Physics 102 Physics 102

Moza M. Al-Rabban

Professor of Physics

Lecture 11RefractionLecture 11Refraction

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RefractionRefraction When light is incident on a smooth boundary between two transparent materials (e.g., air and glass), two things happen:

1. Part of the light reflects from the boundary, obeying the law of reflection.

2. Part of the light crosses the boundary, changes direction, and continues into the second medium. This is called refraction.

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Snell’s LawSnell’s Law

1 1 2 2sin sinn n

In a medium in which light slows down, a ray bends closer to the perpendicular.

Willebrord van Roijen Snell1580 - 1626

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The Index of RefractionThe Index of Refraction Light travels through transparent media at a speed less than its speed c in vacuum.

We define the index of refraction in a transparent medium as:

medium

cn

v

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 Analyzing refractionAnalyzing refraction

1. Draw a ray diagram. Represent the light beam with one ray.2. Draw a line normal to the boundary. Do this at each point

where the ray intersects a boundary.3. Show the ray bending in the correct direction. The angle is

larger on the side with the smaller index of refraction. This is the qualitative application of Snell’s law.

4. Label angles of incidence and refraction. Measure all angles from the normal.

5. Use Snell’s law. Calculate the unknown angle or unknown index of refraction.

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Example 3: Deflecting a Laser BeamExample 3: Deflecting a Laser Beam

A laser beam is aimed at a 1.0 cm thick glass sheet at an angle of 300 above the glass.

(a)What is the laser beam’s direction of travel in the glass?

(b) What is its direction of travel in the air on the other side?

(c) By what distance d is the laser beam displaced?

1 11 12

2

sin 1.0sin 60sin sin 35.3

1.5

n

n

3 2 35.3

1 12 34

sin 1.5sin 35.3sin sin 60

1 1.0

n

n

1 22

(1.0 cm)sin sin sin 24.7 0.51 cm

cos cos35.3

td l

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Example: Measuring theIndex of Refraction

Example: Measuring theIndex of Refraction

A laser beam is deflected at an angle of 22.60 by a 300-600-900 prism. What is the prism’s index of refraction?

1 30

2 30 22.6 52.6

21 2

1

sin sin 52.6(1.0) 1.59

sin sin 30n n

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Total Internal ReflectionTotal Internal Reflection Suppose instead we have a 450-450-900 prism with n=1.50. At what angle will a beam normal to the long side be deflected?

1 45 1 12 1sin 1.5sin sin 1.061

There is no such angle! Therefore, there will be no refracted ray, and the light will be completely reflected. This is called total internal reflection. It occurs only when the 2nd medium has a lower index of refraction than the 1st medium (n2<n1), and when ≥c.

1 2

1

sinc

n

n

When 1=c, 2=900.

902 A critical angle is reached when There is no critical angle and no total internal reflection if

12 nn

9

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A light bulb is set in the bottom of a 3.0 m deep swimming pool. What is the diameter of the ring of light seen on the pool’s surface?

Example: Total Internal Reflection

Example: Total Internal Reflection

1 1.0sin 48.7

1.33c

2 tan 2(3.0 m) tan(48.7 ) 6.83 mcD h

This is the so-called “ring of bright water” seen when looking up from within a pool of water. The outside world is compressed to lie within the ring.

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Fiber OpticsFiber Optics Total internal reflection makes possible fiber optic light pipes, which can transport light and light-encoded signals over long distances without significant loss.

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Clicker Question 1Clicker Question 1 Light travels from medium 1 to medium 3 as shown. Which of the following describes the indices of refraction?

(a) n3 > n1;(b) n3 = n1;(c) n3 < n1;(d) We cannot compare n3 and n1 without knowing n2.

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Image Formation by RefractionImage Formation by Refraction

1 2tan ' tanl s s

1

2

tan'

tans s

2 1 1

1 2 2

sin tan

sin tan

n

n

2

1

'n

s sn

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Example: Air Bubblein a Window

Example: Air Bubblein a Window

A fish and a sailor look at each other through the 5.0 cm thick glass port hole of a submarine. There is a small air bubble half way through the glass. How far behind the glass surface does the bubble appear to the fish? How far behind the glass surface does the bubble appear to the sailor?

n = 1.00Air

n = 1.33Water

n = 1.50Glass

2

1

(1.33)'(fish) (2.5 cm) 2.2 cm

(1.50)

ns s

n

2

1

(1.00)'(sailor) (2.5 cm) 1.7 cm

(1.50)

ns s

n

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Color and DispersionColor and Dispersion

1. What we perceive as white light is actually a mixture of all colors. White light can be disbursed into colors and, equally important, colors can be combined to produce white light.

2. The index of refraction is slightly different for different colors of light. Glass has a slightly higher index of refraction for violet light than for green or red light. Consequently, different colors refract at slightly different angles.

Experiments with light and prisms show the following:

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DispersionDispersion This can be made quantitative by measuring the index of refraction of a transparent material as a function of wavelength and associating wavelengths with colors..

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The Dispersion RelationThe Dispersion Relation The cause of dispersion is the oscillations of bound electrons in the transparent medium. In particular, the index of refraction obeys a dispersion relation of the form:

2

2 20 0

12 ( )e

N en

m

Here N is the number of atoms per unit volume, e is the electron charge, me is the electron mass, is the resonant frequency of the bound electrons, and c/ is the angular frequency of the light.

Notice that if , the index of refraction n can be less than 1, indicating that the phase velocity of light in the medium is greater than c.

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Example: Dispersing Lightwith a Prism

Example: Dispersing Lightwith a Prism

We saw that light incident on a 300 prism is deflected by 22.60 if the prism’s index of refraction is 1.59. Suppose this is the index of refraction for deep violet light, and that deep red light has an index of refraction of 1.54.

(a)What is the deflection angle for deep red light?(b)If a beam of white light is dispersed by the prism, how

wide is the rainbow spectrum on a screen 2.0 m away>

1 violet violet1.54; 52.6 ; 22.6n

1 11 1red

2

sin 1.54sin 30sin sin 50.4

1.00

n

n

red violet red20.4 ; 2.2 0.038 rad

(2.0 m)(0.038 rad) 0.076 m 7.6 cms r

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RainbowsRainbows A rainbow is formed by spherical water droplets in which light is refracted twice and reflected once in the droplet. The maximum reflection angle of red light is 42.50, and that of violet light is 40.80. The net effect is to produce a bright ring, dispersed in wavelength, and centered 1800 away from the direction of the Sun. There is also a two-reflection secondary rainbow with reversed colors outside the main rainbow.

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Colored Filtersand Colored Objects

Colored Filtersand Colored Objects

When white light passes through green glass and emerges as green light, what happens? Does the glass add “greenness” to the light?

No. The glass removes the non-green light from the beam. More precisely, colored glass absorbs all wavelengths except those of one color, and that color is transmitted through the glass without absorption.

For example, leaves are green because of chlorophyll, which selectively absorbs red and blue light while reflecting green.

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Example: Filtering LightExample: Filtering Light

White light passes through a green filter and is observed on a screen.

(a) What happens if a second green filter is placed between the first filter and the screen?

(b) What happens if a red filter is placed between the first filter and the screen?

(a) Nothing happens because both filters transmit the same light.

(b) No light is transmitted, because the red filter blocks what the green filter transmits, and vice versa.

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Scattering: Blue Skiesand Red Sunsets

Scattering: Blue Skiesand Red Sunsets

4scatteredI

4

blue

red

650 nm4

450 nm

I

I

At the atomic level, light passing through the atmosphere undergoes Rayleigh scattering, which depends on frequency to the 4th power (or -

4). This preferentially scatters blue light, making the sky blue and removing blue light from beams of sunlight, particularly during sunset when the distance light travels through air is a maximum.

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Five Things You Should Have Learned from This Lecture

Five Things You Should Have Learned from This Lecture

1. Transparent media transmit light, but also may reduce the velocity of light as it passes through the medium.

2. Snell’s Law describes how a light beam is deflected as it crosses the interface between one transparent medium and another

3. The index of refraction n=c/v quantifies the reduction in speed of light passing through a transparent medium .

4. When light travels through an interface where the velocity increases, there is a critical angle beyond which it cannot be refracted. This produces the phenomenon of total internal reflection, used in light pipes and binoculars.

5. The index of refraction depends on wavelength. This phenomenon, called dispersion, can be used to separate white light into colors.