PHYSICS

OpticsReflection, Refraction & Snell’s

LawLesson 2

Laws of Reflection of Light

• Light rays interact with different media (surfaces) in different ways.

• When light hits a medium (surface) one of 2 things can happen.1. It can totally reflect off the

surface and go back into the medium it came from.

2. It can refract internally.

Laws of Reflection of Light

reflection

interface

Normal

Angle of Incidence

Incident Ray

Reflected Ray

Angle of Reflection

i r’

Laws of Reflection of Light

Reflection - the change in direction of a light ray at an interface (boundary) between two different media so that the light ray returns into the medium from which it originated.

reflection

interface

Laws of Reflection of Light

There are 2 laws of reflection1. The angle of incidence is equal to the

angle of reflection. 2. The incident ray, the normal, and the

reflected ray are coplanar.

reflection

interface

Normal

Angle of IncidenceIncident

RayReflected Ray

Angle of Reflection

i r’

Laws of Refraction of Light

Refraction – the change in direction of a light ray at an interface (boundary) between two different media as a result of

a change in speed of the light ray interface

r’

RefractionAngle of Refraction

Index of Refraction

• Remember that light travels at a speed (c) 3 x 108 m/s (in a vacuum).

• When light travels through another material medium, however, like water, glass or air its speed changes (is less than c).

• Every medium has an index of refraction which tells us how much slower light travels through that medium.

Index of Refraction

Index of Refractionindex of refraction = speed of light in vacuum (c) (n) speed of light in medium

(v)

n = c v

• The index of refraction (n) in a vacuum is always equal to 1 (the index of refraction is so close to 1 that we simply use n = 1 for air as well).

Snell’s Law

• Used to describe the relationship between the angle of incidence and refraction, when light rays pass through a boundary between two different medium, such as

water and glass.• The amount of bending that takes place

when a light ray strikes a refractive

boundary.

where;

n2 = index of refraction (of incidence ray

θ2 = angle of incidence

n1 = index of refraction of refracted ray

θ1 = angle of refraction

If n2 > n1 then θ2 < θ1

If the transmitting medium has a higher index of refraction (n) than the incident medium, then the ray will bend toward the normal.

interface

θ2

Refraction towards the normal

n1 = 0.33

n2 = 0.57

θ1

If n2 < n1 then θ2 > θ1

If the transmitting medium has a lower index of refraction (n) than the incident medium, then the ray will bend away from the normal.

interface

θ2

Refraction away from the normal

n1 = 0.55

n2 = 0.23

θ1

Snell’s Law Problems

A ray of light travelling through air is incident on a piece of glass whose refractive index is 1.5. if the sine of the angle of incidence is 0.6, what is the sine of the angle of refraction?

Step 1: Draw a diagram (trust me this will help)

θ1

θ2

n1

n2

Step 2: Write down the values you are given

n1 = 1

θ1 = 0.6

n2 = 1.5

θ2 = ?

Snell’s Law Problems

Step 3: Write down the formula and rearrange it to solve for the desired value.

n1sinθ1 = n2sinθ2 – we need to solve for sinθ2 so we divide both sides by

n2 to isolate sinθ2

n1sinθ1 = n2sinθ2

n2 n2

n1sinθ1 = sinθ2

n2

Snell’s Law Problems

sinθ2 = (1)(0.6)

1.5sinθ2 = 0.6/1.5

sinθ2 = 0.4

Notice: that sinθ2 is less than sin θ1 – this immediately tells us that θ2 < θ1