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

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

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

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

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’

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 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).

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

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.

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.

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) Step 2: Write down the values you are given n1 = 1 θ1 = 0.6 n2 = 1.5 θ2 = ? θ1 n1 n2 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 n2 n2 n1sinθ1 = sinθ2 n2

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