A right-angled isosceles prism is held on the | vedclass

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(A) Given,$n_{L}=C_{A} n_{A}+(1-C_{A}) n_{B}$.Here,$n_{A}=1.5$ and $n_{B}=1.3$.Substituting these values,we get:$n_{L} = C_{A}(1.5) + (1-C_{A})(1.3) =

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(A) Given,$n_{L}=C_{A} n_{A}+(1-C_{A}) n_{B}$.Here,$n_{A}=1.5$ and $n_{B}=1.3$.Substituting these values,we get:$n_{L} = C_{A}(1.5) + (1-C_{A})(1.3) = 1.3 + 0.2 C_{A}$.Using Snell's law at the prism-liquid interface for the critical angle $	heta_{C}$:$n_{p} \sin 	heta_{C} = n_{L} \sin 90^{\circ}$.Since $n_{p} = 1.5$,we have:$\sin 	heta_{C} = \frac{n_{L}}{1.5} = \frac{1.3 + 0.2 C_{A}}{1.5}$.$	heta_{C} = \sin^{-1} \left( \frac{1.3 + 0.2 C_{A}}{1.5} ight)$.At $C_{A} = 0$,$	heta_{C} = \sin^{-1} \left( \frac{1.3}{1.5} ight) = \sin^{-1} \left( \frac{13}{15} ight) \approx 60^{\circ}$.At $C_{A} = 1$,$	heta_{C} = \sin^{-1} \left( \frac{1.5}{1.5} ight) = \sin^{-1}(1) = 90^{\circ}$.As $C_{A}$ increases from $0$ to $1$,$	heta_{C}$ increases from $60^{\circ}$ to $90^{\circ}$. The function $	heta_{C} = \sin^{-1}(f(C_{A}))$ is a non-linear increasing function. Graph $(A)$ correctly represents this variation.

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