1. 1.To look further at refraction maths 2.To understand dispersion Book Reference : Pages 190-193

2. Incident and emergent rays are parallel On exit from glass block, new incident ray (i 2 ) is equal to first angle of refraction r 1 Similarly for r 2 and i 1 i1i1 i2i2 r1r1 r2r2 Medium 1 Medium 2

3. We found for the ray at point of entry that : 1 n 2 = sin i 1 sin r 1 Where 1 n 2 is the refractive index between medium 1 and medium 2 However, from symmetry, i 2 =r 1 and r 2 =i 1 So for the emergent ray (medium 2 to medium 1) 2 n 1 = sin i 2 =sin r 1 = 1/ 1 n 2 sin r 2 sin i 1 Snell’s Law

4. Refraction occurs because the speed light changes in different materials. (Note. This does not contravene a relativistic view which says that the speed of light is a constant which can never be exceeded... For a particular single material (.e.g. a vacuum) it is a constant which cannot be exceeded. We’re just saying that light in each different material has a particular constant speed which cannot be exceeded for that medium)

5. Consider a wavefront, which crosses a straight boundary between a vacuum and a glass block X Y X’ Y’ i r glass vacuum

6. If c is the speed of light in a vacuum and c s is the speed of light in the substance, then the distance travelled in time t is: For X to X’ is c s t (using s=d/t for d) For Y to Y’ is ct Now consider the triangle XYY’, Since YY’ is the direction of travel for the wave in the vacuum, which is perpendicular to XY then: YY’ = XY’ sin i (substituting from above) ct = XY’ sin i

7. Now consider the triangle XX’Y’, Since XX’ is the direction of travel for the wave in the substance, which is perpendicular to X’Y’ then: XX’ = XY’ sin r (substituting from above) c s t = XY’ sin r Returning to Snell’s law: sin i=n s =c sin r c s

8. The frequency f of the wave does not change during refraction, However since: c = f and c s = f s n s =c= f = n s =c= f = c s f s s

9. When a light ray crosses the boundary between materials where light travels at c 1 and c 2 in each material and they have refractive indices of n 1 and n 2 then: sin i=c 1 sin rc 2 Rearranging and multiply by c: c sin i = c sin r c 1 c 2 Since that c/c 1 = n 1 & c/c 2 = n 2 substituting : n 1 sin i = n 2 sin r“The n sin i rule”

10. We can apply n 1 sin i = n 2 sin r to TIR and the critical angle: icic We know that at i c then the angle of refraction r is 90° then : n 1 sin i c = n 2 sin 90° Since sin 90° is 1 then: Sin i c = n 2 /n 1