Q. In a biprism experiment, 5$^{th}$ dark fringe is obtained at a point. If a thin transparent film is placed in the path of one of waves, then 7$^{th}$ bright fringes is obtained at the same point. The thickness of the film in terms of wavelength $l$ and refractive index $\mu$ will be
A) $\frac{1.5\lambda}{(\mu - 1)}$
B) $1.5(\mu - 1)\lambda$
C) $2.5(\mu - 1)\lambda$
D) $\frac{2.5\lambda}{(\mu - 1)}$
Explanation
(d) For 5$^{th}$ dark fringe, $x_{1} = (2n - 1)\frac{\lambda}{2}\frac{D}{d} = \frac{9\lambda D}{2d}$
For 7$^{th}$ bright fringe, $x_{2} = n\lambda\frac{D}{d} = \frac{7\lambda D}{d}$
$x_{2} - x_{1} = (\mu - 1)t\frac{D}{d}$
$\frac{\lambda D}{d}\left\lbrack 7 - \frac{9}{2} \right\rbrack = (\mu - 1)t\frac{D}{d}$
$t = \frac{2.5\lambda}{(\mu - 1)}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.