midterm 3, overview
This overview highlights the key points of the material.
It is not necessarily complete; you should judge what
additional material should be on your equation sheet.
PHY232
Remco Zegers
zegers@nscl.msu.edu
Room W109 – cyclotron building
http://www.nscl.msu.edu/~zegers/phy232.html
light as waves (22)
in vacuum, the speed of light is 3x108 m/s
in any other substance: vlight=c/n with n the index of
refraction of the material
nair = 1.003 (nearly like vacuum)
v=f with : wavelength, f: frequency
I=Psource/(4R2) I: intensity (W/m2) Psource :power of source. R
distance from source
PHY232 - Remco Zegers
-
2
example
Light with a wavelength of 600 nm (in water) travels from
water (n=1.33) to glass (n=1.5). What are the wavelength,
frequency and speed of light in the water and in the
glass?
In water:
v=c/n=3x108/1.33=2.26x108 m/s
=600x10-9 m (given)
f=v/=3.77x1014 Hz
In glass:
v=c/n=3x108/1.5=2x108 m/s
f: remains unchanged!! = 3.77x1014 Hz
=v/f=2x108/3.77x1014=5.30x10-7 m
PHY232 - Remco Zegers
-
3
reflection/refraction (22)
 reflection:1=r
 refraction:
1 r
 Snell’s law: n1sin1=n2sin2
n1
 two cases:
n2
 1) n1< n2: 1> 2 and no total
internal reflection (but some
reflection can happen)
 2) n1>n2: 1< 2 but
if sin1>n2/n1 total internal
reflection will take place and
=sin-1(n2/n1) is called the
n1
critical angle
n1<n2
2
1 r
n1>n2
n2
2
PHY232 - Remco Zegers
-
4
example
 light travels from glass (n=1.5) to air (n=1) under an angle
of 400. What is the angle of refraction? is there an incident
angle for which total internal reflection will take place? If
so, what is the critical angle?
 Snell’s law: n1sin1=n2sin2 n1=1.5 n2=1 1=400 so 2=74.60
 since n1>n2 total internal reflection can take place and the critical
angle is sin-1(n2/n1)=41.80
PHY232 - Remco Zegers
-
5
mirrors and lenses: general (23)




p: object distance (we always choose this +)
q: image distance (can be + or -)
f: focal length (can be + or -)
lens/mirror equation: 1/p+1/q=1/f note: use of signs is
different for mirrors or lenses
 virtual image: light is NOT actually passing through the
image
 magnification M=-q/p=(size of image)/(size of object)
 negative if inverted (upside-down relative to object)
 |M| < 1 : image is smaller than object
 |M| > 1 : image is larger than object
PHY232 - Remco Zegers
-
6
Mirrors: an overview (23)
type
image
image
direction
M
q
f
concave p>f
real
inverted
|M|>0 M -
concave p<f
virtual
not
inverted
|M|>1 M +
+
-
+
+
convex
p>|f|
virtual
not
inverted
|M|<1 M +
-
-
convex
p<|f|
virtual
not
inverted
|M|<1 M +
-
-




p?
mirror equation 1/p + 1/q = 1/f
f=R/2 where R is the radius of the mirror
magnification: M=-q/p
q negative means: image is on other side of mirror than objects
PHY232 - Remco Zegers
-
7
example
 an object is placed at a distance of two times the focal
length in front of a convex mirror. If the focal length is –5
cm, what is the magnification?





1/p+1/q=1/f
f=-5 cm, p=10 cm, so 1/q=1/(-5)-1/10=-0.3
q=-3.33 cm
M=-q/p=-(-3.33)/10.=0.33
The image is virtual (q is negative, I.e. on the other side of
the mirror as the object), upright (M+) and demagnified
(|M|<1).
PHY232 - Remco Zegers
-
8
lenses, an overview (23)
type
p?
image
image
direction
M
q
f
converging
p>f
real
inverted
|M|>0 M -
converging
p<f
virtual
not
inverted
|M|>1 M +
+
-
+
+
diverging
p>|f|
virtual
not
inverted
|M|<1 M +
-
-
diverging
p<|f|
virtual
not
inverted
|M|<1 M +
-
-




mirror equation 1/p + 1/q = 1/f
magnification: M=-q/p
different meaning
lens makers equation: 1/f=(n-1)(1/R1-1/R2) than for mirrors
q negative: on same side of the lens as object
PHY232 - Remco Zegers
-
9
example
 an object is placed in front of a lens. An image is created
that is situated on the same side of the lens as the object
and that is larger than the object. Is the lens converging or
diverging?
 the image must be virtual (same side as object). This still
allows both types of lenses. But a diverging lens must have
|M|<1 (I.e. image smaller than the object) so that the lens
must be converging.
PHY232 - Remco Zegers
-
10
interference (24)
 constructive interference: path length difference between
two sources: m with m=0,1,2…
 destructive interference: path length difference between
two sources: (m+1/2) with m=0,1,2…
destructive
constructive
PHY232 - Remco Zegers
-
11
double-slit experiment
L
constructive: dsin=m
destructive: dsin=(m+1/2)
m=0,1,2,3…
constructive: bright fringe/maxima
destructive: dark fringe/minima
 if  small: =sin=tan : radians!!
 constructive: ym=mL/d
 destructive: ym=(m+1/2)L/d
 if  not small (but can also be used if  is small):
 constructive: dsin=m
& tan=y/L
 destructive: dsin=(m+1/2) & tan=y/L
PHY232 - Remco Zegers
-
follows
from
12
example
 the distance between the central maximum and first
minimum in the interference pattern created by a double
slit system is 1 cm. If the distance between the slits and the
screen is 10 cm, what is the slit distance d? Given, =550
nm
 central maximum: y=0
 first minimum: dsin=1/2  m=0
 tan=y/L=1/10 so =5.71o
 d=0.5 x 550x10-9/sin(5.71)=2.76x10-6 m (2.76 micrometer)
PHY232 - Remco Zegers
-
13
reflection & interference(24)
1
1
2
n1<n2
1/2 phase change
2
n1>n2
no phase change
 remember that the wavelength of light in a medium with
index of refraction n is is vacuum/n
PHY232 - Remco Zegers
-
14
na
1
thin layers
2
nb
nc
more general
constructive:
even number of phase
shifts: 2d=mb
odd number of
phaseshifts: 2d=(m+1/2)b
destructive:
even number of phase
shifts: 2d=(m+1/2)b
odd number of
phaseshifts: 2d=mb
 constructive interference between ray 1
and 2: a procedure:
 if no phase shifts occur at boundaries a-b
and b-c (na>nb>nc): 2d=mb
 if phase shift occurs at boundary a-b only
(na<nb nb>nc): 2d=(m+1/2)b
 if phase shift occurs at boundaries a-b and
b-c (na<nb nb>nc): 2d=mb
 destructive interference between ray 1 and
2: a procedure:
 if no phase shifts occur at boundaries a-b
and b-c (na>nb>nc): 2d=(m+1/2)b
 if phase shift occurs at boundary a-b only
(na<nb nb>nc): 2d=mb
 if phase shift occurs at boundaries a-b and
b-c (na<nb nb>nc): 2d=(m+1/2)b
PHY232 - Remco Zegers
-
15
example
 at top boundary: no
phase shift since na>nb
 a lower boundary: no
phase shift since nb>nc
na
 so 2d=mb for constructive
nc
interference (even number
of phase shift)
PHY232 - Remco Zegers
1
2
nb
given na=1.5 nb=1.3 nc=1,
for which d will constructive
interference occur?
-
16
single-slit diffraction(24)
diffraction due to a single slit of width a:
• minima (destructive interference) asin=m, m=1,2,3…
• maxima (constructive interference)
asin=(m+1/2), m=1,2,3… and =0
PHY232 - Remco Zegers
-
17
example
 what is the smallest slit width that will produce a diffraction
interference pattern for light with a wavelength of 500 nm?
 an interference pattern is made if there is at least one
minimum (besides the central maximum). for this
minimum <900 so sin<1
 asin=m for minima, m=1 so a>
PHY232 - Remco Zegers
-
18
diffraction grating
constructive interference dsin=m, m=0,1,2,3…
destructive interference dsin=(m+1/2), m=0,1,2,3…
instead of giving d, one usually gives the number of slits per
unit distance: e.g. 300 lines/mm d=1/(300 lines/mm)=0.0033 mm
PHY232 - Remco Zegers
-
19
example
 a diffraction grating has 2000 lines per cm. what is the
difference in angle between the 1st order maxima for light
with a wavelength of 500 nm and 600 nm?





1st order maximum: m=1
d=0.01/2000=5x10-6 m
for 500 nm: sin=/d so =sin-1(500x10-9/5x10-6)=5.70
for 600 nm: sin=/d so =sin-1(600x10-9/5x10-6)=6.90
difference: 6.8-5.7=1.10
PHY232 - Remco Zegers
-
20
polarization (24)
 if unpolarized light with intensity I0 falls on a linear polarizer
with polarization axis , then the intensity after the
polarizer is I0/2
 if polarized light with intensity I1 and polarization axis 1
(relative to a fixed axis, usually the horizontal or vertical)
falls on a linear polarizer with polarization axis 
(measured relative to the same axis), then the intensity of
the light after this polarizer (I2) is:
I2=I1cos2(|1-|)
 light reflected from a surface is completely polarized if:
=tan-1(n2/n1) (the brewster angle)
PHY232 - Remco Zegers
-
21
example
first polarizer: I1=I0/2=970/2=485 W/m2
for every next polarizer: In=In-1cos2100=In-10.96985
This is done 9 times, so I10=I1(0.96985)9=0.759I1
so I10=0.759*485=368 W/m2
PHY232 - Remco Zegers
-
22
optical instruments (25)
 power of a lens P = 1/f (unit: diopters (1/m))
 farpoint: the largest distance at which objects can clearly be seen by
the eye. It is `infinity` for a good eye
 to solve nearsightedness: the lens should be made such that the
image of an object (p) at infinity is projected at the far-point (FP)of
that person: 1/ + 1/(-FP) = 1/flens
 f will be negative (diverging lens), note: take q negative
 nearpoint: the smallest distance at which objects can clearly be seen
by the eye. It is about 25 cm for a good eye
 to solve farsightedness: the lens should be made such that the
image of an object (p) located at the desired nearpoint (I.e. ~25
cm or what is given in the problem) is located at the near point of
the person: 1/NPdesired+1/(-NPperson)=1/flens note: take q negative
 f will be positive (converging lens)
PHY232 - Remco Zegers
-
23
example
 a person has a far point of 2 m and wears corrective
lenses to solve this defect. Using these lenses, where will
the image of an object located at the nearpoint (25 cm)
be located?
 to solve the nearsightedness, the lenses must have a focal
length: 1/ + 1/(-FP) = 1/flens
 so: flens=-FP=-2 m.
 the image of an object at the nearpoint will the be
located at:
 1/(-25 cm)+1/q=1/(-200 cm) solve for q: -22.2 cm
PHY232 - Remco Zegers
-
24
simple magnifier (25)
 angular magnification m=I/0
 m=I/0=(h/p)/(h/NP)=NP/p
with p<f
 best result if p=f: m=NP/f
PHY232 - Remco Zegers
-
25
microscope (25)





magnifying power: m=-NP L/(fOfe)
fo: focal length of objective lens
fe: focal length of eye piece
NP: near point (25 cm or what is given)
L: distance between the two lenses
PHY232 - Remco Zegers
-
26
telescope (25)




magnifying power m= e/o=fo/fe note L=f0+fe
fo: focal length of objective lens
fe: focal length of eye piece
L: distance between the two lenses
PHY232 - Remco Zegers
-
27
rayleigh’s criterion (25)
 two images are just resolved if rayleigh’s criterion is fulfilled.
 Rayleigh’s criterion: the central maximum of image A false into the
first minimum of image B
 first diffraction minimum: sin=/a with A the slitwidth
 images separated by a minimum angle min=/a can just be resolved
 if the aperture is circular with diameter D: min=1.22/D
PHY232 - Remco Zegers
-
28
example
 the diameter of an aperture used for a telescope is 1 m. If
two stars are 400x1015 m away from earth, what is the
smallest separation between these two stars for which
they can still be distinguished by the telescope? Assume
=500 nm
 min=1.22/D=1.22 x 500x10-9/1=6.1x10-7
 tan(min)= min=distance/400x1015 m
 distance=2.44x1011 m
PHY232 - Remco Zegers
-
29
relativity (26)
v

c

1
1  2
Time dilation
tp: proper time is the time interval between two events as measured by
an observer who sees the two events occur at the same positions (the
observer is in the same frame of reference as the clock)
Lorentz contraction: L=Lp/=Lp(1-v2/c2)
Lp: proper length is the length of an object as measured by a person
who is at rest relative to that object (the person is in the same frame of
reference as the object)
PHY232 - Remco Zegers
-
30
example
 you are traveling in a space ship at a speed of 0.9 times
the speed of light relative to MSU campus along the
direction of Shaw Lane. If Shaw lane is 2 miles long as
viewed by a person living on campus, how long is it as
viewed from your space ship?
 proper length: as seen by the persn living on campus.
 L=Lp/=Lp(1-v2/c2)
 =2.3 so L=2 miles / 2.3=0.87 miles.
PHY232 - Remco Zegers
-
31
Relativistic addition of velocities (26)
 frame d is moving in the +x direction relative to frame b
with a velocity vdb. The velocity of an object a is
measured in frame d to be vad. Then the above equation
gives the velocity vab of a in the frame b.
 Note that if vad and vdb are small, vab=vad+vdb which is the
common equation for relative motion.
PHY232 - Remco Zegers
-
32
example
 a spaceship is moving away from earth at 0.5 times the
speed of light. It sends a message towards earth at the
speed of light (aimed directly at earth). What is the speed
of the message as seen from a person on earth?
 vdb=+0.5c (speed of ship relative to earth)
 vad=-1.0c (speed of message relative to ship)
 vab=(-1.0c+0.5c)/(1+(-1.0cx0.5c)/c2)=-1c
PHY232 - Remco Zegers
-
33
relativistic energy and momentum (26)
•relativistic momentum:
•energy-mass equivalence (rest mass):
•kinetic energy
•total energy:
•kinetic energy of a particle with charge |q| accelerated
over a voltage V: qV
note: 1eV=1.6x10-19 J=the amount of energy gained by a
unit charge when accelerated over 1V.
PHY232 - Remco Zegers
-
34
example
 a rocket with a mass of 10 kg is traveling at half the speed
of light? Which is greater, its kinetic energy or its rest
energy?
 rest energy: E=mc2 = 10*(3x108)2=9x1017 J




kinetic energy:
v

=1.15, so
c
2
2
Ekin=(1.15-1)mc =0.15mc
this is only 15% of the rest energy.
PHY232 - Remco Zegers
-

1
1  2
35