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
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.