CH28 interference and diffraction
2 waves out of phase by 90 degrees
1/4 lamda (wavelength)
light with a wavelength of 511 nm forms a diffraction pattern after passing through a single slit of width 2.20x10^-6 m. find the angle associated with (a) the first and (b) the second fringe above the central bright fringe
1st DF @=sin^-1(mλ/W) =sin^-1((1)(511x10^-9m)/(2.2x10^-6 m) = 13.4 degrees
light consists of particles, the final screen should show _________ stripes, one corresponding to each slit
2 thin
width of central max=
2@=2λ/W
a beam consisting of red light (λvacuum=662nm) and blue light (λvacuum=465nm) is directed at right angles onto a thin soap film. If the film has an index of refraction n= 1.33 and is suspended in air (n=1) find the smallest non zero thickness for which it appears red in reflected light
2nt/λv = m , m=0,1,2
condition for constructive interference
2nt/λvacuum - 1/2 = m -- longer λ interfere destructively
the condition for destructive interference
2nt/λvacuum = m - m=0,1,2
an air wedge is formed by placing a human hair between 2 glasses on one end, and allowing them to touch on other end. When this edge is illuminated with red light (λ=771nm), it is observed to have 179 bright fringes. how thick is the hair?
2t/λ = m-1/2 t= λ/2(m-1/2) = 771x10^-9/ 2(179-0.5) = 68.8um
the linear distance separating the headlights of a car is 1.1 m. Assuming light of 460 nm, a pupil diameter of 5.0 nm and an average index of refraction for the eye of `.36, find the max distance at which the headlights can be distinguished as two separate sources of light
@min=1.22λn/D = 1.22(460 nm/1.36)/(5mm)) = 8.3x10^-5 rad L=y/tan@min=13,00 m
the spacing between crystal planes is close enough to the wavelength of the x rays (0.1541 nm for Cu K-a) to allow diffraction patterns to be seen
Bragg's law 2dsin@=λ
is the point where the glass plates touch in an air wedge (a) a dark fringe or (b) a bright fringe?
Dark fringe . 1/2 + 2λ = m +1/2 m = 0,1,2 Since d=0, above eqn satisfied for m=0 therefore, first fringe is dark
this pattern is due to the difference in the path length of rays from different parts of the opening. According to __________________ each point within the slit is a source for new waves
Huygen's principle
Ray 1: effective path length λ/2 Phase change:
Phase change: (divided by λ)
constructive interference
The interference that occurs when two waves combine to make a wave with a larger amplitude - an integral number of wavelengths
destructive interference
The interference that occurs when two waves combine to make a wave with a smaller amplitude - where they differ by a half-odd integral number of wavelengths
the second fringe occurs when:
W/4sin@ = λ/2 Wsin@ = 2λ
single slit diffraction
a diffraction pattern consisting of bright and dark fringes: central bight fringe is surrounded by alternating dark and bright fringes
there is a ____________________ when light reflects from a region with a higher index of refraction, or from a solid surface
a half wavelength phase change
diffraction gratings
a system with a large number of slits is called a diffraction grating. the pattern consists of a set of principal maxima and secondary maxima -- as the number of slits grows, the peaks become narrower and more intense
diffraction
a waving passing through a small opening will diffract. this means that, after opening, there are waves traveling in directions other than the direction of the original wave.
if 2 waves occupy the same space, their __________ add at each point. they may interfere either __________________ or ________________
amplitudes, constructively, destructively
the fringes move closer since λn of light in water is λ/n smaller
because of this, @ is smaller and fringes are closer
blue lasers encode more information
blue ray: λ=405 DVD: λ=650 - the smaller the λ, the smaller the bumps and therefore the DVD can have more information
an interference pattern is seen from 2 slits, now cover one slit with glass, introducing a phase difference of 180 degrees (1/2 wavelength) at the slits. How is this pattern altered?
bright and dark spots interchanged
reflected waces can interfere due to path length differences ___________
bu they can also interfere due to phase changes upon reflection
diffraction through a small __________________ (eye pupil) results in a circular pattern of fringes. this limits our ability to distinguish one object from another when they are very close
circular aperture
1/2 + 2d/λ = m m = 1,2,3
constructive interference
l1 - l2 = mλ
constructive interference m = 0,1,2
the light on the screen has alternating light and dark fringes, ______________ to constructive and destructive interference
corresponding
X ray diffraction is used to determine ____________ structure
crystal
1/2 + 2d/λ = m + 1/2 m=0,1,2
destructive interference
l2 - l1 = (m-1/2)λ
destructive interference (m=1,2,...)
the dark fringes are between the bright fringes, the condition for fringes is
dsin@ = (m-1/2)λ -- m =1,2,3..... above central dsin@ = (m+1/2) --- m = -1,-2,-3..... below central
condition for bright fringes (constructive interference) is:
dsin@ = mλ m = -2,-1,0,1,2
diffraction fringes can be observed by holding your ___________ and ____________ very close together (it helps not to be too farsighted)
finger and thumb
lower index of refraction to higher index of refraction
half wavelength phase change
A wave pulse that reflects from a fixed end
has the same shape but is inverted on the way back
Huygens's principles
if light is a wave, each slit servers as a new source of "wavelets" as show, and the final screen will show the effects of interference.
Rayleigh's criterion of resolution:
if the firs dark fringe of one circular diffraction pattern passes through the center of a second diffraction pattern, the 2 sources responsible for the patterns will appear to be a single source 2 objects can be seen as separate only if their angular separation @>@min
dsin@ = mλ
if λ is increased and d does not change, then @ must increase so the pattern spreads out
2 light sources emit waves of λ = 1 m which are in phase. the 2 waves from these sources meet at a distant point. wave 1 traveled 2 m to reach the point and wave 2 traveled 3 m. when the waves meet they are _________________ of phase
in phase since λ = 1 m, wave 1 has traveled twice this wavelength while wave 2 has traveled three times this wavelength. thus, their phase difference is one full wavelength which means they are still in phasse
Intereference will be constructive where the 2 waves are ____ phase, and desctructive where they are _____ phase.
in, out
CDs and DVDs depend on ________________ for their functioning. the signal is encoded as tiny bumps in the surface of a smooth reflecting surface. the DVD uses a ____________ to read the stored info. the decoding is based on ______________; the reflected laser beam varies in intensity depending on whether it is reflecting from a bump or not. path difference = 2h=λ/2, h=λ/4
interference, red laser, interference
a wave pulse reflecting from an end that is free to move
is not inverted
if the width of the slit through light pass is reduced, does the central bright fringe (a) become wider (b) become narrower or (c) remain the same
it becomes wider w decreases --> 2@ increase width of central max = 2@=2λ/W
blue has a ______________ λ interfere destructively
longer
if t increases m will increase
m = 2t/λ + 1/2 = 2(80x10^-6)/771x10^-9 +0.5 = 208
interference is only noticable if the light sources are _____________________ (so all the light has the same wavelength) and _____________ ( different sources maintain the same phase relationship over space and time)
monochromatic, coherent
higher index of refraction to lower index of refraction
no phase change
there is _______________ when light reflects from a region with a lower index of refraction
no phase change
resolving two point sources
on the left, the angular separation of sources not large enough, diffraction patterns overlap and it appears to be a single source; on the right, 2 sources can be clearly resolved because angular separation is larger
Young's 2 slit experiment
original light sources need to be coherent; it becomes so after passing thru the very narrow slits S1 and S2
Path difference between slits
path difference = d sin0
Ray 2: effective path length length 2t
phase change: 2t/λn = 2tn/λ (since λn=λ/n)
red has a ______________ λ interfere destructively
shorter
the location of the first dark fringe determines the size of the central spot
sin@ = 1.22λ/D for small @, sin @~ @=1.22λ/D D= diameter of aperature
when 546 nm light passes through a particular diffraction grating, a second order principal maximum is observed at an angle of 16 degrees. how many lines per centimeter does this grating have?
sin@=mλd d= mλ/sin@ = 2(546x10^-9)/sin16 = 3.96x10^-6 m
if a thicker hair is used in this experiment, will the number of bright fringes increase, decrease, or stay the same? how many bright fringes will be observed if the hair has a thickness of 80 um?
since 2t/λ +1/2 = m
the smaller the wavelength and larger the aperture, the _____________ and ______________ separation and the greater the resolution
smaller, angular
In a single slit experiment, light passes through the slit and forms a diffraction pattern on a screen 2.31 m away. if the wavelength of light is 632 nm, and the width of the slit is 4.20x10^-5 m, find the linear distance on the screen from the center of the diffraction pattern to the first dark fringe
solution 1. find the angle of the first dark fringe 2. use y=Ltan@ to find the linear distance Wsin@ =mλ, sin@=632x10^-9/4.4x10^-5 = 0.0015 wsin@=2λ sin@=2λ/w = 2x632x10^-9 m/4.2x10^-5 m @=1.73 y=Ltan@=6.93 cm
the asteroid ida is orbited by its own small "moon" called dactyl. if the separation between these 2 asteroids is 2.5 km, what is the max distance at which the hubble space telescope (aperature diameter =2.4 m ) can still resolve them with 550 nm light?
solution: 1. calculate the min angular separation for the asteroid to be resovled 2. express L in terms of Y and @min 3. subsitute numerical values to find L Lmax= 2.5x1-3m/tan@ -- If D incr, @min decr, tan@ decr and L incr
camera lenses (n=1.52) are often coated with a thin film of magnesium fluoride (n=1.38). these "non reflective coatings" use destructive interference to reduce unwanted reflections. find the condition for destructive interference for light in the middle of the visible spectrum
solution: 1. give the phase change for ray 1 2. give the phase change for ray 2 3. set the difference in phase changes equal to one half 4. solve for thickness 5. substitute numerical values 2t/λn = 1/2 t=λn/4 = 102 nm
bending of waves
sound waves will diffract around doors, corners, and other barriers. Diffraction is why we can hear sound even though we are not in a straight line from the source
2 identical microscope slides in air illuminated with light from a laser are creating interference pattern. the space between the slides is now filled with water (n=1.33). what happens to the interference fringes?
spaced closer together
in a double slit experiment, when the wavelength of light is increased, the interference patter
spreads out dsin@ = mλ
Wsin@=mλ (m= -2,-1,0,1,2)
the +ve and -ve values of m account for the symmetry of the pattern around the center
resolution
the ability to visually separate 2 closely spaced objects
to find the first dark fringe, divide the slit into 2 parts. the path difference pairs (1,1') is (W/2)sin@
the first dark fringe occurs when (W/2)sin@=λ/2 Wsin@=λ
a 2 slit experiment is performed in the air. later the same apparatus is immersed in water and the experiment is repeated. when the apparatus is in water, are the interference fringes (a) more closely spaced, (b) more widely spread or (c) spread the same as the apparatus was in the air?
the fringes move closer together since λn of light in water is λ/n smaller because of this, @ is smaller and fringes are closer
films of varying thickness
the rainbow of colors we see is due to the different wavelengths of light (λf>λb)
Newton's rings
the spacing of fringes gets smaller as you move away from the central fringe
interference can also occur when light refracts and reflects from both surfaces of a thin film. _____________________
this accounts for the colors we see in oil stick and soap bubbles
- the path difference between ray 2 and 3 is 2t (in addition, ray 3 experiences a phase change of 180). thus, the dark fringes will occur for: 2t =mλwater where λwater=λair/λ
thus, the water has decreased the wavelength of the light
surface of a CD shows that the information is encoded in ______________
tiny bumps
a telescope is used to resolve 2 distant stars. what will increase the resolution of the telescope?
use a blue light filter - the resolution for a lens is given by the min angle and the formula is : @min=1.22λ/D
Now, we have not only path differences and phase changes on reflection; we must account for the change in ____________________ as the light travels through the film
wavelength
the amount of diffraction depends on the ____________. which is why we can hear corners (λ~1m) but not see around (λ~10^-7m)
wavelength
λ=v/f=c/nf
wavelength formula
to reduce the min angle and thus to increase the resolution you should decrease the _________________ or increase the _________. Using its lower wavelength than the peak of visible light and red light
wavelength, D
the positions of the peaks are different for different _____________ of light
wavelengths - the condition for constructive interference in a diffraction grating dsin@ = mλ, m=0,1,2
the rainbow of colors we see is due to the different _________________
wavelengths of light (λr>λb)
the diffraction pattern below arises from a single slit. if we would like to sharpen the pattern. i.e make the central bright spot narrower, what should we do to the slit width?
widen the slit
linear distance from central fringe to other fringes
y = Ltan@
Wsin@=mλ for dark fringes. If @ increase, _______________
λ will increase λ=wsin@=(2.2x10^-6 m)(sin15degrees) = 569 nm
λn =
λvacuum/n
