Chapter 35– Interference Hints-5/23/11
1. One of the early challenges to a wave theory of light was demonstrating that light could interference. The
definitive proof of light interference was Young’s double slit experiment. The following questions deal with
this experiment and the difficulties in demonstrating light interference.
a) Explain why two flashlights held close together do not produce an interference pattern on a distant
screen. The problems are: light is not monochromatic, is not coherent, and the sources are too far apart.
b) Why is it so much easier to perform interference experiments with a laser than with an ordinary light
source? A laser is a monochromatic and coherent source of light.
c) Why is it necessary to use two slits rather than simply two sources? And why must the slit separation be
very small? Two separate sources of light and generally not coherent. If the separation is too large
relative to the wavelength the interference nodes and antinodes are too close together to be discerned.
d) In Young's double-slit experiment, why do we use monochromatic light? If white light is used, how
would the pattern change? The pattern scale depends on the wavelength; different wavelength light
would have difference nodal separations. If white light is used the colors would become separated as
happens in a prism…
e) If Young's double-slit experiment were performed underwater, how would the observed interference
pattern be affected? Underwater the wavelengths would be shorter and the nodal and antinodal
separations would be smaller.
2. Consider the following two examples of a two-source wave interference pattern. The circles indicate wave
crests. The broken lines are axes and the solid line is a reference line parallel to the source separation
distance d. In the first illustration the sources are two wavelengths apart, and in the second the sources are
3.5 wavelengths apart.
a) Draw the antinodal and nodal “lines” (they are actually curves) in one quadrant, and label them in order
(A1, N1, A2 , etc) from the center antinode (Ao). How do the number of nodes and antinodes in one
quadrant relate to the source separation distance d? If you count the line along the axes as ½ (since they
are shared with another quadrant), the no. of antinodal lines or nodal lines equals d/ (the max order).
b) Assign the points labeled in the diagram to the node or antinode line to which each belongs. The points
are small but visible. In the first picture: B in A1; C in N2; D in Ao; E in A1.
c) Measure the approximate angular direction ø of each A line measured from the center between the
sources. Show that the angles correspond to the formula sinø=∆r/d, where ∆r is the path difference from
the sources to any point on a specific A-line. Angles written into illustration are ~ arcsin(∆r/d)
d) Verify that the distance between A-lines along a line parallel to d (the solid line in the illustrations) is
about the same for lines within ~30º of the center antinode. You should be able to notice that the
separation between A’s or N’s along the reference line are ~ the same within the small angle region.
A2
N4
Reference
line
N2
A3
N3
C
•
B
A2
A1
•
C
B
•
D
•
~30º
E
•
~59º
~32º
~17º
Ao
D
•
E
•
•
N2
A1
N1
Ao
e) If the wavelength were 1.5 cm and the separation were 12 cm. How many nodes and antinodes would
there be in one quadrant? m=8 How many nodes and antinodes within 30º of the central antinode? m=4
f) If the sources were 180º out of phase how would the illustrations change? The nodes and antinodes
would switch places, for example, the central line would be a node.
3. In the analysis of the two-source interference pattern there are two approximations that we generally rely
on:
r1
r2
I. That the pattern is being analyzed far away from the sources (L>>d)
d
ø
so that the ray paths (r1 and r1) from sources to each point are nearly
L
∆r
parallel, which allows us to use the formulas sinø=∆r/d and tanø=y/L.
II. That the angular direction ø is small enough (<~30º) that we can assume sinø~tanø.
A or N
y
In this problem we compare the various approximations. This was done in class…
Two radio antennas separated by 200 m transmit identical signals at the same wavelength. A radio of a car is
at the position of the second maximum 400 m from the center line, which is 1000 m from the sources.
a) Determine the wavelength of the signal using trig, geometry, and the fact that ∆r= 2, without using any
approximation at all. This would be the most correct answer.
r2- r1 =2 =37 m
b) Find the wavelength again using the first approximation mentioned
400 m
above, but not assuming that sinø~tanø. How far off is this result
from the value obtained in (a)? Is this a good approximation?
200 m
ø
1000 m
tanø=y/L and sinø=2 /d =37.1 m~0.3% off, pretty good…
c) Find the wavelength one last time using both approximations
mentioned above including sinø~tanø. How far off is this result
from the value obtained in (a)? Is this a good approximation?
y/L =2 /d =40 m~8% off, not so good…but if L were larger, the approximation would be better.
d) How much farther must the car travel to encounter the next minimum
in reception? Which approximation would you use to find the result?
Using the approximations tanø=y/L and sinø=(2.5) /d (for N3) y=522 m∆y=122 m farther to min.
The following are practice problems with 2-slit interference.
4. On a day when the speed of sound is 350 m/s, a 2500-Hz sound wave falls on two slits 30 cm apart. In all
these the distance L is long enough to use the small angle approximation.
a) At what angle is the first maximum located? =350/2500 =0.14 m sinø=14/30ø=28º
b) The sound wave is replaced by 3.0-cm microwaves, what slit separation gives the same angle for the first
maximum? 14/30=3/dd=6.4 cm
c) If the slit separation is1.00 µm, what light frequency would give the same first maximum angle?
14/30= /1 =0.467 µmf =c/ =6.4 x1014 Hz.
5. A double slit with a spacing of 0.083 mm between the slits is 2.5 m from a screen.
a) Yellow light of wavelength 570 nm strikes the double slit, what is the separation between the 0th and 1st
order maxima on the screen? y=L /d=17 mm
b) If blue light of wavelength 410 nm strikes the double slit, what is the separation between the 2nd and 4th
order maxima? ∆y= L2 /d=25 mm
c) Repeat parts (a) and (b) for the minima. (a) 8.6 mm; (b) 25 mm
6. Monochromatic light illuminates a double-slit system with a slit separation d of 0.30 mm. The second
order maximum occurs at a distance y= 4.0 mm from Ao on a screen 1.0 m from the slits. Determine (a) the
wavelength, (b) the position of the 3rd order maximum, (c) the angular position of the 1st minimum, and (d)
the total number of antinodes in one quadrant.
(a) =yd/mL=4(0.3)/2(103)= 0.6x10-3 mm=600 nm; (b) y=L /d=6.0 mm; (c) sinø=/2d=10-3 ø=0.057º
(very small!); (d) maximum order, m=d/ =500 antinodal lines (counting A’s at 0º and 90º as ½ antinode
each).
7. In a double-slit arrangement d =0.15 mm, L=140 cm, 643 nm, and y =1.8 cm.
(a) What is the path difference ∆r for the two rays from the two slits arriving at the point at y?
m=yd/L=3∆r =3 =1929 nm
(b) Express this path difference in terms of . ∆r =3
(c) Does this point correspond to a maximum, a minimum, or an intermediate condition? It’s a max.
(d) There’s over 200 antinodes in one quadrant in this arrangement, what would be position at L=140 cm of
the m=200 antinode? This order is outside the small angle approximation range, so you need to find the
angular position from sinø=m /d=200(643nm/0.15mm)=0.857ø=59º. Then, tan59º=y/140y=233 cm
8. In a Young's interference experiment, the two slits are separated 0.15 mm, and the incident light includes
light of wavelengths 1 =540 nm and 2 =450 nm. The overlapping interference patterns are formed on a
screen 1.40 m from the slits. Calculate the minimum distance from the center of the screen to the point
where a bright line of the 1 light coincides a bright line of the 2 light.
y1= y2 m11L/d = m22L/d, where m1 and m2 are integers 1/2 = m2/m1=54/45=6/5. So m1 =5 and m2
=6 y1= y2=5(540nm)(1.4m)/0.15mm=2.52 cm
9. To determine the relative brightness (intensity) of the interference pattern in the region between the
maxima and the minima we use phasor techniques to add the two waves generated by the sources. Light is a
wave of electromagnetic fields, so, in a 2-slit interference pattern, the interfering waves can be described
mathematically as E1= Eosin (kx-t) and E1= Eosin (kx-t +). The phase difference  is the result of the
different path lengths (r1 and r2) each wave travels to a point on a distant screen.
a) Use phasor techniques to show that the resultant wave is ER=(2Eo cos )sin(kx-t +).
You should review the phasor addition techniques discussed
2Eocos/2
/2
Eo
in the wave unit…Use trig and geometry on the diagram shown.
/2


Eo
b) Convince yourself that the phase angle  is related to the path difference ∆r by the formula
=2π(∆r/)=2π(dsinø/).
This was discussed in the wave unit. Just as the wavelength  is proportional to a full cycle 2π, the path
difference is proportional to the phase differencehence 2π=(∆r/). Since ∆r
=dsinø=2π(dsinø/).
c) Convince yourself of the validity of the intensity formula for the 2-slit pattern: IR =Io cos 2(πdsinø/).
In the wave unit we learned that the average intensity of a wave (I=Power/Area) is proportional to the
amplitude squared so that Iaver =½[2Eo cos ()] 2 (recall cos2ø aver=½). The antinode maxima intensity
corresponds to =0, so Io =½(2Eo )2. Finally, IR =Io cos 2()=Io cos 2(πdsinø/).
10. At a particular location in a Young's interference pattern, the intensity on the screen is 6.4% of the
closest max.
a) Find the minimum phase difference (in radians) between the sources producing this result.
Io
Here IR /Io =0.064= cos 2()2(arcos√ 0.064)=151º=2.63 rad
b) Determine the path difference for 587-nm light.
2π=∆r=0.42nm, note that this is near a minimum where ∆r=0.5
IR=0.064Io
c) If the slit separation is 6 wavelengths, determine the angular position from the closest maximum to this
particular location in the pattern.
=2π(dsinø/)sinø=2πd=0.42/6ø=4.0º
d) If the pattern is being projected 100 cm from the slits, what’s the distance between that maximum and this
point in the pattern.
y=Ltanø=100(.07)=7.0 cm
11. If the two slits in a 2-slit pattern are unequal in size one source will have more intensity than the other
and the interfering waves will have different amplitudes. Assume the two wave sources are represented by
E1= 12sin t and E2=16sin(t +). Use phasors to find the resulting wave surperposition at
a) a maximum; b) a minimum; c) when the phase difference =90º; d) when =60º.
a) For a max =0,2π, etc.. and the phasors line up E12=28sin t; b) For a min =π, 3π, etc.. and the
phasors subtract E12=-4sin t; c) Here the phasors are perpendicular and the amplitude of the result is
the hypotenuse  E12=20sin (t +π/4); d) Here you need to use the laws of
EoR
sines and cosines…EoR2 =122 +162 – 2(12)(16)cos120º EoR =24.3 units; also

º
sin/16=sin120º/24.3 =arcsin(0.57)= 0.19π rad E12 =24sin(t+0.19π)

12
e) Compare the intensities of the superimposed waves above to the maximum intensity.
Intensities are proportional to amplitudes squared: Imax /Imax=1; Imin /Imax=(4/28)2; I=90º /Imax=(20/28)2;
I=60º /Imax=(24/28)2.
16
12. To generalize from the problem above, sketch a phasor diagram to illustrate the resultant of two
coherent waves with different amplitudes, E1= Eo1sint and E2= Eo2sin(t +). Use the sketch and the law
of cosines to show that the resultant intensity can be written in the form IR=I1 + I2 + 2cos√(I1I2).
Using the law of cosines EoR2 = Eo12 + Eo22 – 2(Eo1)( Eo2)cos
EoR

Eo2


EoR2 = Eo12 + Eo22 +2(Eo1)(Eo2)cos. Intensities are proportional to the

Eo1
amplitudes squared: IR∝ EoR2; I1∝ Eo12; I2∝ Eo22. Since the proportionality constants must be the same for
each intensity term, you can write the expression above as IR=I1 + I2 + 2cos√(I1 I2).
13. An interesting variation of the 2-source interference problem occurs when a reflection from a single
source interferes with the original signal. And added detail is that reflections from a mirror or opaque
surface are 180º out of phase with the incoming ray so that there is an extra half-wavelength added to (or
subtracted from) the reflected path.
(Not to scale) d
The illustration shows a radio wave transmitter and a receiver separated
by d= 600 m and both h= 35 m high. The receiver can receive signals
directly from the transmitter and indirectly from reflections off the ground.
h
Assuming that the ground is level between the two towers and that a/2
phase shift occurs upon reflection, determine the longest wavelengths that
interfere (a) constructively and (b) destructively at the receiver.
Here ∆r = rreflected - rdirect =2[h2+(d/2)2] 1/2 ±/2 – d2= 604 ±/2 - 600 = 4 ±/2. (a) For longest wavelength
max: ∆r =4 +/2=  =8 meters; (b) For longest wavelength min: ∆r =4 - /2=  =4 meters.
14. A single light source near a long mirror can also produce an interference pattern. This is called a Lloyd's
mirror arrangement. Assume that a 606 nm source is a distance h above a mirror, and that interference
antinodes, 1.2 mm apart, are formed on a screen 2.0 m from the light.
a) Find the vertical distance h of the source above the reflecting surface.
Light
h
Screen
Here the separation between the sources d=2h, since ∆y/L=/2h
h=L/2∆y=0.505 m
Mirror
b) How’s this pattern different from one made by two actual light sources
a distance 2h apart? The max and min switch positions.
c) Find the position on the screen of the 3rd antinode. The formula is y=L(m +½)/d, and m=2 for A3
y=3.0 mm
15. Another common interference phenomenon occurs when light going through a thin film reflects rays
both at the front and back of the film. These reflected rays interfere and create some interesting and colorful
patterns. Review the basics of thin-film interference and answer the following.
a) In order to observe interference in a thin film, the film must be very thin (on the order of a few wavelengths). Why is that? In a thick film the reflections would not be coherent enough to create interference.
Also the rays need to be nearly normal to the surface so they will be close enough to interfere.
b) In thin film interference what refraction conditions are present when the reflected rays are in phase due
to the reflections alone?... when the reflected rays are out of phase due to the reflections alone. A
reflection is out-of-phase when the wave slows down, so the light has to be going from a lower to a
higher “n”, otherwise the reflection is in-phase.
c) An oil film on water appears brightest at the outer regions, where it is thinnest. From this information,
what can you say about the index of refraction of oil relative to that of water? When the oil film is
extremely thin (t~0) there’s effectively no path difference between the reflected rays at the top and
bottom of the film. If the interference produces a bright reflection, then the reflected rays must be in
phase with each other. Going from air to oil the ray is reflected out-of-phase, so the reflection in the oilwater transition must also be out-of-phase to “catch-up” in phase to the reflection in the air-oil
transition. So noil< nwater.
d) A soap film on a wire frame held up in air appears black in the thinnest regions when observed by
reflected light, and it shows a variety of colors in thicker regions. Explain these phenomena. In this case
the reflection from the air-soap transition is out-of-phase and the reflection from the soap-air transition
is in-phase. At the thinnest region, there is no effective path difference to undo the “out-of-phaseness” of
the reflected rays and the result is darkness. In the thicker regions, there will be enough path difference
to undo the reflection effect and bright regions will develop on the film. A specific color will be brightly
reflected when the film thickness fits the specific color wavelength…
e) A lens with outer radius of curvature and index of refraction n rests on a flat glass plate and the
combination is illuminated with white light from above. Is there a dark spot or a light spot at the center
of the lens? What does it mean if the observed rings are noncircular? If you understood the previous
question, the answer is that the spot will a dark.
The following are practice problems with thin-film interference.
16. A beam of 580-nm light passes through two closely spaced glass plates, as shown.
For what minimum nonzero value of plate separation, does the transmitted ray and its reflected-transmitted
component interfere constructively? Only the relevant rays are shown; assume they are very close together.
Each reflection produces a ±/2 phase shift. If we call the plate gap “d”, then ∆r =2d ±/2 ±/2 =2d (there
is no effective phase change from the reflections). So for constructive interference 2d=m Therefore the
minimum separation is d=/2=580/2=290 nm. Note that we use the wavelength in air (not glass) because
the “film” in question is made of air.
17. A plastic film (n=1.45) floating on water is illuminated by white light at normal incidence. The film is
280 nm thick. Find (a) the dominant observed color in the reflected light, and (b) the dominant color in the
transmitted light. Explain your reasoning.
a) This film has a higher n than water, so the reflection at the top is out-of-phase, and the reflection at the
bottom is in-phase, so ∆r=2t ±/2=m for constructive interference (note that the path difference has been
adjusted to include the phase change on reflection).  So, film=2t/(m ±½)=air/nair=n2t/(m ±½). This
wavelength must be in the visible range (400700 nm) so, after a bit of trial-and-error, (m
±½)=1.5air=541 nm yellow-green light.
b) Here the problem is more like problem 16 above, ∆r=2t=mfilm air/n air=n2t/m. For the visible
range m=2 and air=406 nm, violet light.
18. A glass plate (1.61) is covered with a thin uniform layer of oil (n=1.20). Non-monochromatic light in air
is incident normally on the oil surface. Observation of the reflected beam shows destructive interference at
500 nm and constructive interference at 750 nm with no intervening maxima or minima. Calculate the
thickness of the oil layer.
Given that both top and bottom reflection are ±/2 out-of-phase, the rays will be in phase with each other
from the reflections, so ∆r=2t. For destructive interference 2t=(m±½)air/n, and for constructive
interference 2t=mair/n. So, (m +½)500=m750 m=1 t=mair/2n=750/2(1.2)=313 nm
19. Two rectangular fIat glass plates (n=1.52) are in contact along one end and separated along the other end
by a sheet of paper 4.0 X 10-3 cm thick. The top plate is illuminated by monochromatic light (=546 nm).
a) Calculate the number of dark parallel bands crossing the top plate (include
the dark band at zero thickness along the edge of contact between the two plates).
For dark bands 2t= mt=mair. For the highest order at paper end, m=2t/air=146.5,
so there would be 146+1=147 dark bands counting the m=0 band at the contact end.
b) How far apart are the dark bands if the plates are 12 cm long? ~12/146.5=0.082 cm
c) The sheet of paper is replaced with a wire and 30 bright fringes are counted in a 2 cm section of the
plates. What is the diameter of the wire? There must be ~30x6=180 dark band over the 12-cm length of
the plates. Solving for t=mair /2=180(546 nm)/2~49 µm.
20. A piece of transparent material having an index of refraction n is cut into
h
the shape of a wedge, as shown. The angle of the wedge is small, and monoy
chromatic light of wavelength  is normally incident from above. (a) If the height
x
L
of the wedge is h and the length is L, show that bright bands occur at positions
x = L(m+½)/2hn, and dark bands occur at x = Lm/2hn, where m=0, 1, 2, ... and x is as shown.
The theory is basically the same is the problem above, where the thickness t=y. For dark bands: 2y=mt =
mair/n , and y=xh/L x= Lmair/2nh. For bright bands: 2y=(m±½)air/n  x= L=(m±½)air/2nh.
(b) Describe any changes in the pattern of dark and bright band under the following conditions: (i) double L,
(ii) double h, (iii) surround wedge with a substance with a lower index of refraction, (iv) surround wedge
with a substance with a higher index of refraction.
(i) same no. of bars but farther apart, (ii) double no. of bands closer together, (iii) & (iv) no change.
21. Another example of thin-film interference occurs when a lens is placed on top
R
of a flat glass plate and light is shown from above. A series of dark and bright rings
t
appear known as “Newton’s rings” after the discoverer (Issac Newton, who else?)
r
a) Show that in a Newton-rings apparatus the dark circles will have radii given by the formula r=(Rm)1/2,
where R is the radius of curvature of the lens,  is the wavelength of the light in the air, and m=0, 1, 2, ...
Explain why we use the wavelength in the air and not in the glass. This will be derived in class. The
basic idea is that there is an air film between the lens and the plate and the geometry requires that
R2=r2+(R-t)2 r~(2Rt)1/2. Since t is the film thickness, 2t=mair for dark bands r =(Rmair)1/2.
b) When a liquid is introduced into the air space between the lens and the plate in the Newton-rings apparatus, the diameter of the tenth ring changes from 1.50 to 1.31 cm. Find the index refraction of the
liquid. The difference is due to the wavelength in the liquid film being (lair/n) r =(Rmlair/n)1/2. We can
set up a ratio: 1.50/1.31=(Rmlair)1/2/(Rmlair/n)1/2 =n1/2 n =1.31
c) Is the contact point between lens and plate a point of constructive or destructive interference? Explain.
Destructive interference. The reflections are out of phase and there is no space to allow the reflections to
get back in phase.
Challenge problems
R2
22. A plano-convex lens having a radius of curvature R1= 4.00 m is
placed on a concave reflecting surface having a radius of curvature
R1
R2 =12.0 m, as shown. Determine the radius of the 100th bright ring
if 500-nm light is incident normal to the flat surface of the lens.
t
Reflective surface
We know that 2t=mfor dark rings. If the reflective surface were flat,
r
then t1 ~(r2/2R1) as in problem 21 above.
But here, both lens and reflective surface curve up from the contact point, so the air film thickness is
reduced by the amount that the reflective surface curves up at the same spot, which is t2 ~(r2/2R2). So the air
film between the lens and surface has thickness ∆t~(r2/2R1) -(r2/2R2) r2[(R2- R1)/ R1R2]= m r=0.017 m
23. Consider the double slit arrangement shown where
the separation d=0.30 mm and the distance L= 1 meter. A
sheet of clear plastic (n=1.5) 0.050 mm thick (about the
plastic
thickness of this page) is placed over the upper slit. As a
sheet
y´
result the central max (=0) of the interference pattern moves d
L
upward a distance y’. Find this distance.
We know that y´/L=∆r/d y´=L∆r/d, where ∆r=m for a max.
What is meant by the phrase “the central max moves a distance y´” is that Ao is y’ above the central line. So
Ao
the total no. of wavelengths along paths r1 and r2 are the equal (=0), even though the actual distances are
different. This requires that the plastic sheet contain the same number of wavelength as in ∆r. Since ∆r =
mait, we must have t = mplastic = mait /nplastic. So y´=Ltn/d=1.45(0.05)/.3=0.24 mm.
