Ch 24 1

Chapter 24

Wave Nature of Light:

© 2006, B.J. LiebSome figures electronically reproduced by permission of Pearson

Education, Inc., Upper Saddle River, New Jersey Giancoli, PHYSICS,6/E © 2004.

Ch 24 2

Huygens’ Principle

•Wave motion is described by some rather complex equations but Huygens’ principle provides a simple way to predict wave propagation.

•Every point on a wave is a source of tiny wavelets that spread out in the forward direction at the speed of the wave itself.

•The new wave is the envelope of all the wavelets.

Ch 24 3

Diffraction

•Diffraction occurs when a wave encounters an obstacle. Applying Huygens principle results in a bending of the wave behind the obstacle.

•Note that if light were a stream of particles, there would be no light in the shadow of the obstacle.

•Note that a small opening (usually it’s a slit) results in a nearly spherical wave.

Ch 24 4

Interference•We will study several situations where a wave is split in half, travels over different paths and then is reunited.

Constructive Interference results when the amplitudes of the two waves add to give a larger amplitude.

Destructive Interference results when the amplitudes of the two waves add to give a smaller amplitude.

Ch 24 5

Young’s Double-Slit Interference

•Waves passing through two slits will diffract and spread out. The two waves will then interfere with each other when they reach the screen.

Ch 24 6

Young’s Double-Slit Interference•In the figure below, the path taken by the lower wave is greater by “d sin ”

•The waves interfere constructively when the path difference is “m ” where m is an integer.

.....2,1,0sin mmd

Condition for Constructive interference

Ch 24 7

Double Slit Pattern

The double slit interference pattern is a series of bright lines with the m = 0 line the brightest. The value of m is often called the order of the interference fringe.

Ch 24 8

Small Angle Approximation

In working these problems it is often convenient to use the small angle approximation which is true when

7o

tansin

Ch 24 9

Example 24-1: If 520-nm and 640-nm light passes through two slits 0.50 nm apart, calculate the angle of the second order fringes for these two wavelengths.

md sin

d

m 1sin

m

m3

91

520 1050.0

)10520)(2(sin

120.0

m

m3

91

640 1050.0

)10640)(2(sin

150.0

How far apart are the second-order order fringes for these two wavelengths on a screen 1.5m away?

md sin

d

m sin

tanLy

2m

For small angles

d

m sintan

d

mLy

m

mmy

3

9

520 1050.0

)10520()2()5.1(

mm12.3

mmy 84.3640

mmmmmm 72.012.384.3 y

yL

Ch 24 10

Diffraction Grating

A diffraction grating has many slits. As additional slits are added, the interference maxima become sharper and narrower as shown below (a) for two slits and (b) for six slits.

If “d” is the distance between adjacent slits, the equation for the grating is the same as for double slits.

md sin

Ch 24 11

Using a Diffraction Grating to Produce an Atomic Spectrum

If the light striking a diffraction grating is not monochromatic (single color) then the light is spread out into its component wavelengths. The resulting pattern is called a spectrum. We can the determine the wavelength from

....3,2,1sin mdm

Ch 24 12

Types of Spectrum

•Continuous Spectrum: includes all wavelengths

•Line Spectrum: contains only certain discrete wavelengths characteristic of the atom. Only emitted by gases.

•Absorption Spectrum: continuous spectrum with dark lines characteristic of the atoms absorbing the light. Example is the solar absorption spectrum shown below. Note that double dark lines correspond to the sodium spectrum.

Ch 24 13

Example 24-2: What is the highest spectral order that can be seen if a grating with 6000 lines per cm is illuminated with white light?

nmviolet 400 nmred 750White light spans the visible range

cm

m

cmlinesd

100

1

/6000

1m610667.1

md sin

sind

m

The limit of the spectra is θ = 90º and in order to see the full spectra, you can figure m for each λ

violetviolet

dm

90sin

m

m9

6

10400

10667.1

2.4

2.210750

10667.19

6

m

mmred So only the second order of red would be

seen and so m = 2

Ch 24 14

Dispersion of a PrismThe index of refraction n depends slightly on wavelength with n being highest for short wavelengths. This allows a prism to produce a spectrum with red being deviated the least.

Ch 24 15

Rainbows

Rainbows are produced by dispersion in spherical water droplets. Red light is bent the least, so it is seen from droplets higher in the sky than violet light. This separation produces a spectrum of the sunlight.

Ch 24 16

Single Slit Diffraction

This picture shows the diffraction pattern caused by monochromatic light on a narrow slit

•The single-slit pattern is produced by interference of light from different parts of the slit as shown below.

Ch 24 17

Single Slit Diffraction II

...,3,2,1sin mmD

•This equation gives the angular position of the minima of the intensity curve.

•D is the width of the slit.

•Note that if the width of the slit is made narrower, the above pattern gets wider.

Ch 24 18

Thin Film Interference•Light reflected off of the air/oil interface interferes with light reflected off of the oil/water interface.

•The situation can be complicated by a phase shift that can occur during reflection.

Ch 24 19

Phase Shift of Reflected Ray

n2 > n1 n2 < n1

180o phase shift no phase shift

•There is no single equation to solve these problems.

•For constructive interference, the path difference ( usually 2 t where t is the thickness) must equal m if there is no phase shift

•If there is a phase shift then for constructive interference

2 t = (m+1/2)

Ch 24 20

Example 24-4: Solar cells are often coated with a transparent thin film such as silicon monoxide SiO, n = 1.45) to minimize losses due to reflection. A silicon solar cell (n = 3.50) is coated with silicon monoxide for this purpose. Determine the minimum thickness of film that will produce the least reflection for light of wavelength 500nm.

nt 2

1)2(

nt 4

1

45.1air

n

45.1

500

4

1 nmt

nm86

t

Note that there is a phase change at each reflection. In order to cancel. The path difference (2 t ) must equal half of the wavelength in SiO.

Ch 24 21

Polarization

•We pictured light as a wave with transverse electric fields (E) perpendicular to transverse magnetic fields (B).

•Most light is a jumble of photons with different orientations of E fields (always with B field perpendicular to E). This is unpolarized light.

•With polarized light, all of the E fields have the same orientation.

Ch 24 22

Polaroid Materials•Plane-polarized light can be made using special materials with oriented long molecules. These molecules act like slits that pass light with E fields in the slit direction but eliminate light with perpendicular E fields.

•The action of polaroid materials is illustrated below

Ch 24 23

Intensity

2

0 cosII

Ch 24 24

First Polarizer

Ch 24 25

Intensity

2

0 cosII Note: The first polarizer reduces the intensity by 1/2 , so the intensity after a pair of polarizers is

20 cos2

II

Ch 24 26

Example 24-5 Unpolarized light passes through two polaroids. The axis of the first polaroid is vertical and the second is at 450 to the vertical. What percent of the light intensity is transmitted through the two polaroids.

223 )45(cos II

01 2

1II

08

1I

A third polaroid is added at 900 to the vertical. What percent is transmitted? (Note that none would be transmitted without the polaroid at 45º.)

Answer = 12.5%

203 )45(cos

4

1 II

212 )45(cos II

02 4

1II

Answer = 25%

The first polaroid eliminates half of the light

Light emerging from the second polaroid is polarized at 45º to the vertical and this is 450 to the third polaroid.

Ch 24 27

Polarization by ReflectionReflected light is partially polarized. It is 100 % plane polarized when the angle between the reflect and refracted ray is 90o.

For light from air on a medium of index of refraction n, reflected

light is 100 % plane polarized when:

Polarized with E page

nP tan