Snell's Law applet

opposite hypotenuse hypotenuse opposite θ adjacent

30˚ 50˚ 50˚ 30˚

For any given angle θ, in a right angled triangle, … Length of opposite side Length of hypotenuse = a constant value Called sine θ

Sine functions Let’s look at this circle having a radius of 1. opposite y Sin 45 = hypotenuse Inscribe a rt. triangle where θ = 45˚. 1 opposite 45˚ x Hypotenuse = 1. Opposite side = sin45 ˚. = 0.707 Sine functions

The sine function Sin x= y 1- 180˚ 360˚ -90˚ 90˚ -1- 270˚

If light is travelling from medium 1 into medium 2, the angles of incidence and the angles of refraction are related by Snell’s Law. n1 sin1 = n2sin 2 or sin 1 = v1 = n2 sin 2 v2 n1

If θi = 36˚, What is θR? nwater = 1.33 nair = 1.00 Answer: 26.2˚

When light crosses an interface into a medium with a higher index of refraction, the light bends towards the normal. Conversely, light travelling across an interface from higher n to lower n will bend away from the normal. normal n2< n1 Critical angle causes 90° angle of refraction 2 1 Total internal reflection

This has an interesting implication: at some angle, known as the critical angle, light travelling from a medium with higher n to a medium with lower n will be refracted at 90° . If the light hits the interface at any angle larger than this critical angle, it will not pass through to the second medium at all. Instead, all of it will be reflected back into the first medium, a process known as total internal reflection.

The critical angle (c ) can be found by using Snell’s law. n1 sinc = n2sin 90˚ sinc = n2 n1

Example #1 Light enters water at an angle of 55˚ Example #1 Light enters water at an angle of 55˚. What is the angle of refraction in water if the index of refraction is 1.33. Example #2: What is the critical angle for light passing from glass (n = 1.50) to air? Answer: 38.0˚ Answer: 41.8˚

Example #3:If the refractive index of medium 3 is 1 Example #3:If the refractive index of medium 3 is 1.50, determine the refractive index of a) medium 2 b) medium 1

Example #4 A beam of light inside a diamond travels out to the air Example #4 A beam of light inside a diamond travels out to the air. The angle of incidence at the air interface is 26 . a) Will part of the beam enter the air or will it be totally internally reflected? Show your calculations. b) The same diamond is immersed in water. What will happen to the beam now?