Reflection and Refraction of Light25922.54From the sketch at the right, observe that the angle of incidence at Ais the same as the prism angle at point O. Given that 60.0θ=°, application of Snell’s law at point Agives ( )1.50sin1.00 sin60.0β35.3or =°From triangle AOB, we calculate the angle of incidence and reflection, γ, at point B: ( ) ( )90.090.0180+°−+°−=°or 60.035.324.7θβ=−=°−°=°Now, we find the angle of incidence at point Cusing triangle BCQ: ( ) ( ) ( )90.090.090.0180δθ−°−+°(+=°or )90.090.084.75.26δθγ−+=°°− =°Finally, application of Snell’s law at point Cgives ( ) ( ) ( )1.00 sin1.50 sin 5.26φ=°or ()1sin1.50sin5.−26=7.91°±²³´µ±±²³³´µ9±²±³´µ´±22.55The path of a light ray during a reflection and/or refraction process is always
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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.