Question

A ray falls on a prism $ABC(AB=BC)$ and travels as shown in figure. If refractive index of glass with respect to air is $1.5$, find $\sin r$

• A. $\frac{3}{\sqrt{2}}$
• B. $\frac{3}{2\sqrt{2}}$
• C. $\frac{\sqrt{2}}{3}$
• D. $\frac{2\sqrt{2}}{3}$

Answer 1

The correct option is B $\frac{3}{2\sqrt{2}}$

$\begin{array}{c}\text{as}AB=BC\\ \angle A={45}^{\circ }=\angle C\\ \text{Here Do}//BC\\ \text{:}\angle AOD={45}^{\circ }\text{(altenate angle with C)}\\ \text{Hence}\angle i={45}^{\circ }\text{)}\\ \text{as}\angle gOA={90}^{\circ }\text{(Normal)}\end{array}$

$\begin{array}{c}\text{By snell's law}\\ \Rightarrow \text{Mglass sini = Uair sinr}\\ \Rightarrow \left(1.5\right)\sin {45}^{\circ }=\sin \gamma \\ \Rightarrow \frac{3}{2}\left(\frac{1}{\sqrt{2}}\right)=\sin \gamma \text{0.}\sin \gamma =\frac{3}{2\sqrt{2}}\\ \text{hence option}B\text{is correct .}\end{array}$