-6}}{6\times {10}^{-7}}=\frac{(2n-1)}{2}$
$\frac{15}{6}=\frac{2n-1}{2}$
$\frac{5}{2}=\frac{2n-1}{2}$
$\frac{6-1}{2}=\frac{2n-1}{2}$
$\therefore \frac{2\times 3-1}{2}=\frac{2n-1}{2}$
Above equation gives,
n = 3
$\therefore 3^{rd}$ dark bands will occur at Q.
$\therefore$ No. of bright bands between Q and central bright band O = 2.
$\therefore$ No. of bright bands between P and Q = 12 + 2 + 1 (central bright band) = 15