Respuesta :

Critical angle is a special case of angle of incidence for which angle of refraction is 90 degree.

Now using Snell's law

[tex]n_1\sin\theta_1=n_2\sin\theta_2[/tex]

Here

[tex]\begin{gathered} n_1=\text{ refractive index of incident medium;} \\ n_2=\text{refract}\imaginaryI\text{ve index of refracting medium;} \\ \theta_1=\text{ angle of incidence;} \\ \theta_2=\text{ angle of refraction;} \end{gathered}[/tex]

Now according to problem

[tex]\begin{gathered} 1.46\sin\theta_1=1.33\sin90\degree; \\ \sin\theta_1=\text{ }\frac{1.33}{1.46}\begin{cases}sin90={1} \\ \end{cases} \\ \theta_1=\sin^{-1}(0.91095)=\text{ 65.6}\degree \end{gathered}[/tex]

Final answer is 65.6°

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