Question

In figure , light 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 surfaces,

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

From figure, we can deduce that,

${\mu }_1\sin \theta ={\mu }_2\sin r={\mu }_3\sin c$

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

Substituting the given data we get,

$1.6\sin \theta =1.3\sin {90}^{\circ }$

$\Rightarrow \sin \theta =\frac{1.3}{1.6}$

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


Hence, option (b) is the correct answer.

Why this question? Note:- Across an interface, product of refractive index and sine of the angle of incidence is constant. $$\mu \sin \theta =\text{constant}$$