To solve this problem, we must recall Snell's Law:
[tex]n_{1}\alpha= n_{2}\beta[/tex]
For us to solve for the critical angle [tex]\alpha[/tex], [tex]\beta[/tex] is 90°. Thus leaving the equation with
[tex]n_{1}\alpha = n_{2}// sin\alpha = \frac{n_{1}}{n_{2}}[/tex]
Given that the glass in air forms a critical angle of 35.5° and that a glass' index of refraction is 1.00, we have
[tex] sin35.5 = \frac{1}{n_{glass}} \\ n_{glass} = \frac{1}{sin35.5} \\ n_{glass} = 1.72 [/tex]
Now, to solve for the critical angle between glass and water, we have
[tex] sin\alpha = \frac{n_{air}}{n_{glass}} \\ sin\alpha = \frac{1.33}{1.72} \\ \alpha = sin^{-1}(\frac{1.33}{1.72}) \\ \alpha= 50.65[/tex]
Answer: θ= 50.65°
