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 $nL$ $=C_A{n}_A+\left(1-C_A\right)n_B$, 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)

$n_p\sin {\theta }_C=n_L\sin 90^{\circ }$
${\theta }_C={\sin }^{-1}\left(\frac{n_L}{n_P}\right)$
${\theta }_C={\sin }^{-1}\left(\frac{C_A{n}_A+\left(1-C_A\right)n_B}{1.5}\right)$
$\to$ Graph between ${\theta }_C$ and $C_A$ will be curve of ${\sin }^{-1}$
Check for $C_A=0.5$, to find most appropriate graph
${\theta }_C={\sin }^{-1}\left(\frac{0.5(1.5)+0.5(1.3)}{1.5}\right)$
${\theta }_C={\sin }^{-1}\left(\frac{14}{15}\right)\simeq 69^{\circ }$