Physics 1C Lecture 25B "There are two ways of spreading light: to be the candle or the mirror that reflects it." --Edith Wharton
Refraction of Light When light passes from one medium to another, it is refracted because the speed of light is different in the two media. The index of refraction, n, of a medium can be defined: n is a unitless ratio For a vacuum, n=1 (exactly). For all other media, n>1.
Indices of Refraction
Frequency Between Media As light travels from one medium to another, its frequency does not change. Both the wave speed and the wavelength do change. The wavefronts do not pile up, nor are created nor destroyed at the boundary, so the frequency must stay the same. λ 1 λ 2 = n 2 n 1
Slow to fast, light bends away from the normal. Snell s Law Fast to slow, light bends towards the normal. Or think of a car axle: one wheel stuck in mud (slow), other wheel on pavement (fast) pivots around. This is the basis for Snell s Law.
Refraction of Light The angle of refraction depends upon the material (by index of refraction) and the angle of incidence: n 1 sinθ 1 = n 2 sin where θ 1 is the angle of incidence (30 o in the diagram). is the angle of refraction. n 1, n 2 are the indices of refraction of the first and second media, respectively.
Snell s Law For example, for this specific case (air to glass with an incident angle of 30.0 o ): n 1 sinθ 1 = n 2 sin sin = n 1 sinθ 1 n 2 n = sin 1 1 sinθ 1 n 2 1.0003 = sin 1 1.52 sin30 = sin 1 0.329 ( ) =19.2
Refraction of Light What if instead of going from air to glass with an incident angle of 30.0 o, it went from glass to air with the same incident angle: n 1 sinθ 1 = n 2 sin sin = n 1 n 2 sinθ 1 n = sin 1 1 sinθ 1 n 2 1.52 = sin 1 1.0003 sin30 = sin 1 0.760 ( ) = 49.4 Here the angle increased (slow to fast).
Refraction of Light What if instead having an incident angle of 30.0 o (from glass to air) it had an incident angle of 60.0 o (still glass to air): n 1 sinθ 1 = n 2 sin sin = n 1 n 2 sinθ 1 n = sin 1 1 sinθ 1 n 2 1.52 = sin 1 1.0003 sin60 = sin 1 ( 1.32) =????? Here the angle doesn t exist. The angle was so great that it refracted the incident ray back to the first medium (reflected?).
Total Internal Reflection This is known as Total Internal Reflection (TIR). It can only occur if you move from a slow medium to a fast medium such that the refracted ray is bent away from the normal compared to the incident ray. Here you can see light ray 5 undergoing total internal reflection. This means that at angle that light ray 5 hits at, no light enters the second medium.
Total Internal Reflection We define the critical angle as a particular angle of incidence that will result in an angle of refraction of 90 o. For angles of incidence greater than the critical angle, the beam is entirely reflected at the boundary (TIR). This ray will obey the Law of Reflection at the surface boundary.
Conceptual Question 25B-1 A light ray travels from medium 1 to medium 3 as shown in the figure below. What can we say about the relationship between the index of refraction for medium 1 (n 1 ) and the index of refraction for medium 3 (n 3 )? A) n 3 > n 1. B) n 3 = n 1. C) n 3 < n 1. D) We can t compare n 1 to n 3 without knowing n 2.
Conceptual Question Answer A light ray travels from medium 1 to medium 3 as shown in the figure below. What can we say about the relationship between the index of refraction for medium 1 (n 1 ) and the index of refraction for medium 3 (n 3 )? Use Snell s Law twice: Equation the two n 2 : 30 o
Prisms Example An incident ray in air is headed straight towards an equilateral plastic prism (n=1.50). The ray is parallel to the bottom of the prism. Use Snell s Law to find the angle (with respect to the normal) that the light ray exits the prism on the right. Answer The diagram is given but it is up to you to draw the normal(s) and path of the ray.
Answer Start with: Prisms =19.47 30 o 60 o 60 o Thus, the incident angle is 30 o from air to plastic. Using Snell s Law we find: n 1 sinθ 1 = n 2 sin n = sin 1 1 1 sinθ 1 = sin 1 n 2 1.5 sin30 = sin 1 0.333 ( )
Answer Now, we need to look at the second boundary. We can now examine the small top triangle created by the ray in the prism. The bottom left angle on this triangle will be: 90 19.47 = 70.53 Prisms 70.53 o 60 o 49.47o This means that the bottom right angle of the triangle will be: 180 70.53 60 = 49.47
Prisms Answer Is the 49.47 o, the angle we will use in Snell s Law? No, it is not with respect to the normal. θ We draw the normal and 3 = 40.53 find: 90 49.47 = 40.53 At the second boundary, will this ray be refracted or totally internally reflected? Check by calculating the critical angle. sinθ c = n 1 1 θ n c = sin 1 = sin 1 0.667 2 1.5 60 o 49.47o θ 3 refracted, barely ( ) θ c = 41.81
Answer Prisms Now we need to calculate the outgoing ray with Snell s Law again: θ 3 = 40.53 θ 3 θ 4 n 2 sinθ 3 = n 1 sinθ 4 n θ 4 = sin 1 2 sinθ 3 n 1 1.5 θ 4 = sin 1 1 sin40.53 = sin 1 0.975 ( ) θ 4 = 77.10 This is the outgoing angle with respect to the normal
For Next Time (FNT) Start reading Chapter 26 Finish working on the homework for Chapter 25