Question

A right-angle isosceles prism is held on the surface of a liquid composed of miscible solvents $A$ and $B$ of refractive index $n _{ A }=1.50$ and $n _{ B }=$ $1.30$, respectively. The refractive index of prism is $n_{p}=1.5$ and that of the liquid is given by $n L$ $= C _{ A } n _{ A }+\left(1- C _{ A }\right) n _{ B }$, where $C _{ A }$ is the percentage of solvent $A$ in the liquid :-

If $\theta_{C}$ is the critical angle at prism-liquid interface, the plot which best represents the variation of the critical angle with the precentage of solvent is:

• A.
• B.
• C.
• D.

Answer 1

Answer: (A)

$n_{ p } \sin \theta_{ C }= n _{ L } \sin 90^{\circ}$
$\theta_{ C }=\sin ^{-1}\left(\frac{ n _{ L }}{ n _{ P }}\right)$
$\theta_{ C }=\sin ^{-1}\left(\frac{ C _{ A } n _{ A }+\left(1- C _{ A }\right) n _{ B }}{1.5}\right)$
$\rightarrow$ Graph between $\theta_{C}$ and $C_{A}$ will be curve of $\sin ^{-1}$
Check for $C_{A}=0.5$, to find most appropriate graph
$\theta_{ C }=\sin ^{-1}\left(\frac{0.5(1.5)+0.5(1.3)}{1.5}\right) $
$\theta_{ C }=\sin ^{-1}\left(\frac{14}{15}\right) \simeq 69^{\circ}$