Question

A monochromatic light of wave length 500 A^o is incident on two identical slits separated by a distance of 5 x 10^-4 m. The interference pattern is seen on a screen placed at a distance of 1 m from the plane of the slits. A thin glass plate of thickness 1.5 x 10^-6 m and refractive index 1.5 is placed between one of the slits and the screen. Find the intensity at the centre of the slit now ?

Answer 1

$$\begin{gathered} \mathrm{The}\mathrm{path}\mathrm{difference}\mathrm{introduced}\mathrm{due}\mathrm{to}\mathrm{introduction}\mathrm{of}\mathrm{glass}\mathrm{plate} \\ o\mathrm{f}\mathrm{thickess}t\mathrm{and}\mathrm{refractive}\mathrm{index}\left(\mu \right)\mathrm{is}\mathrm{given}\mathrm{by} \\ D=\left(\mu -1\right)t \\ \mathrm{Phase}\mathrm{difference}\mathrm{introduced},\varphi =\frac{2\mathrm{\pi }}{\lambda }\left(\mu -1\right)t \\ \Rightarrow \varphi =\frac{2\mathrm{\pi }}{5\times {10}^{-7}}\times \left(1.5-1\right)\times 1.5\times {10}^{-6} \\ \Rightarrow \varphi =\frac{2\mathrm{\pi }}{5\times {10}^{-7}}\times 7.5\times {10}^{-6}=3\mathrm{\pi } \\ \mathrm{Now},\mathrm{according}\mathrm{to}\mathrm{the}\mathrm{formula}\mathrm{of}\mathrm{relative}\mathrm{intensity},\mathrm{we}\mathrm{have} \\ I=I_1+I_2+2\sqrt{I_1{I}_2}\cos \varphi \\ \mathit{\Rightarrow }I=I_0\mathit{+}{I}_0+2\sqrt{I_0{I}_0}\cos \frac{2\mathrm{\pi }}{\lambda }\left(\mu -1\right)t\left(\mathrm{Le}t{I}_1=I_2=I_0\right) \\ \mathit{\Rightarrow }I=\mathit{2}{I}_0+2I_0\cos 3\mathrm{\pi } \\ \mathit{\Rightarrow }I=2I_{\mathit{0}}\mathit{-}\mathit{2}{I}_0=0 \\ \mathrm{Thus},\mathrm{intensity}\mathrm{at}\mathrm{the}\mathrm{centre}\mathrm{of}\mathrm{the}\mathrm{slit}\mathrm{is}\mathrm{zero}. \end{gathered}$$