lecture 17 - ctapsctaps.yu.edu.jo/physics/courses/phys207/chapter5_bw_h2.pdf · lecture 17 © dr....

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Phys. 207: Waves and LightPhysics Department

Yarmouk University 21163 Irbid Jordan

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Lecture 17

© Dr. Nidal Ershaidat

http://ctaps.yu.edu.jo/physics/Courses/Phys207/Lec5-1

© Dr. N. Ershaidat Phys. 207 Chapter 5: Electromagnetic Waves

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In 1870, James Clerk Maxwell's established a mathematical frame based on four equations in electricity and magnetism, in which all phenomena electricity and magnetism can be explained.

Maxwell’s Equations

Electricity and magnetism were thus unified in what we call now Electromagnetism.

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3

dtd

i EΦΦΦΦµµµµεεεε++++µµµµ 000

Ampere-Maxwell Law

Gauss’ Law for electricity

Gauss’ Law for magnetism

Faraday’s Law

Maxwell’s Equations

0

.εεεε

====→→→→→→→→

∫∫∫∫q

AdE

0. ====→→→→→→→→

∫∫∫∫ AdB

dtd

ldE BΦΦΦΦ−−−−====→→→→→→→→

∫∫∫∫ .

====→→→→→→→→

∫∫∫∫ ldB .


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James Clerk Maxwell's showed that electromagnetic energy propagates in vacuum with the speed of light (c).

Light is em Waves

His major conclusion was that light is nothing else but em waves, i.e. a beam of light is a traveling wave of electric and magnetic fields—an electromagnetic waveOptics, the study of visible light, became a branch of electromagnetism.

We shall concentrate on strictly electric and magnetic phenomena, and we build a foundation for optics.

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5

c and MKS

00

1

µµµµεεεε====c

The MKS system (SI) has for base this equation.

All other units are defined starting from the definition of c.


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In Maxwell's time, the visible, infrared, and ultraviolet forms of light were the only electromagnetic waves known.

H. Hertz

Heinrich Hertz discovered what we now call radio waves and verified that they move through the laboratory at the same speed as visible light.

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• Note the overlap between types of waves

•Visible light is a small portion of the spectrum

•Types are distinguished by frequency or wavelength

The Electromagnetic Spectrum1 nm = 10-9 m

1 Å = 10-10 m

Fig. 1


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The relative sensitivity of the average human eye to electromagnetic waves at different wavelengths.

Fig. 3

Visible Light

This portion of the electromagnetic spectrumto which the eye is sensitive is called visible light.

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5-2 The Traveling Electromagnetic Wave,

Qualitatively


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! Any accelerating charge will emit electromagnetic waves.

The Electromagnetic Spectrum

!An electromagnetic wave is a wave that

combines the electric wave “electric field E”

and magnetic wave “magnetic field B’.

! “Oscillating charges” is an example accelerating charges.

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Generation of electromagnetic waves

Fig. 4

An LC circuit is used to produce emoscillations. An energy source is used to compensate for the energy loss in the wires (represented by the resistance R)

Oscillations are transmitted to an antenna which emits the em waves in space.


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Description of the electric and magnetic fields as they pass point P on the Fig.4

Propagation of an em wave

Fig. 5

Propagation direction of the wave

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Properties of the electric and magnetic field in

electromagnetic waves.


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1. EM waves are transverse waves.

2.The electric field is always perpendicular to the magnetic field.

EM Waves are transverse waves

The electric and magnetic fields and are always perpendicular to the direction of travel of the wave., as discussed in Chapter 17.

Er

Br

3.The cross product always gives the direction of travel of the wave.

BErr

××××

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Properties of the electric and magnetic field in electromagnetic waves.

)sin(),(

)sin(),(

txkBtxB

txkEtxE

m

m

ωωωω−−−−====ωωωω−−−−====

4.The fields always vary sinusoidally, just like the transverse waves discussed in Chapter 17 . Moreover, the fields vary with the same frequency and in phase (in step) with each other.

5.The E and B field can be described by:


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Speed of propagation6. All electromagnetic waves, including visible

light, have the same speed c in vacuum Where c is given by:

00

1εεεεµµµµ

============m

m

BE

BEc

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Where n is a parameter which characterizes the medium and called the index of refraction of the medium.

7. Speed of light in dense medium is less than c:

Speed in Material Media

ncv ====



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Lecture 18



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Qualitatively


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! Any accelerating charge will emit electromagnetic waves.

The Electromagnetic Spectrum

!An electromagnetic wave is a wave that

combines the electric wave “electric field E”

and magnetic wave “magnetic field B’.

! “Oscillating charges” is an example accelerating charges.

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Generation of electromagnetic waves

Fig. 4

An LC circuit is used to produce emoscillations. An energy source is used to compensate for the energy loss in the wires (represented by the resistance R)

Oscillations are transmitted to an antenna which emits the em waves in space.


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Description of the electric and magnetic fields as they pass point P on the Fig.4

Propagation of an em wave

Fig. 5

Propagation direction of the wave

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• Two rods are connected to an ac source, charges oscillate between the rods (a)• As oscillations continue, the rods become less charged, the field near the charges decreases and the field produced at t = 0 moves away from the rod (b)

• The charges and field reverse (c)

• The oscillations continue (d)

EM Waves from an Antenna

Fig. 6


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Propagating Oscillations

+

-x

z

y

Current (up and down) creates B field into and out of the page!

Fig. 7

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Properties of electromagnetic waves.


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1. EM waves are transverse waves.EM Waves are transverse waves

The electric and magnetic fields and are always perpendicular to the direction of travel of the wave, as discussed in Chapter 17.

Er

Br

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Speed of propagation2. All electromagnetic waves, including visible

light, have the same speed c in vacuum Where c is given by:

00

1εεεεµµµµ

============m

m

BE

BEc

v = c = λλλλ f

kc ωωωω====

Fig. 8


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Where n is a parameter which characterizes the medium (the index of refraction of the medium).

3. Speed of light in dense medium is less than c:

Speed in Material Media

ncv ====

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1.The electric field is always perpendicular to the magnetic field.

Properties of the electric and magnetic field in electromagnetic waves.

2.The cross product always gives the direction of travel of the wave.

BErr

××××

3.The fields always vary sinusoidally, just like the transverse waves discussed in Chapter 17. Moreover, the fields vary with the same frequency and in phase (in step) with each other.

(((( )))) (((( )))) (((( )))) (((( ))))txkBtxBtxkEtxE mm ωωωω−−−−====ωωωω−−−−==== sin,,sin,

4.The E and B field can be described by:


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Plane Electromagnetic waves.

(((( )))) (((( )))) (((( )))) (((( ))))txkBtxBtxkEtxE mm ωωωω−−−−====ωωωω−−−−==== sin,,sin,

The previous equations define what we call a plane wave

(((( )))) (((( )))) (((( )))) (((( ))))txkim

txkim eBtxBeEtxE ωωωω−−−−ωωωω−−−− ======== ,,,

Which we can write in a complex form as:(((( ))))txkie ωωωω−−−−

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Quantitatively


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Consider a representative plane of area A.

Fig. 9 shows a wavefront at some reference instant t0. After time dt, the wavefront has traveled cdt.

propagates in the y+

direction and in the z+

direction

Em/Bm= c - Induced Electric Field

Fig. 9

Er

Br

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Let us apply Faraday’s law of induction along the dashed ………… path:

Em/Bm= c, Induced Electric Field

dtd

ldE BΦΦΦΦ−−−−====→→→→→→→→

∫∫∫∫ .

dx

h

(((( )))) dEhhEhdEEldE ====−−−−++++====→→→→→→→→

∫∫∫∫ .

dtABd

dtd B )(====

ΦΦΦΦ

dtdBdxhdEh −−−−====

dtdB

dxdE −−−−====

dtdBA====

dtdBdxh====

E

E + dE


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Em/Bm= c

(((( )))) (((( ))))txkBtxkkE mm ωωωω−−−−ωωωω++++====ωωωω−−−−⇒⇒⇒⇒ coscos

ckBE mm ====ωωωω====⇒⇒⇒⇒

(((( )))) (((( ))))txkEtxE m ωωωω−−−−==== sin,Oscillations are described by:

(((( )))) (((( ))))txkBtxB m ωωωω−−−−==== sin,

(((( )))) (((( ))))txkEkx

txEm ωωωω−−−−====

∂∂∂∂∂∂∂∂⇒⇒⇒⇒ cos,

( ) ( )txkBt

txBm ωω −−=

∂∂

cos,

dtdB

dxdE −−−−====

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Let us apply Ampere-Maxwell law of induction along the dashed ………… path:

Em/Bm= c, Induced Magnetic Field

dtd

ldB EΦΦΦΦµµµµεεεε−−−−====→→→→→→→→

∫∫∫∫ 00.

dx

h(((( )))) dBhhBhdBBldB −−−−====−−−−++++−−−−====→→→→→→→→

∫∫∫∫ .

dtAEd

dtd E )(====

ΦΦΦΦ

dtdEdxhdBh 00 µµµµεεεε====−−−−

dtdE

dxdB

00 µµµµεεεε−−−−====

dtdEA====

dtdEdxh====

→→→→→→→→++++ BdB

→→→→B

→→→→E


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Em/Bm= c

(((( )))) (((( ))))txkEtxkkB mm ωωωω−−−−ωωωωµµµµεεεε++++====ωωωω−−−−⇒⇒⇒⇒ coscos 00

(((( )))) ckBE

m

m

0000

11µµµµεεεε

====ωωωωµµµµεεεε

====⇒⇒⇒⇒

(((( )))) (((( ))))txkEtxE m ωωωω−−−−==== sin,Oscillations are described by:

(((( )))) (((( ))))txkBtxB m ωωωω−−−−==== sin,

(((( )))) (((( ))))txkEt

txEm ωωωω−−−−ωωωω−−−−====

∂∂∂∂∂∂∂∂⇒⇒⇒⇒ cos

,

(((( )))) (((( ))))txkBkx

txBm ωωωω−−−−====

∂∂∂∂∂∂∂∂ sin,

dtdE

dxdB

00 µµµµεεεε−−−−====

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εεεε0 µµµµ0 c2 = 1

cBE

m

m ====

We have:

cBE

m

m

00

1µµµµεεεε

====

1200 ====µµµµεεεε c

00

1µµµµεεεε

====c

Which gives:

The speed of em waves is:

5-4 Energy Transport and the Poynting Vector

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From sun light, We all know that an electromagnetic wave can transport energy and deliver it to a body on which it falls.

The rate of energy transport per unit area in such a wave is described by a vector , called the Poynting vector.

Sr

It is called after physicist John Henry Poynting(1852–1914), who first discussed its properties.

Poynting Vector


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The direction of the Poynting vector of an electromagnetic wave at any point gives the wave's direction of travel and the direction of energy transport at that point.

Sr

Poynting Vector

The Poynting vector is defined by:

→→→→→→→→→→→→××××

µµµµ==== BES

0

1

====

µµµµ====

====→→→→→→→→→→→→

→→→→→→→→→→→→

shrSandBE

BES

..,

1

0

The magnitude of the Poynting vector is given by:

BES0

1µµµµ

====

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The units of S is power per unit area or W/m2

Its magnitude S is related to the rate at which energy is transported by a wave across a unit area at any instant.

Remember that for a plane wave:

Power emitted and Poynting

[[[[ ]]]] [[[[ ]]]] [[[[ ]]]] [[[[ ]]]]][][ Area

PowerArea

TimeEnergyS ========

cBE

BE

m

m ========

2

0

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0

2BcSorEc

cES

µµµµ====εεεε====

µµµµ====⇒⇒⇒⇒


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The intensity is defined as:

or:

Intensity Emitted

cE

Ec

SI m

avgavg0

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0 21

µµµµ====

µµµµ========

20

0

2

21

2 mm Ec

BcI εεεε====

µµµµ====

cErms

0

2

µµµµ====

cErms

0

2

µµµµ====

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The energy of electromagnetic waves is related to its momentum p by the following relation:

If an electromagnetic wave is incident normally on an area A as shown in fig. 10 in a time ∆∆∆∆t and it is totally absorbed then there is momentum change given by:

Energy – Momentum of an em Wave

cpU ====

cUp ∆∆∆∆====∆∆∆∆

Fig. 10

Where this momentum change is totally transferred to the area A

A


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The net force on the wall is:

The radiation pressure Pr is:

Radiation Pressure

(((( )))) (((( )))) (((( ))))AI

cor

ttAI

cor

tU

cor

tp

F212121 ====

∆∆∆∆∆∆∆∆====

∆∆∆∆∆∆∆∆====

∆∆∆∆∆∆∆∆====

(((( ))))cIor

AFPr 21========

If the electromagnetic wave is totally reflected then the change in momentum is:

cU

p∆∆∆∆====∆∆∆∆ 2

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Lecture 19



Light Incident on an Object

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Light incident on an object

• Reflects (bounces)

• Refraction (bends)

•Absorbed


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Angle between light beam and normal Angle of incidence = Angle of reflection

Law of Reflection

θθθθi = θθθθf

θθθθi

θθθθf

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49


Refractive index n and wavelength

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Snell’s Law

θθθθ1

When light travels from one medium to another the speed changes v=c/n, but the frequency is constant.

So the light bends: n1 sin(θθθθ1)= n2 sin(θθθθ2)

n1n2

θθθθ2

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Which is true? 1) n1 > n2

2) n1 = n2

3) n1 < n2

Snell’s Law – Example 1

n1

n2

θθθθ1

θθθθ2

θθθθ1 < θθθθ2

sin θθθθ1 < sin θθθθ2

n1 > n2

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A ray of light traveling through the air (n=1) is incident on water (n=1.33). Part of the beam is reflected at an angle θθθθr = 60°°°°. What is θθθθ2?

Snell’s Law – Example 2

1θ

2θ

sin(60°°°°) = 1.33 sin(θθθθ2)

θθθθ2 = 40.6 degrees

rθθθθθ1 =θθθθr =60°°°°

n2 sin(θθθθ2) = n2 sin(θθθθ2)n2=1.33

n1=1.0

normal

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Total Internal ReflectionSnell’s Law: n1 sin(θθθθ1)= n2 sin(θθθθ2)

when n1 > n2 , θθθθ2 > θθθθ1

When θθθθ1 = sin-1(n2/n1) θθθθ2 = 90°°°°This is a critical angle!

Light incident at a larger angle will be completelyreflected θθθθi = θθθθr

normal

θθθθ2

θθθθ1

n2

n1

θθθθc

θθθθiθθθθr

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Total internal reflectioncan occur when light attempts to move from a medium with a high index of refraction to one with a lower index of refraction.

Ray 5 shows internal reflection

Total Internal Reflection

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A particular angle of incidence will result in an angle of refraction of 90°

Critical Angle

This angle of incidence is called the critical angle

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2sin nnfornn >>>>====θθθθ


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Chromatic DispersionThe index of refraction n encountered by light in any medium except vacuum depends on the wavelength of the light.

The dependence of n on wavelength implies that when a light beam consists of rays of different wavelengths, the rays will be refracted at different angles by a surface; that is, the light will be spread out by the refraction. This spreading of light is called chromatic dispersion, in which “chromatic” refers to the colors associated with the individual wavelengths and “dispersion” refers to the spreading of the light according to its wavelengths or colors

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Chromatic Dispersion

12

12 sinsin θθθθ====θθθθ

nn


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red

blue

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Lecture 20



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61

The maximum electric field at a distance of 10 mfrom an isotropic point light source is 2.0 V/m. What are:a) the maximum value of the magnetic field andb) the average intensity of the light there?c) What is the power of the source?

Problem 17

a)

b)

c)

TeslacEB 8

810666.0

103

2 −−−−××××====××××

========

278

0

2

/0106.0104103

4 mWcE

I m ====××××ππππ××××××××

====µµµµ

==== −−−−

WIrP 32.130106.010044 2 ====××××××××ππππ====ππππ====

Solution:


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A plane electromagnetic wave, with wavelength 3.0 m, travels in vacuum in the positive x direction with its electric field , of amplitude 300 V/m, directed along the y axis

(a) What is the frequency f of the wave?

(b)What are the direction and amplitude of the magnetic field associated with the wave?

Problem 25

Hzcf 88 103103 ====××××====λλλλ====

axizzthealongcEB −−−−====××××======== −−−−68 10103300

→→→→E

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Problem 25(c)What are the values of k and ωωωω if

E = Em sin(k x – ωωωω t)?

mk 1.2322 ====ππππ====λλλλππππ====188 1028.61022 −−−−××××====××××ππππ====ππππ====ωωωω sradf

,

(d)What is the time-averaged rate of energy flow in watts per square meter associated with this wave?

(((( ))))(((( )))) (((( )))) 2782

02

3.1191041032300

2

mW

cESI mavg

====××××ππππ××××××××××××====

µµµµ========

−−−−


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Problem 25(e)If the wave falls on a perfectly absorbing sheet

of area 2.0 m2, at what rate is momentum delivered to the sheet and what is the radiationpressure exerted on the sheet?

NcAI

dtdp 8

8105.79

103

23.119 −−−−××××====××××

××××========

291075.39 mNcIpr

−−−−××××========

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Problem 39A beam of partially polarized light can be considered to be a mixture of polarized and unpolarized light. Suppose we send such a beam through a polarizing filter and then rotate the filter through 360° while keeping it perpendicular to the beam. If the transmitted intensity varies by a factor of 5.0 during the rotation, what fraction of the intensity of the original beam is associated with the beam's polarized light?


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Let I0 be the intensity of the incident beam and fthe fraction of it which is polarized.Then the intensity of the polarized portion is f I0:

Problem 39 – Solution

(((( )))) θθθθ++++−−−−====′′′′ 200 cos21 IfIfI

and the intensity of the unpolarized portion is (1-f) I0:Initially the intensity of transmitted portion before rotating the polariod is:

The minimum intensity is :

The maximum intensity is:(((( )))) (((( )))) {{{{ }}}} (((( )))) 21211cos 000

2max IfIfIfII ++++====++++−−−−========θθθθ′′′′====′′′′

(((( )))) (((( )))) {{{{ }}}} (((( )))) 210210cos 002

min IfIfII −−−−====++++−−−−========θθθθ′′′′====′′′′

325

11

min

max ====⇒⇒⇒⇒====−−−−++++==== f

ff

II

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Problem 49Prove that a ray of lightincident on the surface of a sheet of plate glass of thickness t emerges from the opposite face parallel to its initial direction but displaced sideways, as in Fig. 34-47.

Show that, for small angles of incidence θθθθ, this displacement is given by

nn

tx1−−−−θθθθ====

where n is the index of refraction of the glass and θθθθ is measured in radians.


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Problem 59In Fig. 34-52 , light enters a 90°°°° triangular prism at point P with incident angle θθθθ and then some of it refracts at point Q with an angle of refraction of 90°°°°(a) What is the index of refraction of the prism in terms of θθθθ? (b) What, numerically, is the maximum value that the index of refraction can have? Explain what happens to the light at Q if the incident angle at Q is (c) increased slightly and (d) decreased slightly.

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Chapter 6: Interference

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