Lecture 24Lecture 24

Physics 2102Jonathan Dowling

EM wavesEM wavesGeometrical opticsGeometrical optics

The intensity of a wave is power per unit area. If one has a source thatemits isotropically (equally in all directions) the power emitted by thesource pierces a larger and larger sphere as the wave travels outwards.

24 r

PI

s

!=

So the power per unit areadecreases as the inverse ofdistance squared.

EM spherical wavesEM spherical waves

ExampleExampleA radio station transmits a 10 kW signal at a frequency of 100MHz. (We will assume it radiates as a point source). At a distanceof 1km from the antenna, find (a) the amplitude of the electric andmagnetic field strengths, and (b) the energy incident normally on asquare plate of side 10cm in 5min.

2

22/8.0

)1(4

10

4mmW

km

kW

r

PI

s ===!!

mVIcEEc

Imm

/775.022

1 2

==!= 0

0

µµ

nTcEBmm

58.2/ ==

mJSAtUA

tU

A

PS 4.2

/==!"

!==

Receivedenergy:

Radiation PressureRadiation PressureWaves not only carry energy but also momentum. The effect is very small (we don’t ordinarily feel pressure from light). If lightis completely absorbed during an interval Δt, the momentum transferred is given by

c

up

!=!

t

pF

!

!=Newton’s law:

Now, supposing one has a wave that hits a surfaceof area A (perpendicularly), the amount of energy transferred to that surface in time Δt will be

tIAU !=! thereforec

tIAp

!=!

I

A

c

IAF =

)reflection (total 2

),absorption (total c

Ip

c

Ip rr ==

Radiation pressure:

and twice as much if reflected.

Radio transmitter:

If the dipole antennais vertical, so will bethe electric fields. Themagnetic field will behorizontal.

The radio wave generated is said to be “polarized”.

In general light sources produce “unpolarizedwaves”emitted by atomic motions in random directions.

EM waves: polarizationEM waves: polarization

When polarized light hits a polarizing sheet,only the component of the field aligned with thesheet will get through.

)= !cos(EEy

And therefore: !20 cosII =

Completely unpolarized light will have equal components in horizontal and verticaldirections. Therefore running the light througha polarizer will cut the intensity in half: I=I0/2

EM waves: polarizationEM waves: polarization

ExampleExampleInitially unpolarized light ofintensity I0 is sent into a systemof three polarizers as shown.Wghat fraction of the initialintensity emerges from thesystem? What is thepolarization of the exitinglight?

• Through the first polarizer: unpolarized to polarized, so I1=½I0.• Into the second polarizer, the light is now vertically polarized.Then, I2=I1cos260ο = 1/4 I1 =1/8 I0.• Now the light is again polarized, but at 60o. The last polarizer ishorizontal, so I3=I2cos230ο =3/4 Ι2=3/32 I0=0.094 I0.• The exiting light is horizontally polarized, and has 9% of theoriginal amplitude.

Reflection and refractionReflection and refraction

Law of reflection: the angle of incidence θ1equals the angle of reflection θ’1.

Law of refraction:1122 sinsin !! nn = Snell’s law.

When light finds a surface separating two media (air and water, forexample), a beam gets reflected and another gets refracted(transmitted).

n is the index of refraction of the medium. In vacuum, n=1. In air, n~1. In all other media, n>1.

ExampleExampleWater has n=1.33. How much does a beam incident at 45o refracts?

n2 sin θ2= n1 sin θ1

sin θ2= (n1 /n2) sin θ1

=(1/1.33) sin 45o

=0.0098θ2= 32o Actual light ray

Light ray the brainimagines (as if in air)Actual object

Image of the object

Example: an optical illusionExample: an optical illusion

The index of refractiondecreases with temperature:the light gets refracted andends up bending upwards.

We seem to see water on theroad, but in fact we arelooking at the sky!

Chromatic dispersionChromatic dispersionThe index of refraction depends on the wavelength (color) of the light.

Total internal reflectionTotal internal reflectionFrom glass to air, the law of refraction uses n2<n1, so θ2>θ1: it mayreach 90o or more: the ray is “reflected” instead of “refracted”.

For glass (fused quartz) n=1.46,and the critical angle is 43o:optical fibers!

n2 sin θ2= n1 sin θ1

n1

n2~1

θ1

θ2

Polarization by reflectionPolarization by reflectionDifferent polarization of light getreflected and refracted withdifferent amplitudes(“birefringence”).

At one particular angle, the parallelpolarization is NOT reflected at all!This is the “Brewster angle” θB,and θB+ θr=90o.

!!! ="= ### cos)90sin(sin 221 nnno

1

2tan

n

n=!"

Geometrical OpticsGeometrical Optics

• “Geometrical” optics: light rays (“particles”) that travel in straightlines.

• “Physical” optics: electromagnetic waves which have amplitude andphase that can change.

Light is BOTH a particle (photon) and a wave: wave-particle duality.

Plane mirrorsPlane mirrorsLight rays reflect on a plane mirror,and produce a virtual image behindthe mirror.What’s a virtual image? It means thelight rays are NOT coming from areal point, they only seem to comefrom there to our eyes and brain.

i= −p for a plane mirror

object

image

Spherical mirrorsSpherical mirrors• Focal point is at half the curvatuire radius: f= − r/2 .•Rays parallel to the axis, reflect through the focal point.• Rays hitting the mirror after going to the focal point, emerge parallel.• Rays going through the center of curvature, reflect back on themselves.

Concave mirrors: r > 0

Convex mirrors: r < 0

Images fromImages fromspherical mirrorsspherical mirrors

fip

111=+

Consider an object placed between the focalpoint and the mirror. It will produce a virtualimage behind the mirror.

When the object is at the focal point theimage is produced at infinity.

If the object is beyond the focal point, a realimage forms at a distance i from the mirror.

Check the signs!!

p

im !=

lateralmagnification

ExampleExampleAn object 2cm high is located 10cm from a convex mirror with aradius of curvature of 10cm. Locate the image, and find its height.

Focal length: f= r/2= −10cm/2= −5cm.

Image position: 1/i=1/f-1/p= -1/5cm –1/10cm= -3/10cm i= - 10/3cm = -3.33 cm: the image is virtual.

Magnification: m= -i/p= - (-3.33cm)/(10cm)=0.33 (smaller).

If the object image is 2cm, the image height is 0.33 x 2cm=0.67 cm.

How important is it to knowHow important is it to knowwhere the image is?where the image is?

Hubble mirror: The central region ofthe mirror was flatter than it shouldbe - by just one-fiftieth of the widthof a human hair. This is equivalent toonly four wavelengths of visiblelight, but it was enough. One insidersaid that the Hubble mirror was"very accurate, very accurately thewrong shape".

A star seen with a groundtelescope and with “old” Hubble

The same star seenwith the “new” Hubble

Before and after...