Refraction

• As waves move into a new medium they can be refracted- speed, and direction can change

• frequency stays the same- depends on the source

• Thus a change in speed and direction must be due to a change in wavelength

Refraction – Soldier Analogy

• Imagine a line of soldiers marching from a road onto sand

• They will move more slowly on the sand and the line will bend

Direction of Bending

• FST- fast to slow: towards the normal• SFA- slow to fast: away from normal• Draw the estimated refracted ray

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RefractionRefraction

• What Are The Different Media?–Water

–Glass

–Air

Dispersion of light through refraction

• Different wavelengths of light refract by different amounts

• Thus the prism

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Mirages

• Hot air near surface of road causes bending• Your brain interprets this the only way it

knows how- there must be water on the road that is reflecting

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Quantifying Refraction

• n= index of refraction (no units)• n=c/v • c= speed of light in a vacuum• v=velocity of light in the medium

• Remember v=f ?• n= f/ 2f• Since f does not change n=/2

Light at interface between 2 mediums

• When light reaches interface, it generally splits into 2 parts: – Part is reflected (follows

law of reflection)– Part is refracted

• Refracted ray enters new medium and can change speed, wavelength, and direction

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Snell’s Law

• n1sin1=n2sin2

• All incident and reflected are labeled (1) and all refracted are labeled (2)

• Refraction is reversible- you could turn light ray around and it would follow the same path

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Snell’s Law

• In lab, we’ll often use a semicircular tray

• Notice how if you aim the incident ray at the center of the flat side, it will exit the tray at the normal to the curved surface

• Since = 0 there will be no refraction at that interface

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Problem Solving: Snell’s Law

• nair=1

• nwater=1.33

• Find the angle of refraction

• Check your answer using the FST, SFA rule

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Snell’s Law: Multiple Interfaces

• Snell’s law can be used to go through successive interfaces

• Find the angle of refraction within the glass

• Find the angle of refraction when it re-enters air

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nglass=1.52

Snell’s Law: prisms

• Where does the light exit the glass and at what angle?

• Treat it like the previous layer problem but now the layers are not parallel

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Total Internal Reflection (TIR)

• When n1>n2 the refracted ray will bend away from the normal

• As you increase 1 you will reach a point where the r=90

• The refracted ray now doesn’t leave the first medium

• If you set r=90 and solve for 1 you find the critical angle (c) where TIR occurs

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TIR

• At any angle greater than the critical angle, you will have TIR

• Remember it only happens from a higher n into a lower n medium

TIR problems

• Calculate the critical angle for an ethanol-air boundary.

• Draw a diagram of the path of the light ray at the critical angle

• nethanol = 1.36

47.3 light travels from ethanol to air

Total Internal ReflectionTotal Internal Reflection

• DiamondsDiamonds

Total Internal ReflectionTotal Internal Reflection

• Fiber Optic Data CablesFiber Optic Data Cables

Total Internal ReflectionTotal Internal Reflection

• RainbowRainbow

Lenses

• Ray tracing for lenses similar to spherical mirrors except light passes through instead of reflecting

• Lenses have real or virtual image

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Converging Lenses

• Converging: thicker in middle

• Light refracted through real focus

• forms REAL, inverted IMAGE

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• RULES- use the 2 that fit the situation– Incident ray entering parallel is refracted through focus– OR Incident ray entering via the focus is refracted parallel– Ray through center of lens doesn’t bend

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Image Formation: Converging Lens

Diverging Lenses

• Diverging: thinner in the middle

• Light bends AWAY from a virtual focus on the incident side of the lens

• Virtual, upright image

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Image Formation: Diverging

• RULES- use the 2 that fit the situation– Incident ray entering parallel refracted away from virtual

focus– Incident ray entering through virtual focus refracted parallel– Incident ray passing through center of lens doesn’t bend

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Mathematics of Lenses

• Similar to Mirrors- uses same equations• BEWARE OF SIGNS

SIGN RULES• Converging: f is +• Diverging: f is negative

• Si is + for real images

• Si is - for virtual images

Lens Equation Problem

• An object is placed 7.10cm to the left of a diverging lens whose focal length is f=-5.08cm.

• Draw ray diagram.

• Find the image distance and determine is image is real or virtual.

• Find the magnification.

solution

• 1/si=(1/-5.08) - (1/7.10)

• si=-2.96

• Since si is negative, image is virtual and located to the left of the lens

• M=-si/so=-(-2.96)/(7.10)= 0.47

Problems with multiple lenses

• Treat each lens separately- work through them in order

• Use the image from the first lens as the object for the 2nd and continue this process until all lenses used

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