Midterm 3, overview PHY232 Remco Zegers [email protected] Room W109 – cyclotron building...

midterm 3, overview

PHY232Remco [email protected] W109 – cyclotron buildinghttp://www.nscl.msu.edu/~zegers/phy232.html

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 - 2

light as waves (22)

in vacuum, the speed of light is 3x108 m/sin any other substance: vlight=c/n with n the index of refraction of the materialnair = 1.003 (nearly like vacuum)v=f with : wavelength, f: frequencyI=Psource/(4R2) I: intensity (W/m2) Psource :power of source. R distance from source


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 HzIn glass:v=c/n=3x108/1.5=2x108 m/sf: remains unchanged!! = 3.77x1014 Hz=v/f=2x108/3.77x1014=5.30x10-7 m


reflection/refraction (22)

reflection:1=r

refraction: Snell’s law: n1sin1=n2sin2

two cases: 1) n1< n2: 1> 2 and no

total internal reflection (but some reflection can happen)

2) n1>n2: 1< 2 but if sin1>n2/n1 total internal reflection will take place and =sin-1(n2/n1) is called the critical angle

1 r

2

n1

n2

1 r

2

n1

n2

n1<n2

n1>n2


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: n1sin1=n2sin2 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


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


Mirrors: an overview (23)

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

type p? 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 + - -


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).


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 lens makers equation: 1/f=(n-1)(1/R1-1/R2) q negative: on same side of the lens as object

different meaningthan for mirrors


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.


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…

constructive destructive


double-slit experiment

if small: =sin=tan : radians!! constructive: ym=mL/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

L

constructive: dsin=mdestructive: dsin=(m+1/2)m=0,1,2,3…constructive: bright fringe/maximadestructive: dark fringe/minima

follows from


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)


reflection & interference(24)

remember that the wavelength of light in a medium with index of refraction n is is vacuum/n

1 2

n1<n2

1/2 phase change

n1>n2

1 2

no phase change


thin layers 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=mb

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=mb

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=mb

if phase shift occurs at boundaries a-b and b-c (na<nb nb>nc): 2d=(m+1/2)b

na

nc

nb

1 2

more generalconstructive:

even number of phase shifts: 2d=mb

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=mb


example

at top boundary: no phase shift since na>nb

a lower boundary: no phase shift since nb>nc

so 2d=mb for constructive interference (even number of phase shift)

na

nc

nb

1 2

given na=1.5 nb=1.3 nc=1,for which d will constructiveinterference occur?


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


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>


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 perunit distance: e.g. 300 lines/mm d=1/(300 lines/mm)=0.0033

mm


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


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)


example

first polarizer: I1=I0/2=970/2=485 W/m2

for every next polarizer: In=In-1cos2100=In-10.96985This is done 9 times, so I10=I1(0.96985)9=0.759I1

so I10=0.759*485=368 W/m2


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)


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


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


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


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


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

rayleigh’s criterion (25)


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


relativity (26)

21

1

c

v

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)


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.


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.


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


•relativistic momentum:

•energy-mass equivalence (rest mass):

•kinetic energy

•total energy:

•kinetic energy of a particle with charge |q| acceleratedover a voltage V: qV

note: 1eV=1.6x10-19 J=the amount of energy gained by a unit charge when accelerated over 1V.

relativistic energy and momentum (26)


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: =1.15, so Ekin=(1.15-1)mc2=0.15mc2

this is only 15% of the rest energy.

21

1

c

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Midterm 3, overview PHY232 Remco Zegers [email protected] Room W109 – cyclotron building...

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