Refraction Refraction From a previous lab we saw

Refraction

Refraction • From a previous lab, we saw that when a wave reaches a boundary, some of the wave is transmitted and some of it is reflected. • We also noticed that the speed changed when the wave entered the new medium.

Refraction • Similarly, when monochromatic light travels between two media, there is a change in the speed of the light wave. – The frequency of the wave does not change. Therefore, the change in velocity is accompanied by a change in its wavelength. • If the light enters at an oblique angle, it will change direction as it passes into the second medium.

Refraction • Refraction can be illustrated in the form of a ray diagram – Note that θi and θr are measured from the normal, and that the normal is extended into both media – Note that the different media are labeled – Note that the rays have arrowheads indicating their direction – Note that in this case as the ray passes into a more “optically dense” medium, it turns toward the normal.

Refraction • The Dutch scientist Willebrord Snell derived the following mathematical expression for refraction sin θ 1 / sin θ 2 = v 1 / v 2 (Snell’s Law) • The derivation can be found on page 400 of your textbook

Refraction • To simplify refraction problems, a quantity known as the absolute index of refraction is defined as follows: n = c/v (reference tables) • where • n is the absolute index of refraction of a medium • c is the speed of light in a vacuum • v is the speed of light in the medium • Table of Absolute Indices of Refraction in Ref. Tables.

Sample Problem • Determine the speed of light in Glycerol. • Solution n = c/v From reference tables… nglycerol = 1. 47 c = 3. 00 x 108 m/s Solving for v gives… v = c/n = (3. 00 x 108 m/s) / 1. 47 v = 2. 04 x 108 m/s

Refraction and Snell’s Law • We can derive an alternate form of Snell’s Law in terms of the index of refraction… The relationship…. . sin θ 1 / sin θ 2 = v 1 / v 2 Becomes sin θ 1 / sin θ 2 = (c/n 1) / (c/n 2) = n 2 / n 1 Rearranging gives Snell’s Law in a useful form n 1 sin θ 1 = n 2 sin θ 2 Snell’s Law (Reference Tables)

Snell’s Law Sample Problem • A monochromatic light ray in air is incident on a surface boundary between air and corn oil at an angle of 60 0 to the normal. Calculate the refracted angle of the ray in the corn oil. • Solution n 1 sin θ 1 = n 2 sin θ 2 = (n 1 sin θ 1) / n 2 from reference tables and question n 1 = nair = 1. 00 n 2 = ncorn oil = 1. 47 Θ 1 = 600 sin θ 2 = (1. 00) (sin 600) / 1. 47 = 0. 59 Θ 2 = sin-1 (0. 59) = 360

Refraction of Light Activity

• The “Broken Pencil Trick” Explained

Total Internal Reflection • The maximum possible angle of refraction is 90 -degrees.

Critical Angle • In the case of the laser beam in the water, there is some specific value for the angle of incidence (we'll call it the "critical angle") which yields an angle of refraction of 90 -degrees. • This particular value for the angle of incidence could be calculated using Snell's Law

Snell’s Law & Critical Angle • Let air be medium 2, water be medium 1 • If monochromatic light passes from medium 1 in which its speed is slower to medium 2 in which its speed is faster, we can draw a ray diagram as below…

Snell’s Law & Critical Angle • There is one unique angle in medium 1 that will produce an angle of 900 in medium 2, as illustrated below… • This unique angle is called the critical angle ( Θ c)

Snell’s Law & Critical Angle • We can derive an equation for the critical angle using Snell’s Law: • n 1 sin Θc = n 2 sin 900 since sin 900 = 1, we get sin Θc = n 2 / n 1 In the event that medium 2 is a vacuum or air (n 2=1), the above equation simplifies to… sin Θc = 1 / n 1

Sample Problem • Determine the critical angle between water and • air. Solution sin Θc = 1 / n 1 from reference tables, n 1 = nwater = 1. 33 sin Θc = 1 / 1. 33 Θc = sin – 1 (1/1. 33) = 48. 70

Physics Demo’s Fiber Optics