Unit 4 Electromagnetic Waves (EMR) - Weeblymrsunland.weebly.com/uploads/1/3/2/9/13290043/unit... ·...

Physics 30 Workbook Unit4:EMR

p.67

UNIT 4: ELECTROMAGNETIC

WAVES (EMR)

"The speed of Hght it is 186,000 miles

per second. The speed of Ute i's only 100j000 miles per second."

Einstein discovers that

the speed of light is 65 MPH.

Unit4:EMR Physics 30 Workbook

p.68

A. TRANSVERSE WAVES

Types of Waves

A wave is a transfer of energy, not matter, from a source to a receiver. It is a travelling

"disturbance" in a medium, where the medium is the material that the wave travels through.

There are two types of waves:

1. Mechanical waves

-waves that MUST travel through matter

-the matter is disturbed (vibrated), but it does not move with the wave

e.g. spring, water, dominoes, sound, earthquakes

2. Electromagnetic Waves

-waves that can travel through a vacuum (no medium is required)

e.g. light, ultraviolet, X-rays

Source Vibrations

The source of all waves is a vibration.

e.g. Sound - vibrating guitar string or vocal chord

Water - dropped rock or ocean floor

The source vibration determines two qualities of a wave:

• it provides the initial energy of a wave

- this energy is then transferred through the medium

• it determines the frequency of the wave (fsource = fwave)

-once the wave is created, the frequency of the wave will never change

(assuming that the source and the receiver are stationary- no Doppler effect)

Frequency and Period

1. Frequency of vibration (f):

-the number of cycles (oscillations, repetitions) in one second

f = total # of cycles

total time

cycle 1 Units: Hertz (Hz) where 1Hz 1

s s

2. Period of vibration (T)

-the time it takes for one cycle (repetition)

T = total time

total # of cycles Units: Seconds (s)

3. Frequency and Period:

- both are reciprocals of each other


p.69

-

Anatomy of a Transverse Wave (Ideal)

- in a transverse wave, particles move in a direction perpendicular to the wave motion

Crest

Equilibrium

Source

Vibration

Trough

wave motion

where A is the amplitude (energy) of a wave

A is the wavelength (length of one cycle)

-particles that are one wavelength apart will move identically (they are in phase)

-due to friction, transverse waves will lose their energy (amplitude), but the frequency will never

change

Speed of a Wave I Universal Wave Equation

If the medium doesn't change, the speed of a wave is constant. Thus, you can change the speed of

a wave only by changing the medium (e.g. stretching the spring, different depths of water, etc.). The

frequency and amplitude of source vibration have no effect on the speed of the wave.

However, the frequency of the source vibration does affect the wavelength, as described by the

universal wave equation :

where f is the frequency of the source vibration (in Hz)

A is the wavelength of the wave created (in m)

Note: Wavelength and frequency have an inverse relationship

Based on the equation A = j'

we see that A oc 1

f

(assuming v remains constant)

Thus, if the source frequency is high, the wavelength is short

if the source frequency is low, the wavelength is long

If it is a light wave, then it travels at the speed of light. In air /vacuum, light travels at

v = c = 3.00 x 108

m/s


p.70

B. ELECTROMAGNETIC RADIATION (EMR)

B 1. Maxwell's Laws and EMR

Faraday believed that there was a relationship between electricity, magnetism , and light. This was established by James Maxwell in the 1860's.

Maxwell summarized electromagnetism in 4 laws, described as follows :

1. The distribution of electric charge is related to the net electric field it produces

2. Magnetic field lines are continuous (closed loops), whereas electric field lines begin

and end on electric charges.

3. Electric current can produce a magnetic field (discovered by Oersted).

Also, a changing electric field can produce a magnetic field. ( E---+ B)

4. A changing magnetic field can produce an electric field. (B---+ E) And thus, it can induce a current in a conductor (based on Faraday's law)

Note: An electric field can accelerate a charge from rest, and thus induce current,

but a magnetic field can 't. It follows, then, that by changing the magnetic field,

an electric field is created, and this induces a current.

Based on Laws 3 and 4, Maxwell predicted the existence of a brand new kind of wave, which he

called an electromagnetic wave (or electromagnetic radiation, EMR).

• The source of all electromagnetic waves is an accelerating charge

• This creates a changing magnetic field ( B),based on Law #3

• This B creates a changing electric field ( E),based on Law #4

• This E creates a B (Law #3), and the cycle continues indefinitely

Illustrated as:

e- I B

L1E E B E

velocity

----------- ---11>-

Source Electromagnetic Wave

The magnetic field, electric field, and the velocity are always perpendicular to each other.

j


p.71

t .. •• ' .

Verified by Heinrich Hertz (1886):

switch

inductJon coil

e-l B E B E

primary coil receiver

Hertz used a step-up transfonner to induce a high, alternating voltage in the secondary

coils. This high voltage created an accelerating charge across the spark gap, which according

to Maxwell, would create electromagnetic waves. The EMR then travelled across a distance

and induced a current in the receiver wire. The current was detected by a galvanometer.

This verified the existence of electromagnetic waves. In fact, these waves were later

called radio waves.

B2. The Electromagnetic Spectrum

-using his device, Hertz was able to show that E-M waves had the same characteristics as light waves

e.g. speed= c in vacuum, reflection, refraction, diffraction, interference, and polarization

- this was convincing proof that light and radio waves are both electromagnetic waves

- we have now discovered that there are far more types of electromagnetic waves

----+ they differ only in frequency (determined by the source vibration) and wavelength ( A, = c If)

lo-s .5x 10,.0 10"8 10-10 10·12 Wavelength (m)

About the size oL

Buildings Humans Honey Bee Pinpoint Protozoans Molecules Atoms Atomic Nuclei

1o1s 1Ql6 1Q18 1Q20 Frequency (Hz)


p.72

p

HOMEWORK (EMR)

For the questions below, allforms of EMR travel at the speed of light.

Both A and B

1. Arrange the following forms of EMR in order of increasing wavelength :

ultraviolet radio x-rays visible infrared microwaves

A. 2. A source vibration completes 2.53 x 1010

cycles in 3.00 minutes. What is the wavelength

of the EMR that it produces?

3. For the EMR wave shown,

determine the frequency of its

source electrical vibration. d(km)

4. A DC transformer is set up to create radio waves. Describe the nature of the EMR when

the switch is closed just once. Why?

B. 5. Consider the carrier waves of two radio stations: 820kHz (AM) and 96.3 MHz (FM).

Compare these waves in terms of wavelength, speed, and time to travel in 45.0 km (in JlS).

determine the resulting wavelength.

6. For the source vibration shown,

vC\vQv 8.40 t(JlS)

7. If a radio tower's transmitter is oriented vertically , how should the antenna on your

home radio be oriented for the best reception? Why?

SOLUTIONS

1. x-rays , ultraviolet, visible, infrared, microwaves, radio

2. f= 1.4056 x 108

Hz ; A,= 2.13 m 3. A-=4.5km; f=6.7 x 104

Hz

4. When the switch is closed, there is a brief increase in current, which creates a brief change in

magnetic field. This will result in a short burst of EMR . But if the switch remains closed, the

current will become constant, which will result in a constant magnetic field (constant flux).

Constant magnetic fields (currents) do not create EMR, so no EMR will be created.

5. f= 820kHz: v = 3.00 x 108

m/s ; A= 366m ; t4s Ian = 150 JlS

f= 96.3 MHz: Same speed ; A= 3.12 m (shorter wavelength, since the frequency is higher)

Same time to travel 45 km (150 JlS)

6. T= 2.8 JlS ; f= 3.571 x 105

Hz ; A-= 840 m

7. The antenna should also be oriented vertically (same as transmitter) . The changing electric field

is parallel to the accelerating electrons in the antenna, thus it is vertical as well. If it is to generate

current in a receiver , the conductor must also be vertical.

p.73


,

,

I

I

I

I

C. THE SPEED OF LIGHT (EMR)

C 1. Early Methods

Galileo's Lanterns

The first serious attempt to determine the speed oflight was done by Galileo (1564- 1642). He

and his assistant stood on two distant hilltops with shuttered lanterns . Galileo opened his lantern first,

and when his assistant saw the flash of light, he was to immediately open his lantern . Galileo then used

his pulse to determine the time from the initial opening of the shutter to when he saw his assistant's

flash of light. Finally, knowing the distance between the hills, he would calculate the speed of light.

However, it turned out that light travelled so fast that their reaction times were way too slow to

accurately measure the time . (In fact, light would have taken about 70 !J.S to travel the distance, which

would have been impossible even with digital timers!)

Roemer and Huygens

Olaus Roemer (1644- 1710), a Danish astronomer, observed the motions of the moons around

Jupiter. Although other scientists had already determined their periods of revolution , Roemer wanted

to determine the exact moment that one of the moons would experience an eclipse - when Jupiter

would be in front of the moon. He noticed that as Earth got closer to Jupiter, the times of the eclipse

came earlier and earlier than expected; as Earth moved further away from Jupiter, the times came later

and later.

Roemer concluded that as the Earth got further away, light had to travel further to reach the Earth

and thus the eclipse was seen later. This is illustrated in the diagram below :

diameter of Earth's orbit

------- ,,

, I

I

I

I I ,.... T ... ,

' I B

Jupiter

I

I

I \

'\

'''

I I

sun I

,I

,,

'... _._ .... ' I

I

I

(and moon)

''' ... I

I I

He was able to calculate that when Earth was furthest from Jupiter (B), it took 22 min (or 1320 s)

longer than when Earth was closest (A). He concluded that this was (approximately) the time light

took to travel across the diameter of Earth's orbit around the sun.

A few years later, the Dutch mathematician and scientist Christiaan Huygens calculated the

diameter of Earth 's orbit (3.0 x 1011

m), and using Roemer 's data, calculated the speed of light as follows:

d 3 O x l0 11

v= = · m = 2.3 x 108 m/s 11! 1320 s

Interestingly , the speed they determined was so fast that most scientists of their day rejected it. It was

not accepted until after their death.

p.74


<

C2. Michelson

In 1905, the American scientist Albert Michelson used a rotating mirror to make a very accurate

measurement of the speed of light. His device is as shown below:

Light source

Rotating

8-sided

Mirror

Fixed

mirror

telescopeD

very large

distance (D)

When the mirror is rotated at the minimum frequency (about 32,000 rpm) to see continuous light:

Time for 1/8 revolution = Time for light to travel to the fixed mirror and back

Equation: Speed of light ::::; d

t

2D

Ysr

where D is the distance between the mirrors (in m)

T is the period of rotation of rotating mirror (in s)

Note:

• For ann-sided mirror, usae the formula

Speed of light ::::; d

2D

Ynr

• Today, we use lasers to calculate the speed of light through a vacuum. The currently

accepted value is c = 2.997 924 562 ± 0.000 000 011 x 108

m/s.

In Physics 30, we round this value to 3.00 x 108

m/s

p.75


Homework (Speed of light)

Both A and B

1. In a Michelson experiment, the 8-sided mirror is rotating at 166,000 rpm.

If the distance between the mirrors is 5.8 km, then what is the measured speed oflight?

A.

2.

It takes a total time of83.0 ms (milliseconds) for a microwave signal to travel to a satellite

and reflect back to the Earth, then how far is the satellite from the Earth (transmitter)?

3.

During a lightning storm, the temperature of the air is 28° C (speed of sound is 347.8 m/s).

If lightning strikes 16 km away, then compare the times for sound and light to reach you.

4.

In a Michelson-like experiment, a 6-sided mirror is rotating with a period of355 J.lS. If the

measured speed oflight was 3.2 x 108

rn/s, then fmd the distance between the mirrors (in km).

B.

5.

A radar tower sends out a signal at an angle

of 41 .0° above the horizontal. The signal

reflects off of a plane and returns to the

transmitter in a total time of 7.50 J.lS .

Determine the height of the plane.

6.

In a Michelson experiment, the two mirrors are located 4.7 km apart. If the speed oflight was

measured at 2.9 x 108

m/s, then at what frequency is the 8-sided mirror rotating?

7.

In a Michelson-like experiment, ann-sided mirror rotates at a minimum frequency

of 435.5 Hz for the light to be seen in the telescope . If the distance between the two mirrors

is 31.0 km and the measured speed of light is 2.7 x 108

rn/s, then determine the

number of sides n.

8.

A light year (ly) is the distance that light travels in one year (365 days). If the width of

our galaxy is 5.4 x 1020

m, then how many light years is this?

SOLUTION

1. T= 3.6145 X 10-4

s ; v = 2.6 X 108 rn/s

2. Time one-way: 41.5 ms ; d= 1.25 x 107

m = 1.25 x 104

km

3. Sound: t = 46 s ; Light: t = 5.3 X 10-5 s

4. n=6;D=9.5x10 3

m=9.5km

5. t1-way = 3.75 J.lS ; 1-way distance : 1125 m ; h = 1125 sin 41.0° = 738 m

6. T= 2.5931 X 10-4

s ; f= 3.9 X 103 Hz

7. T= 2.296 X 10-3

s ; n = 10 (i.e. 10-sided mirror)

8. lly = 9.4608 x 1015

m ; The galaxy is 5.7 x 104

ly wide

p.76


h

ray !Y

Light as a Wave or Particle?

Throughout history , the nature oflight has been debated. Many scientists (such as Newton)

thought light was composed of particles, while other scientists (such as Huygens) were convinced that

light was a wave .

We will now investigate the many properties of light that were well known in the times of

"classical physics", which was before the 201

century. Many of these properties gave convincing

evidence that light behaved like a wave.

D. REFLECTION OF LIGHT (EMR) IN PLANE MIRRORS

Dl. Law of Reflection (Particles and waves)

normal Angle of

Incidence

Angle of

Reflection

inciden : /'eflected

ray

**Angles are always measured with respect to the normal**

- this law is true for both particles and waves

mrrror

D2. Images in a Plane Mirror

Consider how we see the image of an object:

Object

hnage

• The image is located where the

reflected rays appear to converge

• The eye cannot tell that the rays

have been reflected

- assumes they travel in straight lines

: ............................IN·············.......... .......... ......... .......................t···· ········.... ···-· ··· ·:;:;,ft - '""... "

...---------:::- irtual Image

_.· -formed by

dotted lines

Characteristics of the hnage

• Same size as object

• Vertically upright (but laterally inverted)

• Virtual (not real)

• The image is located the same distance from the mirror as the object

p.77


Loving

ii Q).

the

Physics

mw J

ffi 2 ffFt

Homework (Plane Mirrors)

A. 1. If the angle of the incident ray is 3go with respect to the mirror surface, then what is the angle

of reflection?

2. For the mirror shown,

predict the angle of

reflection off of mirror B.

97°

B

Both A and B

3. The objects below are placed in front of plane mirrors. Roughly sketch the image.

a)

plane mirror

b)

Loving

the

Physics

plane mirror

B. 4. If light bounces off all

three mirrors, find the

angle of reflection off

of mirror C.

B

5. For the diagram shown,

fmd the angle of incidence

on mirror A.

B

SOLUTIONS (Plane Mirrors)

3. a) b)

p.78


E. REFRACTION

Refraction the change in direction of light (or any other wave) as it passes

at an angle from one medium to another

El. Index of Refraction (n)

- if the medium changes, the speed of light changes

-when light travels slowly through a medium, we say that the medium is optically dense

- to determine the optical density of a material, we use the formula

where c is the speed oflight in a vacuum (3.00 x 108

m/s)

v is the speed of light in the medium

Note:

• There is an inverse relationship between n and v

-thus, the higher then (the more optically dense), the slower the speed oflight

• For air I vacuum, n = 1

- this is the smallest value for n possible, and thus the highest possible speed of light

E2. Patterns of Refraction (Waves only- not particles)

1. Less to More Dense (Low n to High n)

Partial Reflection

ei = 8r Low n (fast)

normal

Partial Refraction

- the light ray bends

towards the normal

High n (slow)

2. More to Less Dense (High n to Low n)

Partial Reflection

ei = 8r High n (slow)

Partial Refraction

- the light ray bends

away from the normal Low n (fast)

p.79


Homework (Refraction)

1. If the index of refraction for a medium is 1.78, then find the time it takes for light

to travel through 250 krn of this medium. Answer in milliseconds.

2. If light takes 76.3 !J.S to travel through 9.0 km of a medium, then what is the index of refraction

for the medium?

3. If the index of refraction for a medium is 2.40, then find the distance that light can travel

through it in 7.35 nanoseconds.

4. Predict the direction of the refracted ray out of the glass.

a) b)

5. Predict the direction of the incident ray, if the refracted ray is given.

a) b)

SOLUTIONS (Refraction)

1. v = 1.6854 x 108 m/s ; t = 1.48 x 10-3

s = 1.48 ms

2. v= 1.1796 x 108

m/s; n=2.54

3. v= 1.25 x 108

m/s; d=0.919m

4. a) b)

R

5. a) b)

p.80


E3. Snell's Law

or, in general

sin 81 = = = .?::!_ sin 82 n1 v2 'A 2

Note: Frequency does not change (i.e. f1 = 6)

E4. Total Internal Reflection (TIR)

-recall, when light travels from a slow to fast medium (high n to low n), there are two rays:

• a reflected ray (where 8i = 8r)

• a refracted ray, which bends away from the normal (8R > 8i)

-but what happens when the angle of incidence is increased?

Critical Angle (8c)

- if we increase the angle of incidence enough, eventually the angle of refraction is 90 o

If 8R = 90°, then 8i is called a critical angle (8i = Elc)

Total Internal Reflection (TIR)

-if the angle of incidence is greater than the critical angle (i.e. 8i > 8c), then there is

no refracted ray (there is only reflection)

R

Note: Critical angle Total internal reflection

TIR cannot occur when the ray goes from a fast medium to a slow medium

Why?

-the ray bends towards the normal (8R < 8i)

- so, even if 8i is 89°, 8R will still exist

p.81


Homework (Snell's Law I Critical Angle)

A. 1. Light travels into a medium X with an angle of 25° with respect to the surface.

If the angle of refraction is 17°, then find the index of refraction for medium X.

2. Light travels from water (n = 1.34) into air, and the angle of refraction is 72.0°.

If the wavelength of the light in water is 510 nm, then find:

a) the angle of incidence

b) the wavelength of light in air

c) the frequency of light in water

3. For the diagram shown,

the ray of light will

reflect off the plane mirror

and then enter into the

medium (n = 1.77).

What would be the angle

of refraction when it enters

into the medium?

Plane

muror

4. Light travels into an air-glass interface, where the index of refraction for the glass is 1.51.

If the wavelength of light is the glass is 720 nm, fmd:

a) the critical angle

b) the frequency of the light in air

B. 5. Light with a wavelength of 660 nm travels into glass with an angle of 30.0° with respect

to the surface. If the index of refraction for glass is 1.57, then fmd:

a) the angle of refraction inside the glass

b) the wavelength of light inside the glass

c) the frequency of the light inside the glass

6. Light in medium A (n = 1.81) has a frequency of2.50 x 1014

Hz. When it enters into

medium B (n = 1.27), the angle of refraction is 56.0°. Find:

a) the wavelength of the light in medium B

b) the angle of incidence

7. For the diagram shown, the ray oflight

will reflect off the plane mirror

and then enter into the medium (n = 2.06).

If the angle of refraction

is 12.0°, then fmd e.

mirror

p.82


8. If the critical angle for a liquid-air interface is 37°, then fmd the time it takes for

light to travel through 2.1 km of the liquid.

9. The index of refraction for medium A is 1.92, while the speed oflight in medium B

is 2.50 x 108

m/s. Find the critical angle for the A-B interface.

10. A horizontal ray oflight enters a piece of glass

(n = 1.52) shaped as an equilateral triangle.

What is the angle of refraction when it leaves

the glass on the other side?

SOLUTIONS (Snell's Law)

1. B1=65o ; n2=3 .1

2. a) 45.2°

b) 683 nm

C) Vj = 2.2388 X 108

rn/s ; ji = 4.39 X 1014

Hz

3. The angle of incidence as it enters the medium is B1 = 33o ; Then, B2 = 17.9° 14

4. a) 41.5° b) A-2 = 1087.2 nm ; f2 = 2.76 x 10 Hz

5. a) 33.5°

b) 420 nm

c) Method 1: Ji = 4.55 x 1014

Hz ; f2 =Ji = 4.55 x 1014

Hz

Method 2: v2 = 1.91 x 108

rn/s ; f2 = 4.55 x 1014

Hz

6. a) fs = 2.5 X 1014

Hz ; VB= 2.3622 X 108

m/s ; As = 9.45 X 10-7

m

b) 35.6°

7. The angle of incidence as it enters the medium is B1 = 25.36° Then, B= 17.4°

8. n1 = 1.6616 ; VJ = 1.8054 X 108 rn/s ; t= 1.2 X 10-S S

9. ns = 1.20 ; Be= 38.7°

10. The angle of refraction as it enters the glass is 19.205°

The angle of incidence as it leaves the glass is 40.7951 o

The angle of refraction leaving the glass is 83.3°

p.83


>

'

F. CURVED MIRRORS I LENSES

Fl. Types of Curved Mirrors I Lenses

There are two types of curved mirrors I lenses:

1. Converging Mirrors (concave)

PA

< f+

2. Diverging Mirrors (Convex)

Converging Lenses (convex)

Diverging Lenses (Concave)

......................... ................................................................ · -----··

PA V _,.--_,/p C

< ) !-

....

< f-

Note : Spherical Aberration

if the mirror is spherical, the parallel

incident rays will not converge on

the same focus

the further the rays are away from the

principal axis, the further they go

away from the principal focus

a better focus is obtained using

parabolic mirrors

- similar with lenses

p.84


F2. Images in Curved Mirrors I Lenses

Step 1: Sketch the following reflected I refracted rays:

• Ray parallel to the PA reflects I refracts through F (or diverges from F)

- c F

-----

: ;< F

• Ray through centre reflects back I refracts on same path

..............----- ---------

...... ................................... ................... .. ················· ···········•······················ ········ ............................

F

• Ray through F reflects I refracts parallel to PA

.........................

C F

.......... ..----· ··.. . .............................. ...............................................

C F

Step 2: Locate the image

• image is always located where the reflected I refracted rays converge (intersect)

• sketch the image from the PA to the intersection

Step 3: Describe the image

• Size - larger I smaller I same size as the object

• Attitude -inverted (flipped vertically) or upright

• Type- real (formed by real rays) or virtual

p.85


do : d;

}

SUMMARY FOR CURVED MIRRORS I LENSES

1. Converging Mirrors (for lenses, C = 2F)

Image Characteristics Location of Image

Beyond C Smaller, Inverted, Real Btw F and C

OnC Same size, Inverted, Real OnC

Between C and F Larger, Inverted, Real Beyond C

OnF No image ----------

Between F and V Larger, Upright , Virtual Other side of mirror

2. Diverging Mirrors I Lenses

All locations Smaller, Upright, Virtual Other side of mirror

In general: • Real images are always inverted.

• Virtual images are always upright.

F3. Equations for Curved Mirrors I Lenses <------->· ------------->

1 1 . 1 -+-=- d o d i J

Also, R = 2f

Sign Convention

h;

Virtual } d + Real Measured from V I optical centre

f + Converging

Diverging

h , Mag + Upright

Inverted

Measured from PA

Note:

If Mag < 1, then the image is smaller than the object

If Mag= 1, then the image is the same size as the object

If Mag > 1, then the image is larger than the object

p.86


Homework (Curved Mirrors I Lenses)

Both A and B

1. For each, sketch the image, describe its characteristics , and show where the eye is located.

a) Converging mirror / lens: Object between C and F (between F and 2F)

I

-- - ----- ---------------------- - --- ------------------------------- . I

Image characteristics:

b) Diverging mirror / lens

----------------- .l ------------------------------ - --------- --

I

Image characteristics:

c) Converging mirror / lens : Object beyond C (2F)

------ --- ------- --------------- - ---------------- ,---- -------- I

I

I

------ . ----------------------------------------- 1-------- ----

-------- ---------- --- ----- -- ----- ---- --- ------- -- ,--- - -------- I

I

I

--- ------------------------------ I----------------------------- ------

Image characteristics :

d) Converging mirror / lens: Object inside focal length

----- -------- - ----------- - -· - · --- + ---------- --- - --------- -- 1 I

--------- - ----------------- ---- " I --------- --- - ------------- ---- ----

1

I

---------------------------------- ·t · ---- - ----------- ----- - --

I -------------- - ------ -- ------- -- T ------------------ ----------- -- - ·· ··

I

Image characteristics : _

p.87


2. Based on the descriptions below, identify the type of mirror I lens and the

approximate location of the object, if possible.

a) image is virtual and smaller

c) image has the same size

e) image has a magnification of -0.25

b) image is bigger and inverted

d) image is virtual and the magnification is 3

f) there is no image

For the rest of the questions, be certain to interpret all signs.

A. 3. A 19 em high object is located 24 em in front of a concave (converging) mirror.

If the mirror's radius of curvature is 36 em, then fmd:

a) the distance to the image b) the height ofthe image

4. An 8.0 em high object is placed in front of a convex (diverging) mirror. If the virtual image

is 3.0 em high and is located 5.0 em from the mirror, then find:

a) the distance to the object b) the mirror's radius of curvature

5. A convex (diverging) mirror has a focal length of 17 em. If an object is placed 30 em in front

of the mirror, then what is the magnification of the image?

6. An 11 em high object is placed 7.0 em in front of a concave (converging) mirror. If the virtual

image has a height of 55 em, then what is the mirror's radius of curvature?

7. A convex (diverging) mirror has a radius of curvature of 42 em. If a 72 em object

is placed in front ofthe mirror and the resulting image is located 16 em from the mirror,

then what is the height of the image?

B. 8. An object is placed 21 em in front of a concave (converging) mirror. If the image is real and

is 4.0 times larger than the object, then find:

a) the distance to the image b) the mirror's focal length

9. A converging (convex) lens has a radius of curvature of 12 em. If an object is placed 15 em in

front of the lens, then what is the magnification of the image?

10. An object is placed 17 em in front of a concave (diverging) lens. If the image is 20% of

the size of the object, then what is the lens's radius of curvature?

11. A converging (convex) lens has a radius of curvature of 40 em. If the virtual image produced

is 18 em tall and 50 em from the lens, than fmd the height ofthe object.

12. A 30 em high object is located in front of a concave (diverging) lens. Ifthe height ofthe

image is 11 em and it is located 5.0 em from the lens, then what is the lens's focal length?

*13. A very large compound microscope is composed of two converging lenses that are 1.65 m

apart. The focal length of the first lens is 20 em, while the focal length for the second lens is

15 em. If the 10 em object is placed 23 em in front ofthe first lens, then determine the total

magnification. Hint: The image for the first lens becomes the object for the second lens

p.88


SOLUTIONS (Curved Mirrors)

1.

Image characteristics: Larger, real, inverted

Image characteristics: Smaller, virtual, upright

c)

Image characteristics : Smaller, Real, Inverted


p.89

.

d)

: - ;;;:-;:_:_: =-

'... -=- ....

-- ")...(_:_:: ----- - ' .

.

Image characteristics: Larger, Virtual, Upright

2. a) Diverging mirror I lens ; Object can be anywhere

b) Converging mirror I lens ; Object is between C and F (between 2F and F)

c) Converging mirror I lens ; Object is on C (on 2F)

d) Converging mirror I lens ; Object is between V and F (between 0 and F)

e) Converging mirror I lens ; Object is beyond C (beyond 2F)

f) Converging mirror I lens ; Object is on F

3. f= 18 em (real) a) d; = 72 em (real) b) h; =-57 em (inverted)

4. d; = -5.0 em (virtual) a) do= 13 em (real) b) f= -8.0 em ; R = -16 em (virtual)

5. f= -17 em (virtual) ; d; = -10.85 em (virtual) Mag=0.36 (upright, smaller)

6. d; = -35 em (virtual) ; f= 8.75 em , R = 18 em (real)

7. f=-2lcm (virtual); d;= -16cm (virtual); do = 67 .2cm (real); h;=l7cm (upright)

8. Mag= -4.0 (inverted) a) d; = 84 em (real) b) f= 17 em (real)

9. f= 6 em (real) ; d; = 10 em (real); Mag= -0.67 (smaller, inverted)

10. d; = -3.4 em (virtual) ; f= -4.25 em (virtual) ; R = -8.5 em (virtual)

11. f= 20 em (real) ; do= 14.2857 em ; h0 = 5.1 em (upright)

12. d; = -5.0 em (virtual) ; do= 13.636 em (real) ; f= -7.9 em (virtual)

13. Lens 1: d; = 153.33 em ; h; = -66.667 em

Lens 2: d; = -52.5 em ; h; = 300 em Mag : 30x


p.90

G. Diffraction , Interference, and Polarization of Light (EMR)

Gl. Diffraction (Waves only- not particles)

When waves go through an opening or around a comer, they bend . This bending is called

diffraction. The longer the wavelength of the incident waves , the more bending takes place .

Around comers: short A long A

---;1- less bending more bending

Through openings: The smaller the opening, the greater the diffraction

Wide opening: Narrow opening: Narrow opening, shorter A

Nearly

straight

Significant

Diffraction

Less

Diffraction

G2. Interference (Waves only- not particles)

Waves can pass right through each other. However , when they occupy the same position

(superposed), they interfere with each other.

Principle of Superposition

- the resultant displacement of a particle is equal to the sum of its separate displacements

There are two types of interference:

1. Constructive Interference (both+, both -)

- add to a higher amplitude (brighter)

light wave

2. Destructive Interference (one+, one-)

- produce a smaller amplitude

- if equal, they cancel each other

u (\ + no wave


p.91

X

G3. Double-Slit Interference (Young's Double-Slit Experiment) (Waves only- not particles)

When a monochromatic wave (i.e. a wave undergoing SHM; a wave of one frequency) passes

through two adjacent, narrow slits, there is diffraction and interference on the opposite side. When a

screen is placed beyond the barrier, we observe bright dots I lines on the screen.

Constructive Interference

2 crests 2 troughs

I"\

' '

Destructive Interference (dim)

crest and trough

ent Jr---------------------------·

front ---------------------------- ---------------------------· trough

crest

Maxima - This is a bright dot I line. The difference in the distances from the two slits is a multiple

of the wavelength, so the waves arrive in phase and undergo constructive interference.

Minima - This is a dark region. The difference in the distances from the two slits is a multiple

of0.5 'A, so the waves arrive out of phase and undergo destructive interference.

Mathematics ofYoung's Double-Slit

T d

1

Equations:

L

-------·-····

where tan () = -

Min 1 (n = 0.5)

Max 1 (n = 1)

Min2 (n = 1.5)

L Max2 (n = 2)

A= dx nL

if () < 10°


p.92

G4. Dispersion of Light (EMR) using Prisms and Diffraction Grating

When light is shone through a triangular prism or diffraction grating, the different wavelengths of

light bend by different amounts. This causes the light to separate into its different wavelengths, which

is called dispersion.

Dispersion using a Triangular Prism (Waves only)

When light passes through a triangular prism, the

different wavelengths of light refract by different amounts.

The shorter the wavelength,

the greater the angle of refraction. R 0 y

When the light disperses into its component G

wavelengths, all wavelengths are observed with no gaps in B between them. This is called a continuous spectrum. I

v Dispersion using Diffraction Grating (Waves only)

Diffraction grating is a transparent glass or plastic that has many fme wires within it, all parallel to

each other and a very small distance (in j..Lm) apart from each other. This results in many small slits, all

side by side.

When light passes through the diffraction grating, the light behaves much like a double-slit

experiment. Each wavelength of light will have maxima and minima shown on the screen. The location

of the maxima are given by the equations

A.= d sine

n

and A.= dx nL

Based on the first equation, it follows that:

The shorter the wavelength, the smaller the angle of deviation

The dispersion pattern (spectrum) for diffraction grating is as shown below:

llight

diffraction

ROYGBIV ROYGBIV ROYGBIV VIBGYOR VIBGYOR VIBGYOR

In reality, these

would overlap

Note: If the light is monochromatic (i.e. one wavelength of light), then you would see only that

wavelength on the screen. This would give a pattern like Young's double-slit experiment.


p.93

G5. Polarization of Light (EMR) (Waves only)

When a wave travels through a filter that allows vibrations only in one plane, then the resulting

wave that is produced will vibrate in that plane only. We call this light polarized.

For example, imagine a rope that is moved up and down and then it is moved side-to-side. If this

rope passes through a vertical slit, then only the vertical waves will pass through; the side-to-side

waves would be absorbed or reflected. If these vertically polarized waves were then passed through a

second slit that is horizontal, then the waves cannot pass through at all. They would be completely

absorbed or reflected and the transmission would almost be entirely stopped.

Vertical

slit

Horizontal

slit

Light waves radiate in all directions, and this light is referred to as unpolarized. Much like the

rope, when it goes through a horizontal polarizing filter, it is plane-polarized (specifically, in this case,

it is horizontally polarized), which means it oscillates entirely on one plane. If it then goes through a

vertical polarizing filter ("cross-polarized "), then the light energy is almost completely absorbed.

unpolarized light plane polarized

light

near zero intensity

horizontal

polarizer

vertical

polarizer

Light can also be partially plane-polarized when it reflects off of a surface or refracts into a new

medium.

Applications of Polarization

• Polarizing filters are used in cameras and sunglasses to remove some of the light that

reflects off of surfaces or refracts (scatters) through the atmosphere, which has been

partially plane-polarized. This reduces the glare.

• LCD displays (in digital watches and calculators) have polarizing film within them.

Millions of microscopic crystals float between electrodes. When the electrodes are charged,

the electric field created causes the crystals to line up "cross-polarized" with the film.

This results in dark areas (since the light cannot exit), which is seen as numbers and text.

• Many insects (such as bees) use polarized sky light for navigation, since it is always

perpendicular to the direction of the sun.

• Radio waves transmitted from towers are typically polarized (defined by the orientation of the

electric field). AM and FM are vertically polarized, while television is horizontally polarized.


p.94

= -- ;

Homework (Young's Double Slit Experiment)

A. 1. Monochromatic EMR is shone through diffraction grating (distance between the slits is

2.50 !liD). If the angle to the 2nd-order maximum is 11.0°, then determine the EMR frequency.

2. 620 nm EMR is shone through diffraction grating (distance between the slits is 9.9 !liD). If the

distance from the grating 80 em, determine the distance between maxima. Assume B < 10°.

3. 470 nm EMR is shone through diffraction grating that is 36 em from the screen and the

distance between the maxima is 11 em. Determine the distance between the slits.

Both A andB

4. Which type of EMR (AM or FM) would have better reception on the other side of a hill?

5. Two microwave transmitters are placed side by side, each emitting EMR in phase with a

frequency of 7.5 GHz. If a receiver is placed 12 em from one transmitter and 22 em from the

other transmitter, would there be a strong signal? Explain.

6. Unpolarized light is sent through a vertical polarizing filter and it then travels to a second

polarizing filter. Describe the light you would see as the second filter is rotated .

B. 7. Monochromatic, 5.90 x 1014

Hz EMR is shone through diffraction grating which is rated

at 200,000 slits/m. Determine the angle to the third order maximum.

8. For a Young's double-slit experiment, the diffraction grating has 800 slits per mm. When

monochromatic light with a wavelength of 600 nm is shone through the grating, the distance

between the maxima is 7.00 em. What is the distance (in em) from the grating to the screen?

9. A Physics 30 student manipulates the distance between the slits and measures the resulting

distance between maxima . For this relationship , determine the equation for the straightened

curve, determine the significance of the slope, and provide the units for the slope.

SOLUTIONS

1. IL = 2.3851 x 10-7

m ; f= 1.26 x 1015

Hz

2. n = 1 ; x = 5.0 em

3. B = 16.99° ; n = 1 ; d= 1.6 x 10-6

m = 1.6!-lm

4. In order to have good reception, the waves must be able to diffract significantly around the hill.

Since AM has a longer wavelength, it refracts more. So, AM would have better reception .

5. A= 4.0 em ; The difference in distances between the receivers is 10 em. Since this distance is

a multiple of 0.5 A , the waves will arrive out of phase . This results in destructive interference,

which means there will be no signal.

6. The light that leaves the first filter will be vertically polarized . Thus, when the second filter is

oriented vertically , the light will go through and be bright. However, as it is rotated towards the

horizontal , less light will make it and it will go dimmer. When the second filter is horizontal, the

light is cross-polarized and no light will make it through .

7. d = 5.0 x 10-6 m ; A= 5.0847 x 10-7

m ; B= 17.8°

8. d=l.25 x 10-6

m; n=1 ; 0=28.7° ; L = 12.8cm x

slope 2

9. Equation: x Slope = IL L ; Units for slope: m d

1

d

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