# Question #f3082

Apr 14, 2017

${61}^{o}$

#### Explanation:

Find the refractive index of the plastic:
${n}_{1} \times \sin {\theta}_{1} = {n}_{2} \times \sin {\theta}_{2}$

${n}_{1} \times \sin {30}^{o} = 1 \times \sin {35}^{o}$
${n}_{1} = \frac{\sin {35}^{o}}{\sin {30}^{o}} = 1.147$

The critical angle is found when the angle of incidence is large enough to make the angle of transmission equal to ${90}^{o}$.

Therefore, going from plastic to air:
$1.147 \times \sin \theta = 1 \times \sin {90}^{o}$
$\sin \theta = \frac{1}{1.147}$
$\theta = {\sin}^{-} 1 \left(\frac{1}{1.147}\right) = {60.7}^{o}$