Question

In figure, a light ray refracts from material $1$ into a thin layer of material $2$, crosses that layer, and then is incident at the critical angle on the interface between materials $2$ and $3$. The angle $\theta$ is

• A. ${\sin }^{-1}\left(\frac{13}{15}\right)$
• B. ${\sin }^{-1}\left(\frac{13}{16}\right)$
• C. ${\cos }^{-1}\left(\frac{13}{15}\right)$
• D. ${\cos }^{-1}\left(\frac{13}{16}\right)$

Answer 1

The correct option is B ${\sin }^{-1}\left(\frac{13}{16}\right)$


We know that, for any two surface, where refraction occurs

$\mu \sin \theta =\text{constant}$

Now, applying this at interface of $1-2$ and $2-3$, we get

${\mu }_1\sin \theta ={\mu }_2\sin r=\mu \sin e\left[\text{fron figure}\right]$

$\Rightarrow {\mu }_1\sin \theta ={\mu }_3\sin e$

$\Rightarrow 1.6\sin \theta =1.3\sin {90}^{\circ }[\text{from question,}e={90}^{\circ }]$

$\Rightarrow \sin \theta =\frac{13}{16}$

$\Rightarrow \theta ={\sin }^{-1}\left(\frac{13}{16}\right)$

Hence, $\left(b\right)$ is the correct option.

$\begin{array}{c}\text{Why this question}?\\ \text{It gives you concept that there is no need to apply}\\ \text{Snell's law separtely for all the interfaces one -by- one}\\ \text{of a composite layer, instead we can take}\mu \sin \theta \\ \text{for the concern layers and equate them}\end{array}$