Reflection and Refraction of Light265(b) Observe from the figure above that sinLdφ=. Thus, the distance the light travels inside the plastic is sindL=, and if 50.0 cm0.500 mL==, the time required is ( )()81.20 0.500 m010 m s sin36.1×°9sin3.4010s3.40 nssin3.0LdnLtvcnc−∆= ====×=22.62 (a) At the upper surface of the glass, the critical angle is given by 1sinaircglassnnnθ==Consider the critical ray PBB′: 4tan4cddttBut, also, 22222221sinin1tancossin111cccnθθ= =2s1−−−Thus, 22141dtn=−or 2241tnd−=giving 241tnd=+±±±²²±³´µ±²¶(b) Using the result from Part (a) and solving for dgives 241tdn=−Thus, if , then 1.52 and 0.600 cm
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This note was uploaded on 12/09/2011 for the course PHYS 2020 taught by Professor Staff during the Fall '10 term at FIU.