Physics - Wave Optics Question with Solution | TestHub

Wavelength $\lambda =500\mathrm{nm}=500\times 10^{-9}\mathrm{m}$ Thickness of the film, $t=2\mu \mathrm{m}=2\times 10^{-6}\mathrm{m}$ Refractive index $\mu =1.5$ When there is no thin film placed in the path of any one of the two beams, the path difference between them is given by $dy/D$ (considering the two beams meetings at a point $P)$. In case we put a thin film in the path of one of the beams, then the optical path
of that beam gets longer. Now for the central maximum, path difference in absence of the film is $\Delta x=0$. But when we put the film, the path difference becomes $$\begin{array}{rl}\Delta x& =\mu t-t=(\mu -1)t\\ & =(1.5-1)\times 2\times 10^{-6}=10^{-6}\mathrm{m}=1\mathrm{mm}.\end{array}$$ Now, $\Delta x=\frac{dy}{D}\Rightarrow y=\frac{D}{d}\Delta x=\frac{D}{d}\times 1\mu \mathrm{m}$ And the fringe width is given by $$\begin{array}{rl}& W=\frac{D\lambda }{d}=\frac{D}{d}\times 500\times 10^{-9}\mathrm{m}=\frac{D}{d}\times 0.5\times 1\mu \mathrm{m}\\ \therefore & y=\frac{D}{d}\times 1\mu \mathrm{m}=2\times \frac{D}{d}\times \frac{1}{2}\times 1\mu \mathrm{m}=2W.\end{array}$$ As the film is placed in the path of the upper beam, the central maximum will shift upward by nearly two fringes.