Q. A light of wavelength $5890\ \mathring{\mathrm{A}}$ falls normally on a thin air film. The minimum thickness of the film such that the film appears dark in reflected light is
A) $2.945 \times 10^{- 7}m$
B) $3.945 \times 10^{- 7}m$
C) $4.95 \times 10^{- 7}m$
D) $1.945 \times 10^{- 7}m$
Explanation
(a) If thin film appears dark
$2\mu t\cos{r = n\lambda}$ for normal incidence $r = 0{^\circ}$
$\Rightarrow 2\mu t = n\lambda \Rightarrow t = \frac{n\lambda}{2\mu}$
$\Rightarrow t_{\min} = \frac{\lambda}{2\mu} = \frac{5890 \times 10^{- 10}}{2 \times 1} = 2.945 \times 10^{- 7}m$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.