wave optics - michigan state university...• spherical wavelets emitted from a distribution of...

April 8, 2014 Chapter 34 1

Wave Optics


Announcements

!  Remainder of this week: Wave Optics !  Next week: Last of biweekly exams, then relativity !  Last week of the semester: Review of entire course, no exam !  Final exam Wednesday, April 30, 8-10 PM

•  Location: WH B115 (Wells Hall) •  Comprehensive, covers material from the entire semester •  Email me if you have an exam conflict

a) at 8pm, b) on Wednesday •  Make-up exam: To be announced


Thin-Film Interference !  Another way of producing interference

phenomena is partial reflection of light from the front and back layers of thin films

!  A thin film is an optically clear material with thickness on the order few wavelengths of light

!  Examples of thin films include the walls of soap bubbles and thin layers of oil floating on water

!  When the light reflected off the front surface interferes with the light reflected off the back surface of the thin film, we see the color corresponding to the particular wavelength of light that is interfering constructively


Thin-Film Interference

!  Consider light traveling in an optical medium with an index of refraction n1 that strikes a second optical medium with index of refraction n2

!  The light can be transmitted through the boundary •  In this case the phase of the light is not changed

!  The light can be reflected off the boundary •  In this case, the phase of the light can be changed depending on the

index of refraction of the two optical media •  If n1 < n2, the phase of the reflected wave will be changed by half

a wavelength •  If n1 > n2 then there will be no phase change


Thin-Film Interference


Thin-Film Interference !  Assume a thin film of thickness t with

index of refraction n and with air on both sides of the film

!  Monochromatic light is incident perpendicular to the surface of the film

!  An angle of incidence is shown for the light waves in the figure for clarity

!  When the light wave reaches the boundary between air and the film, part of the wave is reflected and part of the wave is transmitted

soap bubble


Thin-Film Interference !  Part of the light is reflected by the first surface – phase shift !  Part of the light is reflected at the second surface – no phase

shift !  Criterion for constructive interference: !  The wavelength λ refers to the wavelength of the light

traveling in the thin film, which has index of refraction n !  The wavelength of the light traveling in

air is related to the wavelength of the light traveling in the film by λ = λair/n

Δx = m+ 1

2⎛⎝⎜

⎞⎠⎟ λ = 2t m = 0,m = ±1,m = ±2,...( )


Thin-Film Interference !  We can then write

!  The minimum thickness tmin that will produce constructive

interference corresponds to

!  Note that this result applies only to the case of a material with index of refraction n and air on both sides, like a soap bubble

m+ 1

2⎛⎝⎜

⎞⎠⎟λair

n= 2t m = 0,m = ±1,m = ±2,...( )

tmin =

λair

4n


Lens Coating

!  Many high quality lenses are coated to prevent reflections !  This coating is designed to set up destructive interference

for light that is reflected from the surface of the lens !  Assume that the coating is MgF2, which has ncoating = 1.38

and the lens is glass with nlens = 1.51 PROBLEM

!  What is the minimum thickness of the coating that will produce destructive interference for light with a wavelength in air of 550 nm? SOLUTION

!  We assume that the light is incident perpendicularly on the surface of the coated lens


Lens Coating !  Light reflected at the surface of the coating will undergo

a phase change of half a wavelength because nair < ncoating !  The transmitted light has no phase change !  Light reflected at the boundary between the coating and the

lens will undergo a phase change of half a wavelength because ncoating < nlens

!  This reflected light will travel back through the coating and exit (no additional phase change)

!  Thus both the light reflected from the coating and from the lens will have suffered a phase change of half a wavelength


Lens Coating !  Thus the criterion for destructive interference is

!  Thus the minimum thickness for the lens coating to provide destructive interference corresponds to m = 0

!  Lens coatings that fulfill this destructive interference condition are referred to as “quarter-wave” coatings

!  Because they cause destructive interference for reflected light, they prevent flares

m+ 1

2⎛⎝⎜

⎞⎠⎟

λair

ncoating= 2t m = 0,m = ±1,m = ±2,...( )

tmin =

λair

4ncoating= 550 ⋅10−9 m

4 ⋅1.38= 9.96 ⋅10−8 m = 99.6 nm


Interferometer

!  An interferometer is a device designed to measure lengths or changes in length using interference of light

!  An interferometer can measure lengths or changes in lengths to a fraction of the wavelength of light using interference fringes

!  Here we will describe an interferometer similar to one constructed by Albert Michelson and Edward Morley at the Case Institute in Cleveland, Ohio in 1887

!  Our interferometer is simpler than the one used by Michelson and Morley but is based on the same physical principles


Interferometer !  A photograph and drawing of a commercial interferometer

used in physics labs are shown below


Interferometer !  Transmitted light is totally reflected from m2 back to m1 !  Reflected light is totally reflected from m3 back toward m1 !  Part of the light from m2 is

reflected by m1 toward the viewing screen and part of the light is transmitted and not considered

!  Part of the light from m3 is transmitted through m1 and part is reflected and not considered

!  The light from mirrors m2 and m3 will interfere based on the path length difference between the two ways of arriving at the viewing screen


Interferometer !  Both paths undergo two reflections, each resulting in a

phase change of half a wavelength !  The condition for constructive

interference is

!  The two different paths will have a path length difference

!  The viewing screen will display concentric circles or vertical fringes corresponding to constructive and destructive interference depending on the type of de-focusing lens

Δx =mλ m = 0,m = ±1,...( )

Δx = 2x2 − 2x3

Δx = 2 x2 − x3( )


Interferometer !  If the movable mirror m2 is moved a distance of λ/2, the

fringes will shift by one fringe !  Thus this type of interferometer

can be used to measure changes in distance on the order of a fraction of a wavelength of light, depending on how well one can measure the shift of the interference fringes


LIGO !  The LIGO experiment uses a Michelson interferometer to

detect small movements of one mirror caused by gravitational waves. Gravitational waves have never been seen before, but LIGO hopes to be the first

!  Each arm of the interferometer is 4 km long

!  The sensitivity is ΔL/L~1020


Single-Slit Diffraction !  Diffraction of light through a single slit of width a that is

comparable to the wavelength of light that is passing through the slit

!  Approach: Consider suitable individual rays •  Spherical wavelets emitted from a distribution of points in the slit •  Light emitted from these points will superimpose and interfere based

on the path length difference for each wavelet at each position

!  At a distant screen we observe an intensity pattern characteristic of diffraction

!  Bright and dark fringes !  For diffraction we will

analyze the dark fringes as destructive interference


Single-Slit Diffraction

!  To study the interference let’s expand and simplify our previous figure for single slit diffraction

!  We assume coherent light with wavelength λ incident on a slit with width a that produces an interference pattern on a screen at distance L

!  Just like in the double-slit experiment, we analyze pairs of light waves emitted from points in the slit

!  We start with light emitted from the top edge of the slit and from the center of the slit

!  To analyze the path difference we show an expanded version of our figure to the right


Single-Slit Diffraction !  Here we assume that the point P on the screen is far enough

away that the rays r1 and r2 are parallel and make an angle θ with the central axis

!  Therefore the path length difference for these two rays is

!  The criterion for the first dark fringe is !  Although we chose one ray originating

from the top edge of the slit and one from the middle of the slit, we could have used any two rays that originated a/2 apart inside the slit

sinθ = Δx

a / 2 ⇒ Δx = asinθ

2

Δx = asinθ

2= λ

2 ⇒ asinθ = λ


Single-Slit Diffraction !  Now let’s consider four rays instead of two

!  Here we choose a ray from the top edge of the slit and three more rays originating from points spaced a/4 apart


Single-Slit Diffraction !  We can expand the previous drawing to

represent the case of the screen being far away as shown to the right

!  The path length difference between the pairs of rays (r1,r2), (r2,r3), (r3,r4) is given by

!  The dark fringe is given by

!  All wavelets could be grouped into four rays

sinθ = Δx

a / 4 ⇒ Δx = asinθ

4

asinθ

4= λ

2 ⇒ asinθ = 2λ


Single-Slit Diffraction !  Each group destructively interferes, so an overall dark fringe

is visible on the screen !  The condition asinθ = 2λ describes the second dark fringe !  At this point we could take six pairs and eight pairs and

describe the third and fourth dark fringes, etc. !  The result is that the dark fringes from single slit diffraction

can be described by

!  If the screen is placed a large distance from the slits, the angle θ will be small and we can make the approximation

!  We can express the position of the dark fringes as

asinθ =mλ m =1,2,3,...( )

sinθ ≈ tanθ = y / L

ayL=mλ ⇒ y = mλL

a m =1,2,3,...( )


Single-Slit Diffraction !  The intensity of light passing through a single slit can be

calculated but we will not present the derivation here !  The intensity I relative to Imax if there were no slit is

!  We can see that this expression for the intensity is zero for sinα = 0 (unless α = 0), which means α = mπ for m = 1, 2, …

!  For α = 0, we use

!  So

I = Imax

sinαα

⎛⎝⎜

⎞⎠⎟

2

where α = πaλ

sinθ

limα→0

sinαα

⎛⎝⎜

⎞⎠⎟ =1

mπ = πa

λsinθ ⇒ asinθ =mλ m =1,2,3,...( )


Single-Slit Diffraction !  If the screen is placed a large distance from the slits, θ is

small and can be approximated as sinθ ≈ tanθ = y/L and !  If the screen is L = 2.0 m away from the slit, the slit has a

width of a = 5.0·10–6 m, and the wavelength of the incident light is λ = 550 nm, we get the intensity pattern

α = πay

λL

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