Question

When the plastic thin film of refractive 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 will be ,if the wavelength of light is`5890Å`
A. `6.544xx10^(-4)cm`
B. `6.544xx10^(-4)m`
C. `6.54xx10^(-4)cm`
D. `6.5xx10^(-4)cm`

Answer 1

Answer:
`therefore X_(0)=(beta)/(lambda)(mu-1)trArr5beta=(beta(0.45)t)/(5890xx10^(-10))`
`:.t=(5xx5890xx10^(-10))/(0.45)=6.544xx10^(-4)cm`