refraction questions and answers
d} \] \[ 0.5 = 1.52 \times \sin \theta_{refracted} \] Solve for \(\sin \theta_{refracted}\): \[ \sin \theta_{refracted} = \frac{0.5}{1.52} \approx 0.3289 \] Find \(\theta_{refracted}\): \[ \theta_{refracted} = \arcsin(0.3289) \approx 19.2° \] Answer: Approximately 19.2° inside th