2. Refraction and Snell’s Law

3. Refraction: bending of light at the interface of 2 different materials

4. Refraction  In the Refraction lab you saw the laser light change angles when it traveled from air into water or oil.  Light changes directions because it changes speed: it actually travels slower in water, glass, oil, etc.

5. Snell’s Law  Snells Law: the “equation” for refraction  n in sinΘ in = n out sinΘ out

6. Snell’s Law  n = index of refraction = property of the material  n air = 1.0  n water= 1.3  n glass = 1.5  n diamond = 2.4  n = speed of light in air / speed of light in other material

7. Example  Light traveling in air hits a glass window at an incident angle of 30 degrees. What is the refracted angle in the glass?  n in sinΘ in = n out sinΘ out  1.0 sin 30 = 1.5 sin Θ out  Θout = 19.5 degrees

8. Total Internal Reflection Total internal reflection occurs when light attempts to pass from a more optically dense medium to a less optically dense medium at an angle greater than the critical angle. When this occurs there is no refraction, only reflection. Total internal reflection can be used for practical applications like fiber optics. n1n1 n2n2  n 2 > n 1

9. Total Internal Reflection, idea of the “critical angle”   The incident angle that causes the refracted ray to skim right along the boundary of a substance is known as the critical angle,  c.   The critical angle is the angle of incidence that produces an angle of refraction of 90º.   If the angle of incidence exceeds the critical angle, the ray is completely reflected and does not enter the new medium.   A critical angle only exists when light is attempting to penetrate a medium of higher optical density than it is currently traveling in.

10. Critical Angle Sample Problem Calculate the critical angle for the diamond-air boundary.  c = sin -1 (n r / n i ) = sin -1 (1 / 2.42) = 24.4  Any light shone on this boundary beyond this angle will be reflected back into the diamond. cc air diamond