1-Refraction - PPT slides

Angle of refraction

Dispersion

Partial reflection & refraction

Index of refraction

Snell’s Law

Recall: Speed of Light

3.00 x 108 m/s

(in a vacuum)

However, it is impossible for light to move at this speed

Speed of Light

 _{Speed of light changes}

depending on the medium

 _{Different media slow down}

light by different amounts

 _{Speed of light in a vacuum}

 _{= 3.00x108 m/s}

 _{Speed of light in water}

 _{= 2.26x108 m/s}

 _{Speed of light in acrylic}

Index of Refraction

 _{The amount by which a transparent medium}

decreases the speed of light

 _{A ratio of the speed of light in a vacuum}

compared to the speed of light in a medium

c = speed of light in a vacuum = a constant of 3.00x108 _m/s

v = speed of light in the medium

Think

 _{Does the index of refraction (n) have a unit?}

Explain.

 _{What does a refractive index of 1 mean?}

 _{What does a refractive index greater than 1}

mean?

Index of

Refraction for

various media

Media Index of Refraction Vacuum 1.00

Air 1.0003

Carbon dioxide gas 1.0005

Ice 1.31

Recall GRASP

 _{Given: List the information given to you}

using symbols and numbers. Include units.

 _{Required: List the item that needs to be}

solved. Use symbols.

 _{Analysis: Write mathematical equation(s)}

that will be used.

 _{Solution: Replace the equation with values}

listed in the Given. Solve the equation.

 _{Phrase: Write the answer to the question in}

Given:

v = 1.91 x 108 _{m/s (glass)}

c = 3.00 x 108 _m/s

Required: n (glass)

Analysis: n = c

v

Solution: = 3.00 x 108 _{m/s = 1.57}

1.91 x 108 _m/s

Given: n = 1.33

c = 3.00 x 108 _m/s

Required: v (water)

Analysis: n = c , therefore v = c .

v n

Solution: v = 3.00 x108 _{m/s = 2.26 x 10}8 _m/s

1.33

Refraction

 _{When the speed of light}

is slowed down by

Refraction

 _{Refraction is the change}

in the direction of light (bending)

 _{only occurs when light}

crosses a boundary between two media (substances)

 _{i.e. when it is entering}

Refraction

 _{Refracted ray: the}

ray after crossing a boundary between media

 _{Angle of refraction:}

Refraction

Normal

Angle of refraction Angle of

How Light Refracts

 _{The part of the light beam that hits the}

Refraction Car Analogy

 _{Car travelling at an angle}

towards a muddy surface

 _{One front wheel hits}

muddy surface and slows down

 _{Other wheels continue}

to move at a higher speed

Angle of Refraction

 _{When light is slowed:}

 _{from low to high}

refractive index

 _{e.g. air}  _glass

 _{Light bends towards}

the normal

 _{Angle of refraction is}

Angle of Refraction

 _{When light is sped up:}

 _{from high to low}

refractive index

 _{e.g. glass}  _air

 _{Light bends away}

from the normal

 _{Angle of refraction is}

Snell’s Law

 _{The index of refraction can also be calculated}

using sines of angles (trigonometry)

n

sine i =

n

sine R

i is the angle of incidence R is the angle of refraction

n₁ is the refractive index of the medium that the ray is leaving

Snell’s Law

n

sine i =

n

sine R

Actually written as:

n

sin



= n

sin



 1 is the first medium where the incident ray is leaving

 2 is the second medium where the incident ray is entering

Willebrord Snell (1580-1626)

 _{Identified the relationship}

between the angle of incidence and angle of refraction

 _{Though Snell’s Law is named and}

credited to him, it is now known that this refraction equation was first accurately described by Ibn Sahl in the year 984

 _{Ibn Sahl was a Muslim Persian,}

A light ray moves from water to glass. The angle of incidence in water is 260. Calculate the refracted angle given the index of

refraction for water is 1.33 and glass is 2.04.

Given: Required: Analysis:

Solution:

A light ray moves from water to glass. The angle of incidence in water is 260. Calculate the refracted angle given the index of

refraction for water is 1.33 and glass is 2.04.

Given: ₁ = 26o_{, n}

1 = 1.33, n2 = 2.04

Required: ₂

Analysis: n₁ sin₁ = n₂ sin₂ therefore n₁ sin₁ = sin₂

n₂

Solution: sin₂ = 1.33 x sin 26o _{= 0.497}

2.04

₂ = sin-1 _{0.497 = 16.6}o

Given the diagram below, calculate the angle of refraction and draw the refracted ray.

 _{Step 1: draw the normal}  _{Step 2: measure the angle}

of incidence

 _{Step 3: calculate the angle}

of refraction using Snell’s Law

 _{Step 4: Draw the refracted}

ray using the angle calculated in step 3

Glass n = 1.52

Application: Refraction in Water

 _{Light rays change direction at the surface of}

the water

 _{Example: The image of the chest appears to}

Air Water

Apparent Depth

the depth that an object appears to be at due to the refraction of light in a transparent medium

Air

Apparent Depth

 _{Explains why it’s not as easy}

Partial Reflection & Refraction

 _{Refraction is often accompanied by}

reflection.

 _{Some of the light hitting the surface of a} body of water reflects, and some refracts.

 _{Example: Pond water}

 _{Reflection: sky}

Partial Reflection & Refraction

 _{Ray diagram}

 _{i = angle of incident ray}  _{r = angle reflected ray}

 _{R = angle of refracted ray}

 _{Emergent ray}

 _{any ray of light that is}

leaving a medium

 _{includes both reflected}

Partial Reflection & Refraction

 _{Angle of reflection (r)}

 _{Law of Reflection}  _{i = r}

 _{Angle of refraction (R)}

Partial Reflection & Refraction

Example: Silvered two way mirrors

 _{Plastic or glass with a special coating that reflects} most light, but allows some to be refracted

 _{Results in a mirrored surface that you can see}

Partial Reflection & Refraction

 _{“Don’t Miss A Sec” an architectural artwork by Monica Bonvicini, Dec 2003}  Installed at a construction site across from London’s Tate Britain museum

HW

 _{Explain why the pencil looks bent}

 _{Textbook p.519 #2-4}