1) When a plastic thin film of refractiive index 1.45 is placed in the path of one of the interfering waves then the central fringe is displaced through width of five fringes . The thickness of the film , if the wavelength of light is 5890 Å , will be A) $6.544\times 10^{-4}$ cm B) $6.544\times 10^{-4}$ m C) $6.54\times 10^{-4}$ cm D) $6.5\times 10^{-4}$ cm Answer: Option AExplanation: $X_{0}=\frac{\beta}{\lambda}(\mu-1)t\Rightarrow5\beta=\frac{\beta (0.45)t}{5890\times 10^{-10}}$ $\therefore$ $t= \frac{5\times 5890\times 10^{-10}}{0.45}=6.544\times 10^{-4}cm$
