25_Chapter 22 HomeworkCH22 Reflection and Refraction of Light

25_Chapter 22 - 257 Reflection and Refraction of Light 22.50 nglass sin 2 = nair sin 1 Applying Snells law to this refraction gives If 1 = 2 2 this

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Reflection and Refraction of Light 257 22.50 Applying Snell’s law to this refraction gives 21 sin sin glass air nn θ = If 1 2 2 = , this becomes ( ) 222 sin sin 2 2sin cos ss 2 gla n θθ == or 2 cos 2 glass n = Then, the angle of incidence is 11 12 1.56 2 2cos 2cos 22 glass n −−  =   77.5 = ° 22.51 In the Figure at the right, observe that 1 90 β and 1 90 α . Thus, = . Similarly, on the right side of the prism, 2 90 δ and 2 90 γ , giving = . Next, observe that the angle between the reflected rays is ( ) ( ) B βγ + + =+ , so ( ) 2 B + = . Finally, observe that the left side of the prism is sloped at angle from the vertical, and the right side is sloped at angle . Thus, the angle between the two sides is A = + , and we obtain the result () = BA αγ ± ± ² ³ ± ´ µ ² ³ ² ³ ± ³ ± ³ ²
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This document was uploaded on 10/20/2011.

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