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  • Consider a ray of light incident from air onto a slab of glass (refractive index n ) of width d, at an angle <img alt="\mathrm{\theta}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5C

Consider a ray of light incident from air onto a slab of glass (refractive index n ) of width d, at an angle \mathrm{\theta}. The phase difference between the ray reflected by the top surface of the glass and the bottom surface is:

Option: 1

\mathrm{\frac{4 \pi \mathrm{d}}{\lambda}\left(1-\frac{1}{\mathrm{n}^2} \sin ^2 \theta\right)^{-1 / 2}+\pi}


Option: 2

\mathrm{\frac{4 \pi \mathrm{d}}{\lambda}\left(1-\frac{1}{\mathrm{n}^2} \sin ^2 \theta\right)^{1 / 2}}


Option: 3

\mathrm{\frac{4 \pi \mathrm{d}}{\lambda}\left(1-\frac{1}{\mathrm{n}^2} \sin ^2 \theta\right)^{1 / 2}+\frac{\pi}{2}}


Option: 4

\mathrm{\frac{4 \pi \mathrm{d}}{\lambda}\left(1-\frac{1}{\mathrm{n}^2} \sin ^2 \theta\right)^{1 / 2}+2 \pi}


Answers (1)

best_answer

Consider the diagram, the ray (P) is incident at an angle \mathrm{\theta} and gets reflected in the direction \mathrm{\mathrm{P}^{\prime}} and refracted in the direction \mathrm{\mathrm{P}^{\prime \prime}}. Due to reflection from the glass medium, there is a phase change of \mathrm{\pi.}

Time taken to travel along OP"

\mathrm{ \Delta \mathrm{t}=\frac{\mathrm{OP}}{\mathrm{v}}=\frac{\mathrm{d} / \cos \mathrm{r}}{\mathrm{c} / \mathrm{n}}=\frac{\mathrm{nd}}{\mathrm{c} \cos \mathrm{r}} }

\mathrm{\text { From Snell's law, } \mathrm{n}=\frac{\sin \theta}{\sin \mathrm{r}} \Rightarrow \sin \mathrm{r}=\frac{\sin \theta}{\mathrm{n}}}

\mathrm{\cos r=\sqrt{1-\sin ^2 r}=\sqrt{1-\frac{\sin ^2 \theta}{n^2}}}

\mathrm{\therefore \Delta t=\frac{n d}{c\left(1-\frac{\sin ^2 \theta}{n^2}\right)^{1 / 2}}=\frac{n^2 d}{c}\left(1-\frac{\sin ^2 \theta}{n^2}\right)^{-1 / 2}}

\mathrm{\text { Phase difference }=\Delta \phi=\frac{2 \pi}{\mathrm{T}} \times \Delta \mathrm{t}=\frac{2 \pi \mathrm{nd}}{\lambda}\left(1-\frac{\sin ^2 \theta}{\mathrm{n}^2}\right)^{-1 / 2}}

\mathrm{\text { So, net phase difference }=\Delta \phi+\pi=\frac{4 \pi \mathrm{d}}{\lambda}\left(1-\frac{1}{\mathrm{n}^2} \sin ^2 \theta\right)^{-1 / 2}+\pi}

Posted by

Rishabh

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