1. Optics Reflection, Refraction & Snell’s Law Lesson 2
PHYSICS
Optics
Reflection, Refraction & Snell’s Law
Lesson 2

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.
It can totally reflect off the surface and go back into the medium it came from.
It can refract internally.

3. Laws of Reflection of Light
Normal
Angle of Incidence
Angle of Reflection
Reflected Ray
Incident Ray i r’
interface
reflection

4. 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.
interface
reflection

5. Laws of Reflection of Light
There are 2 laws of reflection
The angle of incidence is equal to the angle of reflection.
The incident ray, the normal, and the reflected ray are coplanar.
Normal
Angle of Incidence
Angle of Reflection
Incident Ray
Reflected Ray
interface i r’
reflection

8. 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
Angle of Refraction
Refraction r’

9. 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.

10. Index of Refraction Index of Refraction
index 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).

11. 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.

12. where;
n2 = index of refraction (of incidence ray
θ2 = angle of incidence
n1 = index of refraction of refracted ray
θ1 = angle of refraction

13. n1 = 0.33 θ1
interface
n2 = 0.57
Refraction towards the normal θ2
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.

14. n1 = 0.55 θ1
interface
Refraction away from the normal
n2 = 0.23 θ2
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.

15. Snell’s Law Problems
A ray of light travelling through air is incident on a piece of glass whose refractive index is 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)
Step 2: Write down the values you are given
n1 = 1
θ1 = 0.6
n2 = 1.5
θ2 = ? θ1 n1 n2 θ2

16. 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
n n2
n1sinθ1 = sinθ2 n2

17. Snell’s Law Problems sinθ2 = (1)(0.6) 1.5 sinθ2 = 0.6/1.5 sinθ2 = 0.4
Notice: that sinθ2 is less than sin θ1 – this immediately tells us that θ2 < θ1