Question

What is the critical angle θc for total internal reflection if the material is clear acrylic, with a refractive index of 1.5?

62°

90°

42°

49°

Answer 1

\[\color{green}{\frac{\sin i_c}{\sin 90^{\circ}} = \frac{1.0}{1.5}}\]

Answer 2

i honestly dont know how to finish solving that :/ @ghuczek

Answer 3

@helpme1.2 if you're not busy (and want to) of course, could you maybe help me with this?

Answer 4

\[\color{green}{\begin{align*}\frac{\sin i_c}{\sin 90^{\circ}} &= \frac{1.0}{1.5} \\ \\ \text{cross-multiply:} \quad 1.5 \sin i_c &= 1.0 \sin 90^{\circ} \\ i_c &= \arcsin\left(\frac{1 \times 1.0}{1.5}\right) \\&\approx 42^{\circ} \end{align*}}\]