Consider a clear liquid in an open container. We determine that the liquid-air critical angle is 44°. If light is shined
Posted: Wed Jul 06, 2022 11:41 am
Consider a clear liquid in an open container. We determine that the liquid-air critical angle is 44°. If light is shined from above the container at varying values of the angle of incidence ₁, an orientation ₂ = 0, will be found where r = 0. Find 0p. • Show your work in the below text box. • To write √ use this notation in the below text box: "sqrt(x)" Remarks: • You must type all the steps of your detailed calculation in the below text box. • A final answer will NOT be accepted if the detailed calculation steps are missing in the below text box. • Files submitted by email are NOT accepted; your detailed answer must be exclusively written in the below text box.