Physics 6C

Physics 6C

Interference of EM Waves

Prepared by Vince ZacconeFor Campus Learning Assistance Services at UCSB


Constructive Interference:

Waves add - larger amplitude.These waves are “In Phase”

Destructive Interference:

Waves cancel - smaller amplitude.These waves are “Out of Phase”They are out of sync by ½λ

Interference in action:http://phys23p.sl.psu.edu/phys_anim/waves/embeder1.203.html

http://phys23p.sl.psu.edu/phys_anim/waves/embeder1.203.html

Thin Film Interference

Basic idea is that we will compare the two reflections.If they are in phase we have constructive interference (bright).If they are out of phase we have destructive interference (dark).

Important details:1) When light reflects from a higher-index medium it is phase-shifted by ½ of a wavelength.

If both reflected rays have this shift we can ignore it, but if only one of them is shifted, we have to switch the formulas for constructive/destructive interference.

2) The wavelength in the formulas is the wavelength in the film, so we have to divide the vacuum wavelength by the index of the film.

edestructiv)m(t2veconstructimt2

shiftrelativenoFORMULAS

21

Thin film demo Prepared by Vince ZacconeFor Campus Learning Assistance Services at UCSB

http://mysite.verizon.net/vzeoacw1/thinfilm.html

Here’s a sample problem:

a) What is the minimum soap-film thickness (n=1.33) in air that will produce constructive interference in reflection for red (λ=652nm) light? (assume normal incidence)

b) Which visible wavelength(s) will destructively interfere when reflected from a soap film of thickness 613nm? Assume a range of 350nm to 750nm for visible light.






1 2

Air n=1

Water n=1.33

Air n=1





1 2

The outgoing rays will interfere, and there is a relative phase shift, since ray 1 reflects from a higher index, but ray 2 does not.This yields the following formulas:

veconstructit2n)m(

edestructivt2nm

021

0

Air n=1

Water n=1.33

Air n=1





1 2


veconstructit2n)m(

edestructivt2nm

021

0

For part a) we use the constructive formula, with m=0 (we want the thinnest film possible)

nm123tt2)33.1nm652( min2

1

Air n=1

Water n=1.33

Air n=1





1 2


veconstructit2n)m(

edestructivt2nm

021

0


nm123tt2)33.1nm652( min2

1

For part b) we use the destructive formula, with different values of m (we want visible wavelengths)

Air n=1

Water n=1.33

Air n=1





1 2


veconstructit2n)m(

edestructivt2nm

021

0


nm123tt2)33.1nm652( min2

1


mnt2t2nm 00

Air n=1

Water n=1.33

Air n=1





Air n=1

Water n=1.33

Air n=1

1 2


veconstructit2n)m(

edestructivt2nm

021

0


nm123tt2)33.1nm652( min2

1


mnt2t2nm 00

nm4074mnm5443m

00

Other values of m give wavelengths that fall outside of the visible range


A thin layer of magnesium fluoride (n=1.38) is used to coat a flint-glass lens (n=1.61).

What thickness should the MgF2 coating be to suppress the reflection of 595nm light?



Air n=1

MgF2 n=1.38

Glass n=1.61

1 2





Air n=1

MgF2 n=1.38

Glass n=1.61

1 2

We need destructive interference (no reflection). In this case both outgoing rays reflect from a higher index, so there is no relative phase shift.Our formulas are:

edestructivt2n)m(

veconstructit2nm

021

0





Air n=1

MgF2 n=1.38

Glass n=1.61

1 2


edestructivt2n)m(

veconstructit2nm

021

0

nm108tt238.1nm595)0( min2

1




We can use any integer m>0, so start with m=0 and solve for t.This will give the minimum thickness.


Air n=1

MgF2 n=1.38

Glass n=1.61

1 2


edestructivt2n)m(

veconstructit2nm

021

0

We can use any integer m>0, so start with m=0 and solve for t.This will give the minimum thickness.

nm108tt238.1nm595)0( min2

1




To get other possible thicknesses that will work, just use larger values for m:

...etcnm539t2mnm323t1m


A thin film of oil (n=1.40) floats on water (n=1.33). When sunlight is incident vertically, the only colors that are absent from the reflected light are blue (458nm) and red (687nm).

Estimate the thickness of the film.






Air n=1

Oil n=1.40

Water n=1.33

1 2





Air n=1

Oil n=1.40

Water n=1.33

1 2

In this case ray 1 reflects from a higher index, but ray 1 reflects from a lower index, so there is a relative phase shift.Our formulas are:

veconstructit2n)m(

edestructivt2nm

021

0





Air n=1

Oil n=1.40

Water n=1.33

1 2


veconstructit2n)m(

edestructivt2nm

021

0

We will use the destructive interference formula for each given wavelength. Since they are the only visible wavelengths that are absent, we can deduce that they correspond to consecutive values for m in the formula.





Air n=1

Oil n=1.40

Water n=1.33

1 2


veconstructit2n)m(

edestructivt2nm

021

0


t240.1nm458)1m(

t240.1nm687m





Air n=1

Oil n=1.40

Water n=1.33

1 2


veconstructit2n)m(

edestructivt2nm

021

0


t240.1nm458)1m(

t240.1nm687m

At this point we have some algebra to do. My preference is to find the integer value of m that fits the formulas, then plug that in to find t.





Air n=1

Oil n=1.40

Water n=1.33

1 2


veconstructit2n)m(

edestructivt2nm

021

0


t240.1nm458)1m(

t240.1nm687m


2m458m229)458)(1()458)(m()687)(m(

)458)(1m()687)(m(40.1nm458)1m(40.1

nm687m





Air n=1

Oil n=1.40

Water n=1.33

1 2


veconstructit2n)m(

edestructivt2nm

021

0


t240.1nm458)1m(

t240.1nm687m


2m458m229)458)(1()458)(m()687)(m(

)458)(1m()687)(m(40.1nm458)1m(40.1

nm687m

nm491tt240.1nm6872

Young’s Double Slit Experiment

In Young’s double-slit experiment light comes from the left and passes through the slits, illuminating the screen some distance R away. The light rays from the 2 slits will travel different distances to get to the screen (except in the center). Depending on the path length difference the waves will be in phase or out of phase when they arrive at the screen. If they are in phase, they combine to give constructive interference (a bright region). Out of phase means destructive interference (dark region). Some geometry gives us a formula for this difference in path length: dsin(θ). This yields the following formulas:

edestructiv)m()sin(dveconstructim)sin(d

21

Ry)tan(

m can be any integer (m=0,±1,±2,±3,±4,…)

y = actual distance on screen (from center) Prepared by Vince ZacconeFor Campus Learning Assistance Services at UCSB

Monochromatic Light


Light with wavelength 546nm passes through two slits and forms an interference pattern on a screen 8.75m away. If the linear distance on the screen from the central fringe to the first bright fringe above it is 5.16cm, what is the separation of the slits?




This should be a straightforward problem. We are given enough information to just use our formulas. Start with the formula involving the distance on the screen:


Ry)tan(




338.0

10897.5)tan( 3m75.8

m1016.5 2


Ry)tan(




338.0

10897.5)tan( 3m75.8

m1016.5 2

Use this angle in the formula for bright fringes, with m=1 m)sin(d


Ry)tan(




Ry)tan(

338.0

10897.5)tan( 3m75.8

m1016.5 2

Use this angle in the formula for bright fringes, with m=1 m)sin(d

m1026.9dm10546)338.0sin(d

5

9


Since the angle was small we could have used the approximate formula: dmRym

Multiple Slits (diffraction gratings)

These work just like the double slit experiment (same formula), but the bright spots are narrower, and the dark spots are wider. If the grating has more slits the result is a sharper image, with narrower bright fringes.



Diffraction

• When light encounters an obstacle it will exhibit diffraction effects as the light bends around the object or passes through a narrow opening.

• Notice the alternating bright and dark bands around the edge of the razor blade. This is due to constructive and destructive interference of the light waves.

Single Slit Diffraction

• Similar to the double-slit experiment.• The formulas are opposite (the geometry just comes out that way).• Notice that the central maximum is double-width compared to the

others.• This is how you can tell a single-slit pattern from a multiple-slit pattern.


a)m(Ry

)m()sin(a21

m

21

a)m(Ry

)m()sin(a

m

Formulas for Constructive Interference (bright fringes)

Formulas for Destructive Interference (dark fringes)

These approximate formulas work when the angle is small


How many dark fringes will be produced on either side of the central maximum if green light (λ=553nm) is incident on a slit that is 2µm wide?




This is a single-slit problem, so the formula for the dark fringes is: m)sin(a




Let’s find the angles to the first few dark fringes. We get a new angle for each value of m.

m θ1 16°2 34°3 56°4 ???



When we try to calculate with m=4 we get a calculator error. Why doesn’t it work?




Recall the single-slit diffraction diagram.For the fringes to show up on the screen, the angle must be less than 90°.Of course the pattern gets very dim near the edges, but mathematically the formula will break down when sin(θ)>1.



R

m θ1 16°2 34°3 56°4 ???





Recall the single-slit diffraction diagram.For the fringes to show up on the screen, the angle must be less than 90°.Of course the pattern gets very dim near the edges, but mathematically the formula will break down when sin(θ)>1.



R

m θ1 16°2 34°3 56°4 ???


So it looks like we will get 3 dark fringes.

Circular Aperture


Light passing through a circular opening gives a circular pattern.

A formula to find the first dark fringe is: 𝑠𝑖𝑛𝜃=1.22 𝜆𝐷

This can be taken as the angular resolution of the aperture. When two light sources are close together this angle limits our ability to “resolve” them as separate objects.

Circular Aperture


Example: You are driving at night on a long straight highway in the desert as another vehicle approaches. What is the maximum distance at which you can tell that it is a car rather than a motorcycle by seeing its headlights, which are separated by a distance of 1.5m?

a) Assume your visual acuity is limited only by diffraction. Use 550 nm for the wavelength, and pupil diameter 6.0mm.

b) What answer do you get if you use a more realistic, typical visual acuity with θmin=5x10-4 rad?

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Physics 6C