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incident on the boundary between medium 1 and medium 2. The frequency with 1 ! f 2 ! f, we see that v 1 ! f ( 1 and v 2 ! f ( 2 (35.5) Because v 1 ! v 2 , it follows that ( 1 ! ( 2 . We can obtain a relationship between index of refraction and wavelength by dividing the first Equation 35.5 by the second and then using Equation 35.4: (35.6) This gives ( 1 n 1 ! ( 2 n 2 If medium 1 is vacuum, or for all practical purposes air, then n 1 ! 1. Hence, it follows from Equation 35.6 that the index of refraction of any medium can be expressed as the (35.7) where ( is the wavelength of light in vacuum and ( n is the wavelength of light in the medium whose index of refraction is n. From Equation 35.7, we see that because n ) ( . We are now in a position to express Equation 35.3 in an alternative form. If we replace the v 2 /v 1 term in Equation 35.3 with n 1 /n 2 from Equation 35.6, we obtain (35.8) The experimental discovery of this relationship is usually credited to Willebrord Snell Snell’s law of refraction. We shall examine this equation further in Sections 35.6 and 35.9. n 1 sin &
1 ! n 2 sin &
2 n ! ( ( n ( 1 ( 2 ! v 1 v 2 ! c/n 1 c/n 2 ! n 2 n 1 S E C T I O N 3 5 . 5 • Refraction 1105 Snell’s law of refraction ▲ PITFALL PREVENTION 35.3 An Inverse Relationship The index of refraction is 2 differs from & 1 in Equation 35.8. Quick Quiz 35.3 Light passes from a material with index of refraction 1.3 into one with index of refraction 1.2. Compared to the incident ray, the Quick Quiz 35.4 As light from the Sun enters the atmosphere, it refracts due to the small difference between the speeds of light in air and in vacuum. The |