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(A) According to Snell's Law,the refractive index of the liquid with respect to air $(n_{la})$ is given by:$n_{la} = \frac{\sin 	heta}{\sin \alpha}$B

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$\frac{\sin \alpha}{\sin 	heta}$

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(A) According to Snell's Law,the refractive index of the liquid with respect to air $(n_{la})$ is given by:$n_{la} = \frac{\sin 	heta}{\sin \alpha}$By definition,the critical angle $	heta_c$ is the angle of incidence in the denser medium (liquid) for which the angle of refraction in the rarer medium (air) is $90^{\circ}$.The refractive index of air with respect to the liquid $(n_{al})$ is:$n_{al} = \frac{1}{n_{la}} = \frac{\sin 	heta_c}{\sin 90^{\circ}}$Since $\sin 90^{\circ} = 1$,we have:$\sin 	heta_c = \frac{1}{n_{la}}$Substituting the value of $n_{la}$:$\sin 	heta_c = \frac{1}{\frac{\sin 	heta}{\sin \alpha}} = \frac{\sin \alpha}{\sin 	heta}$

$A$ beam of light is incident from air on the surface of a liquid. The angle of incidence is $\theta$ and the angle of refraction is $\alpha$. If the critical angle for the liquid when surrounded by air is $\theta_c$,then $\sin \theta_c$ is:

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