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Part 3: Optics

3.1: Electromagnetic Waves

An electromagnetic wave (light wave) consists of oscillating electric and magnetic fields. The

directions of the electric and magnetic fields are perpendicular. The wave travels in the third

perpendicular direction. It is a transverse wave. The fields store energy and transport the energy

in the wave.

For example, the picture below shows the wave traveling in the +x direction. The electric field

oscillates in the y direction and the magnetic field oscillates in the z direction.

The individual electric and magnetic waves are in phase. The fields peak at the same position at

the same time.

In a harmonic light wave, the fields can be written using a cosine or sine function.

𝐸(𝑥, 𝑡) = 𝐸𝑀 cossin

(𝑘𝑥 ± 𝜔𝑡 + 𝜑) and 𝐵(𝑥, 𝑡) = 𝐵𝑀 cossin

(𝑘𝑥 ± 𝜔𝑡 + 𝜑)

EM = electric field amplitude of wave [V/m]

BM = magnetic field amplitude of wave [T]

k = angular wave number [rad/m]

= angular frequency [rad/s]

= phase angle [rad]

2

= wavelength = distance between consecutive crests [m]

k = 2 / [rad/m]

T = period = time for one oscillation or cycle [s]

f = 1 / T = frequency = number of cycles per time [cycles/s = Hz]

= 2 / T = 2f [rad/s]

The sign between kx and t determines the direction the wave travels along the x-axis.

+ wave travels to left (in the direction of decreasing x)

- wave travels to right (in the direction of increasing x)

The phase angle shifts the cosine or sine function left or right. This can be used to match some

initial condition for the wave function. (We won’t have to worry about it.)

Wave Speed and Index of Refraction

The speed of the light wave in a medium is

𝑣 =𝐸𝑀

𝐵𝑀= 𝑓 =

𝜔

𝑘=

1

√𝜀𝜇

where and are the electrical permittivity and magnetic permeability of the medium,

respectively.

In free space (vacuum),

𝑣 =𝐸𝑀

𝐵𝑀= 𝑜𝑓 =

𝜔

𝑘𝑜=

1

√𝜖𝑜𝜇𝑜

= 𝑐 = 3x108 m/s

This is the fastest that light can travel and only light can travel at this speed. In any other

medium, the speed of light is

𝑣 =𝑐

𝑛

where n is the refractive index or index of refraction of the medium.

For vacuum, n = 1. For any other medium, n > 1. Air’s refractive index is approximately 1.

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Power and Intensity

The power is the rate that energy is carried in the wave.

𝑃 ∝ 𝐸𝑀𝐵𝑀𝑓2𝑣 ∝ 𝐸𝑀2 𝑓2

The power is spread out over an area A. The intensity is power per area. It determines the

“brightness” of the light.

𝐼 = 𝑃

𝐴

Example: The Sun emits power in all directions. At Earth, the received intensity is

approximately 1370 W/m2. This is known as the solar constant.

Electromagnetic Spectrum

Frequency of light determines the part of the spectrum. You should know the order of the

spectrum and of the colors in the visible part of the spectrum (ROYGBV).

Light in Different Media

When light is traveling in Medium 1 and enters a different media (Medium 2) the frequency is

constant. The speed and wavelength change.

𝑛1𝜆1 = 𝑛2𝜆2

Light in vacuum has a free-space wavelength of o. Wavelength in medium with index n is

𝜆 = 𝜆𝑜 𝑛⁄

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Example:

A laser pointer delivers a red beam in air with a wavelength of 650 nm. What is the wavelength

and color of the beam underwater (n ~ 1.33)?

Ans. 488 nm and the beam remains red because the frequency is the same

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3.2 Reflection & Refraction

Law of Reflection

𝜃1 = 𝜃𝑟

Law of Refraction (Snell’s Law)

𝑛1sin 𝜃1 = 𝑛2sin 𝜃2

Case 1: n1 < n2 (lo index into hi index) 1 > 2 (light bends towards normal like in picture)

Case 2: n1 > n2 (hi index into lo index) 1 < 2 (light bends away from normal)

Critical Angle (only for Case 2)

sin 𝜃𝑐 =𝑛2

𝑛1

If 1 > c then all light reflects for Total Internal Reflection (TIR)

Normal Dispersion

Index decreases as wavelength increases.

For light going from air into medium, violet bends most, red least.

1

2

r n1

n2

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3.3 Image Formation

p = object distance (from object to element)

q = image distance (from element to image)

h = object height (from axis to tip)

h’ = image height (from axis to tip)

Lateral Magnification

𝑀 =ℎ′

ℎ

Plane Mirror

q = p

h’ = h

M = 1

virtual, non-inverted image

Single Refracting Surface (Apparent Depth)

|q| < p

h’ = h

M = 1

virtual, non-inverted image closer to surface

for small angles, |𝑞| ≈𝑛2

𝑛1𝑝

p

h h’

q

eye

eye

p

q

h h’

n1 n2

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Thin Lens

Positive (Converging) Lens Negative (Diverging) Lens

Lens Maker’s Equation

1

𝑓= (

𝑛𝑙𝑒𝑛𝑠 − 𝑛𝑠𝑢𝑟

𝑛𝑠𝑢𝑟) (

1

𝑅1−

1

𝑅2)

F = focal point

f = focal length

nlens = index of lens material

nsur = index of medium surrounding lens

R1 = radius of curvature of left surface (front surface)

R2 = radius of curvature of right surface (back surface)

R > 0 for ( -shaped surface

R < 0 for ) -shaped surface

Thin Lens Equation

1

𝑓=

1

𝑝+

1

𝑞

Lateral Magnification

𝑀 =ℎ′

ℎ=

−𝑞

𝑝

f

F

f

F

p

h

h’

q

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Simple Optical Systems

(You should know how to do the ray traces for these systems.)

Camera

positive lens, object outside focal point and far from lens

real, inverted image

image smaller than object

Projector

positive lens, object outside focal point but close to lens

real, inverted image

image bigger than object

Magnifer (Magnifying Glass)

positive lens, object inside focal point

virtual, non-inverted image

image bigger than object

Angular Magnification

𝑚 =25 cm

𝑝

Largest useful angular mag occurs when q = -25 cm. Then, and only then, m = M.

𝑚𝑚𝑎𝑥 =25 cm

𝑓+ 1

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Example:

You use a lens with a focal length of +5 cm as a magnifying glass to look at a 1 cm-long bug.

(a) How far is the lens from the bug if you get the maximum useful angular mag?

(b) What is the maximum useful mag?

(c) What is the lateral mag?

(d) How long does the bug look to you?

Suppose you now move the lens so that it is 4.8 cm from the bug.

(e) How far is the image of the bug from you?

(f) What is the angular mag now?

(g) What is the lateral mag now?

(h) How long does the bug look to you?

Ans. (a) 4.17 cm (b) 6x (c) 6x (d) 6 cm (e) 120 cm (f) 5.2x (g) 25x (h) 5.2 cm

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Eye: Correcting Vision Problems with Lenses

Myopia (Near-Sightedness)

can’t see far-away objects

eye has an unhealthy far point (F.P.) of less than infinity

need negative lens to fix

𝑓 = −F.P.

Hyperopia (Far-Sightedness)

can’t see near objects

eye has an unhealthy near point (N.P.) greater than 25 cm

need positive lens to fix with p = 25 cm and q = -N.P.

1

𝑓=

1

25 cm−

1

N.P.

Lens Power (L.P.)

𝐿. 𝑃. =1

𝑓

mks units [1/m = diopter]

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3.2. Interference

Young’s Double Slit

𝑑 sin 𝜃 = {𝑚𝜆

(𝑚 + 1 2⁄ )𝜆 maxima minima

and if L >> d

𝑦 = {𝑚 𝜆𝐿 𝑑⁄

(𝑚 + 1 2⁄ ) 𝜆𝐿 𝑑⁄ maxima minima

where m = 0, ±1, ±2,…

Spacing between consecutive maxima = Spacing between consecutive minima = y = L / d

Diffraction Grating

Grating Equation

𝑑 sin 𝜃 = 𝑚𝜆

where order number m = 0, ±1, ±2,…

The grating constant is the reciprocal of the line spacing d. [# lines / length]

Thin Films

If reflection is lo n off of hi n, then phase shift is π.

If reflection is hi n off of lo n, then phase shift is 0.

Find phase shifts for surfaces 1 and 2. Subtract the two phase shifts to

get Δ.

For near-normal incidence,

Δ = 0 Δ =

2𝑛𝑡 = {𝑚𝜆

(𝑚 + 1 2⁄ )𝜆 maxima minima minima maxima

θ d

L

y

0

θ

d

cover nc

film n

substrate ns

t 1

2

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Example:

A lens with an index of 1.5 has an anti-reflection coating with an index of 1.4 that minimizes the

reflection of 580 nm-light. (a) What is the thinnest that this layer can be? (b) What is the

longest wavelength that is strongly reflected?

Ans. (a) 103.6 nm (b) 290 nm so the transmission is enhnaced for all visible light

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3.5. Diffraction

Single Slit

Intensity minima given by

𝐷 sin 𝜃 = 𝑚𝜆

and if L >> D

𝑦 = 𝑚 𝜆𝐿 𝐷⁄

where m = ±1, ±2,…

Width of central diffraction peak (CDP)

𝑊𝐶𝐷𝑃 = 2 𝜆𝐿 𝐷⁄

Circular Aperture

First minimum given by

𝐷 sin 𝜃1 = 1.22𝜆

and if L >> D

𝑦 = 1.22 𝜆𝐿 𝐷⁄

Diameter of Airy disk (CDP)

𝑊𝐶𝐷𝑃 = 2.44 𝜆𝐿 𝐷⁄

where m = 1, 2,…

θ D

L

y

0

θ D

L

y

0 D

WCDP

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Diffraction-Limited Resolution

Two points on the object are just resolvable as two distinct

points on the image when

𝐷 sin 𝜃1 = 1.22𝜆

The angle θ1 is the resolution angle and since it is very

small, sinθ1 ~ θ1 so the above equation can be written as

𝜃1 =1.22𝜆

𝐷 [rads]

The minimum separations of these two points on the object and image are given by

𝜃1[rads] =𝑑𝑚𝑖𝑛

𝑝 =

𝑑′𝑚𝑖𝑛

𝑞

Useful angle conversions: π rad = 180°

1° = 60’= 60’

1 arcminute = 60 arcseconds = 60”

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Example:

The pupil of a person’s eye is opened to a diameter of 2 mm. Find (a), the resolution angle of

this eye, and (b), the closest that two points can be separated bye on the page of a book that is 25

cm from the eye. Use a free-space wavelength of 560 nm which is in the middle of the visible

spectrum. (c) These two points are imaged on the retina which is 20 mm from the pupil. How

close are the two imaged points on the retina?

Ans. (a) 2.57x10-4

rad = 53” (b) 64 μm (c) 5 μm

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p

dmin

d’min

q

θ1 D

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3.6 Polarization

Linear Polarization of Light by Linear Polarizer

Law of Malus

𝐼 = 𝐼𝑜 cos2𝜃

transmission axis

incident electric

field

The energy from this perpendicular

component is absorbed.

This parallel component

gets through.

Intensity I

Intensity Io

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Linear Polarization by Reflection (Brewster’s Law)

As the incident angle θ gets closer to the polarization angle (Brewster angle) θP, more of the

reflected light is linearly polarized with the electric field axis parallel to the surface.

When θ = θP, all of the reflected light is linearly polarized with the electric field axis parallel to

the surface.

tan 𝜃𝑃 =𝑛2

𝑛1

*Review the Powerpoint slides on “polarization” located on the course page:

http://facstaff.cbu.edu/~jvarrian/252/252PolWeb/252PolAbs.ppt

3.7 Scattering and Absorption

*Review the Powerpoint slides on “polarization” located on the course page:

http://facstaff.cbu.edu/~jvarrian/252/252PolWeb/252PolAbs.ppt

There are a few slides at the end of the presentation that discuss scattering and absorption.