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Please help me solve From a previous question, the index of refraction of the liquid is 1.37

Please help me solve From a previous question, the index of refraction of the liquid-example-1
Please help me solve From a previous question, the index of refraction of the liquid-example-1
Please help me solve From a previous question, the index of refraction of the liquid-example-2

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ANSWER

Step-by-step explanation

From the previous part, we have that the index of refraction of the liquid is 1.37, so we have to replace this in the equation and solve,


\sin\theta_c=(n_(air))/(n_(liquid))=(1.00)/(1.37)\approx0.73

And then, take the inverse of the sine to find the critical angle,


\theta_c=\sin^(-1)0.73\approx46.9\degree

Hence, the critical angle is 46.9°, rounded to the nearest tenth.

Please help me solve From a previous question, the index of refraction of the liquid-example-1
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