The index of refraction of air is 1.0003. This value is so small we ignore it in problem solving and assume that the speed of light in air is equal to that in a vacuum. The slowing of light in a medium is of great practical importance. It provides a way of changing the direction of light, and in this way manipulating it to serve practical ends.

Here is why light is bent. Light, being a wave, travels always at right angles to its wave front. (Ever see a water wave scooting sideways?) When the light beam enters glass at an angle, part of the beam is slowed before the rest, and this alters the wave front and therefore the direction of the beam.

step 1:	step 2:
step 3:	step 4:
The part of the beam closest to the glass slows down while the part that has not yet reached the glass continues to move fast.	After the second part of the beam reaches the glass, motion is perpendicular to the new wave front.

We see, then, that light is bent towards the normal when it moves into an optically denser medium. Likewise, when light passes back into air it is bent away from the normal. In the pictures below try to determine which medium has the lowest light speed and therefore the highest index of refraction.

example 1:	example 2:
example 3:	example 4:
Answers:(1) B; (2) A; (3) B; (4) neither. Look these over if you missed any of them.

Snell's law tells us mathematically the amount a ray is bent when crossing from one medium into another. The proof requires a bit of geometry. A light beam passes from one medium of index n1 to another of index n2. In so doing it changes from an incident angle Ø1 to an angle of refraction Ø2.

Now for a close-up of the point at which the beam strikes the interface. We are interested in the time interval, Çt, that begins when the beam first strikes medium 2, and ends when it completely enters medium 2. During this time it travels a distance s = v1Çt in medium 1 and v2Çt in medium 2. The distance AB is identical for both triangles. Make sure you agree that Ø1 and Ø2 are correctly identified.

From trigonometry, we know that solving both form AB and equating, we get v₁Çt = v₂Çt. But the t's cancel and v₁ = c/n₁ and v₂ = c/n₂. Thus we derive Snell's Law:

n₁sinØ₁ = n₂sinØ₂

Here are a few indices of refraction:

material	index of refraction
air	1.0003
water	1.33
alcohol	1.36
quartz	1.46
glass	1.5 to 1.8
diamond	2.42

14. Using a good protractor, measure the angles of incidence and refraction, and calculate the unknown indices of refraction. Don't forget to measure from the normal.

a)	b)
Angle of incidence = 37° Angle of refraction = 29°	Angle of incidence = 43° Angle of refraction = 56°
a) Angle of incidence = 37° Angle of refraction = 29° n1sinØ1 = n2sinØ2 1.1sin37 = n2sin29 n2 = 1.37	b) Ø1 = 43° Ø2 = 56° n1sinØ1 = n2sinØ2 n1sin43 = 1.3sin56 n1 = 1.58

15. The angle of incidence of light entering glass from air is 38°. What is the angle of refraction?

nair = 1.00 n1sinØ1 = n2sinØ2 (1)sin38 = 1.52sinØ2 Ø2 = 23.9°

16. Light leaves a diamond and enters water. If the angle of incidence is 22°, what is the angle of refraction?

n1sinØ1 = n2sinØ2 2.42sin22 = 1.33sinØ2 Ø2 = 43.0°

17. Given the angle shown, calculate Ø.

Ø1 = 90 - 47 Ø1 = 43° Ø2 = 90 - Ø n1sinØ1 = n2sinØ2 1.2sin43 = 1.5sin(90 - Ø) Ø = 56.9°

18. Find the angle at which light leaves the prism.

Consider each refraction separately: n1sinØ1 = n2sinØ2 (1)sin62 = 1.6sinØ2 Ø2 = 33.5° Ø3 = 90 - 33.5 Ø3 = 56.5° By geometry: 50 + 56.6 + Ø4 = 180 Ø4 = 73.5° Ø5 = 90 - 73.5 Ø5 = 16.5° nsinØ5 = (1)sinØ 1.6sin16.5 = sinØ Ø = 27.0°

19. Light passes through a block of glass with parallel sides. At what angle does it exit?

(1)sinØ1 = nsinØ2 Ø2 = sin-1(sinØ/n) Ø2 = Ø3 (alternate interior angles) sinØ3 = sinØ1/n nsinØ3 = (1)sinØ4 Combine and solve: n(sinØ1/n) = sinØ4 Ø1 = Ø4 note: the exit ray is parallel to the entrance ray

20. Light enters the prism shown, and reflects off a silvered side. At what angle does it leave?

Since the light enters at a right angle, it's path is not bent. On the mirrored surface Øi = Ør. So: Øi + Ør = 30° which is the angle of incidence at the top of the surface. Finally: 1.4sin30 = (1)sinØ Ø = 44.4°