62. (a) The condition (in Eq. 33-44) required in the critical angle calculation is θ3= 90°. Thus (with 2= c, which we don’t compute here), nnn112 233sin==leads to 1= = sin–1n3/n1= 54.3°. (b) Yes. Reducing leads to a reduction of 2so that it becomes less than the critical angle; therefore, there will be some transmission of light into material 3. (c) We note that the complement of the angle of refraction (in material 2) is the critical
This is the end of the preview. Sign up
access the rest of the document.