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3. Interference and Diffraction (Ch. 35, 36) 

Due: 11:59pm on Monday, February 18, 2013 

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Interference in Soap Films 

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Rutgers Analytical Physics 750:228, Spring 2013 ( RUPHYS228S13 ) 

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3. Interference and Diffraction (Ch. 35, 36) [ Edit ] 

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Description: Short quantitative problems on destructive interference in thin films. Based on Young/Geller Quantitative Analysis 26.2. 

The sketch shows a thin film of soapy water of uniform thickness and index of 

refraction suspended in air. Consider rays 1 and 2 emerging from the film: Ray 

1 represents the reflected wave at the air-film interface and ray 2 represents the wave 

that reflects off the lower surface of the film. 

Part A 

Which of the following statements correctly describes rays 1 and 2 and their path difference? 

Check all that apply. 

Hint 1. Phase shifts from reflection 

When light reflects off a surface with a higher index of refraction, it undergoes a phase shift of half a cycle, which corresponds to a shift 

of half a wavelength. However, no phase shift occurs if light reflects off a surface with a lower index of refraction. 

Hint 2. Path length 

The light that reflects off the film-air interface has to pass through the film, where it will have a different wavelength. The total "extra" 

distance it travels is approximately twice the thickness of the film (since the light first travels through the film toward the film-air 

interface and then reflects back out into the air). 

ANSWER: 

Instructor Resources eText Study Area 

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Part B 

Ray 1 undergoes a half-cycle phase shift. Ray 2 does not undergo any phase shift. 

Both rays 1 and 2 undergo a half-cycle phase shift. 

Ray 2 undergoes a half-cycle phase shift. Ray 1 does not undergo any phase shift. 

The path difference between rays 1 and 2 is about . 

There is no path difference between the two rays. 

The path difference between rays 1 and 2 is about . 

Therefore, when calculating interference effects between rays 1 and 2, keep in mind that the actual phase difference of the corresponding 

waves will be given by both the path difference and the phase shift undergone by ray 1. 

When red light is directed normal to the surface of the soap film, the soap film appears black in reflected light. What is thinnest the film can 

be? Let be the wavelength of red light in air. 

Hint 1. How to approach the problem 

If the film appears black in reflected light, it means that the reflected light has undergone destructive interference. Using the information 

found in the previous part, apply the correct condition for destructive interference and solve for the thickness of the film. 

Hint 2. Find the correct condition for destructive interference in this thin film 

Which of the following relations (where is a positive integer) represents the correct condition for destructive interference occurring in 

a thin film of soapy water suspended in air and of thickness ? Assume that light is directed normal to the surface of the film and let 

be its wavelength in the film. 

Hint 1. Destructive interference in thin films 

Consider a thin film of thickness . Light is directed normal to the film surface. When either one of the two waves resulting from 

reflection at the upper and lower surface of the film undergoes a phase shift, destructive interference occurs when 

where is the wavelength of light in the film and is a positive integer ( ). 

Similarly, when neither or both of the two waves resulting from reflection at the upper and lower surface of the film undergo a 

phase shift, destructive interference occurs when 

where, again, is the wavelength of light in the film and . 

ANSWER: 

Since you need to find the thinnest film that can cause destructive interference, use this condition with and solve for . 

Recall that the wavelength of light in water is different than in air. 

Hint 3. Find the wavelength of red light in the soap film 

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, 

,


If the wavelength of red light in air is , what is the wavelength of the same light in soapy water of refractive index ? 

Express your answer in terms of . 

Hint 1. Wavelength of light in a material 

The wavelength of light traveling in a material of index of refraction is less than the wavelength of the same light in 

vacuum. In particular, the wavelength of light in the material is 

where is the wavelength in vacuum. Note that the wavelength of light in air is essentially the same as that in vacuum. 

ANSWER: 

ANSWER: 

If we use white light instead of red, this film will reflect all colors that interfere constructively, but red will not be one of them. 

Thin Film (Oil Slick) 

Description: This problem explores thin film interference for both transmission and reflection. 

A scientist notices that an oil slick floating on water when viewed from above has many different rainbow colors reflecting off of the surface. She 

aims a spectrometer at a particular spot, and measures the wavelength to be 750 nanometers (in air). The index of refraction of water is . 

Part A 

= 

The index of refraction of the oil is . What is the minimum thickness of the oil slick at that spot? 

Express your answer in nanometers, to three significant figures. 

Hint 1. Thin film interference 

In thin films, there are interference effects because light reflects off the two different surfaces of the film. In this problem, the scientist 

observes the light that reflects off the air-oil interface, and off of the oil-water interface. Think about the phase difference that is created 

between these two rays. The phase difference will come from differences in path-length, as well as differences that are introduced by 

certain types of reflection. Recall that if the phase difference between two waves is (a full wavelength) then the waves interfere 

constructively, while if the phase difference is (half of a wavelength) the waves interfere destructively. 

Hint 2. Path-length phase difference 

The light that reflects off the oil-water interface has to pass through the oil slick, where it will have a different wavelength. The total 

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,


Part B 

Part C 

"extra" distance it travels is twice the thickness of the slick (once as it moves toward the oil-water interface, and once as it reflects back, 

out into the air). 

Hint 3. Phase shift due to reflections 

Recall that when light reflects off a surface with a higher index of refraction, it gains an extra phase shift of radians. What used to be 

a maximum is now a minimum! Be careful, though; if two beams each reflect off a surface with a higher index of refraction, they will 

both get a phase shift, canceling out that effect. 

ANSWER: 

= 313 

Suppose the oil had an index of refraction of 1.50. What would the minimum thickness be now? 


Hint 1. Phase shift due to reflections 

Keep in mind that when light reflects off a surface with a higher index of refraction, it gains an extra phase shift. What used to be a 

maximum is now a minimum! Be careful, though; if two beams reflect, they will both get a phase shift, canceling out that effect. Also, 

reflection off a surface with a lower index of refraction yields no reflection phase shift. 

ANSWER: 

= 125 

Now assume that the oil had a thickness of 200 and an index of refraction of . A diver swimming underneath the oil slick is looking at 

the same spot as the scientist with the spectromenter. What is the longest wavelength of the light in water that is transmitted most easily 

to the diver? 


Hint 1. How to approach this part 

For transmission of light, the same rules hold as before, only now one beam travels straight through the oil slick and into the water, 

while the other beam reflects twice; once off the oil-water interface, once again off of the oil-air interface, before being finally 

transmitted to the water. 

Hint 2. Wavelength of light in air 

Find the wavelength of the required light, in air. 

Express your answer numerically in nanometers. 

ANSWER: 

Hint 3. The wavelength of light in water 

There is a simple relationship between the wavelength of light in one medium (with one index of refraction) and the wavelength in 

another medium (with a different index of refraction). 

ANSWER: 

= 600 

= 451 

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This problem can also be approached by finding the wavelength with the minimum reflection. Conservation of energy ensures that 

maximum transmission and minimum reflection occur at the same time (i.e., if the energy did not reflect, then it must have have been 

transmitted in order to conserve energy), so finding the wavelength of minimum reflection must give the same answer as finding the 

wavelength of maximum transmission. In some cases, working the problem one of these ways may be substantially easier than the other, 

so you should keep both approaches in mind. 

Using a Michelson Interferometer 

Description: A Michelson interferometer is used to find the wavelength of a laser and then the index of refraction for an unknown fluid. 

You are asked to find the index of refraction for an unknown fluid, using only a laser and a Michelson interferometer. A Michelson interferometer 

consists of two arms--paths that light travels down, which end in mirrors-- attached around a beam splitter. The beam splitter separates the 

incoming light into two separate beams and then recombines them once they return 

from the ends of the arms. The recombined beams are sent to a telescope, where their 

interference pattern may be observed in detail. 

Part A 

Part B 

First, you must find the wavelength of the laser. You shine the laser into the interferometer and then move one of the mirrors until you have 

counted fringes passing the crosshairs of the telescope. The extremely accurate micrometer shows that you have moved the mirror by 

millimeters. What is the wavelength of the laser? 

Express your answer in nanometers, to four significant figures. 

Hint 1. Relating wavelength and distance in a Michelson interferometer 

Since the change in path difference for a Michelson interferometer comes from both the increased distance that light travels to the 

moving mirror and the distance traveled back from it, the equation for how far the mirror has moved takes the form , where 

is the distance traveled by the mirror, is the number of fringes that have moved past the crosshairs, and is the wavelength of the 

light. 

ANSWER: 

= 632.8 

You now immerse the interferometer in a tank filled with some unknown liquid and carefully align the laser into the interferometer. You move 

the mirror until you count 100.0 fringes passing the crosshairs of the telescope. The micrometer indicates that the mirror has moved 

millimeters. What is the mystery fluid? 

Hint 1. Find the index of refraction 

What is the index of refraction for this liquid? 

Express your answer to four significant figures. 

Hint 1. Find the wavelength of the light in the fluid 

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Find the wavelength of the light in the fluid. To do this, solve the equation from Part A for . 

Express your answer in nanometers, to four significant figures. 

ANSWER: 

Hint 2. Relating wavelengths and the index of refraction 

Recall that the wavelength of light in a medium is related to its wavelength in air by the relation , 

where is the index of refraction for the medium. 

ANSWER: 

ANSWER: 

Understanding Fraunhofer Diffraction 

Description: The equations for Fraunhofer diffraction are derived. Then, a simple numerical problem is worked. 

Learning Goal: 

= 1.329 

= 476.2 

water ( ) 

methanol ( ) 

ethanol ( ) 

acetone ( ) 

isopropyl alcohol ( ) 

saline ( ) 

To understand the derivations of, and be able to use, the equations for Fraunhofer diffraction. 

Diffraction is a general term for interference effects related to edges or apertures. Diffraction is more familiar in waves with longer wavlengths than 

those of light. For example, diffraction is what causes sound to bend around corners or spread as it passes through a doorway. Water waves 

spread as they pass between rocks near a rugged coast because of diffraction. Two different regimes for diffraction are usually identified: Fresnel 

and Fraunhofer. 

Fresnel diffraction is the regime in which the diffracted waves are observed close (as compared to the size of the object causing the diffraction) to 

the place where they are diffracted. Fresnel diffraction is usually very complicated to work with. The other regime, Fraunhofer diffraction, is much 

easier to deal with. Fraunhofer diffraction applies to situations in which the diffracted waves are observed far from the point of diffraction. This 

allows a number of simplifying approximations to be used, reducing diffraction to a very manageable problem. 

An important case of Fraunhofer diffraction is the pattern formed by light shining through a thin slit onto a distant screen (see the figure). 

Notice that if the light from the top of the slit and the light from the bottom of the slit 

arrive at a point on the distant screen with a phase difference of , then the electric 

field vectors of the light from each part of the slit will cancel completely, resulting in a 

dark fringe. To understand this phenomenon, picture a phasor diagram for this scenerio 

(as show in the figure). 

A phasor diagram consists of vectors (phasors) with magnitude proportional to the 

magnitude of the electric field of light from a certain point in the slit. The angle of each 

vector is equal to the phase of the light from that point. These vectors are added 

together, and the resultant vector gives the net electric field due to light from all points in 

the slit. In the situation described above, since the magnitude of the electric field vectors 

is the same for light from any part of the slit and the angle of the phasors changes 

continuously from to , the phasors will make a complete circle, starting and ending 

at the origin. The distance from the origin to the endpoint of the phasor path (also the 

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origin) is zero, and so the magnitude of the electric field at point is zero. 

Part A 

Part B 

Part C 

One reason that Fraunhofer diffraction is relatively easy to deal with is that the large distance from the slit to the screen means that the light 

paths will be essentially parallel. 

Therefore, the distance marked in the figure is the entire path-length difference 

between light from the top of the slit and light from the bottom of the slit. What is the 

value of ? 

Express your answer in terms of the slit width and the angle shown in the 

figure. 

Hint 1. A useful triangle 

Observe that the triangle with sides of length and in the figure is a right triangle. Since you know the length of the hypotenuse and 

one of the angles, you can use basic trigonometry to find the value of . 

ANSWER: 

= 

As described in the problem introduction, a criterion for a dark band to appear at point is that the phase difference between light arriving at 

point from the top of the slit and light arriving at point from the bottom of the slit equal . What length of path difference will give a 

phase difference of ? 

Express your answer in terms of the wavelength . 

ANSWER: 

= 

Combining your answers from Parts A and B gives the criterion for a dark band in the diffraction pattern as . 

Consider the phasor diagram from the introduction. The magnitude of the electric field at a point will equal zero as long as the endpoint for the 

phasor diagram is the origin. Thus, a point with a phasor diagram that goes around a circle twice, for example, ending at the origin, will be 

another location for a dark band. This idea can be used to modify the equation for the location of a dark band by introducing a variable : 

What is the complete set of values of for which this equation gives criteria for dark bands? 

ANSWER: 

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.


Part D 

Part E 

Part F 

The value corresponds to , which is the center of the diffraction pattern. The center of the diffraction pattern is a bright band. 

To see why, notice that if the phase difference from top to bottom is zero, then the phasor diagram will just be a straight line segment 

pointing away from the origin. This gives the maximum possible intensity in the diffraction pattern. 

What are the angles for the two dark bands closest to the central maximum. 

Express your answers in terms of and . Separate the two angles with a comma. 

ANSWER: 

The equation for the angles to dark bands is valid for any angle from to . In practice, the bright bands at large angles are usually so 

dim that the diffraction pattern appearing on a screen is invisible for such angles. For small angles, it is easy to find the distance from the 

center of the diffraction pattern to the dark band on the screen corresponding to a particular value of . 

For small angles, . Since , the small-angle approximation yields . By solving the dark-band 

criterion, you obtain . Setting the two expressions for equal gives the formula for the position (i.e., distance from the 

center of the diffraction pattern) of dark bands: 

or equivalently, 

Assuming that the angle between them is small, what is the distance between the two dark bands closest to the center of the diffraction 

pattern? 

Express your answer in terms of , , and . 

ANSWER: 

, , , 

, , , , 

, , , 

, , , , 

any rational number 

, 

Also accepted: , 

= 

Suppose that light from a laser with wavelength 633 is incident on a thin slit of width 0.500 . If the diffracted light projects onto a 

screen at distance 1.50 , what is the distance from the center of the diffraction pattern to the dark band with ? 

Express your answer in millimeters to two significant figures. 

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, 

.


ANSWER: 

± Single-Slit Diffraction 

Description: ± Includes Math Remediation. The problem describes a typical single-slit diffraction setup, including a picture of the diffraction 

pattern produced. The student is asked to calculate (numerically) the slit width. 

You have been asked to measure the width of a slit in a piece of paper. You mount the paper centimeters from a screen and illuminate it from 

behind with laser light of wavelength nanometers (in air). You mark two of the intensity minima as shown in the figure, and measure the 

distance between them to be millimeters. 

Part A 

Part B 

= 3.80 

What is the width of the slit? 

Express your answer in micrometers, to three significant figures. 

Hint 1. The equation for single-slit diffraction 

The equation for the angle between a line perpendicular to the screen and a line connecting the slit to the th dark fringe is 

, where is a nonzero integer, is the wavelength of the light, and is the width of the slit. Use this equation, together 

with some small-angle approximations, to solve this problem. 

Hint 2. Small-angle approximations 

Recall that for a small angle , . For this problem, find the value of , in terms of the distance to the 

screen and the distance from the central maximum to the dark fringe, and then substitute this expression for in the equation 

for location of the dark fringes. 

ANSWER: 

= 170 

If the entire apparatus were submerged in water, would the width of the central peak change? 


If you are puzzled, look at each variable in your equations and ask yourself whether it would be changed by placing the whole 

apparatus underwater. If you identify any of the variables as changing, think about how such a change would affect the width of the 

central fringe (if at all). 

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ANSWER: 

Doorway Diffraction 

Description: Use Fraunhofer diffraction to calculate the angle within which a person would not hear a sound from a room as it passes through 

a door. 

Part A 

The width would increase. 

The width would decrease. 

The width would not change. 

Sound with frequency 1280 leaves a room through a doorway with a width of 1.18 . At what minimum angle relative to the centerline 

perpendicular to the doorway will someone outside the room hear no sound? Use 344 for the speed of sound in air and assume that the 

source and listener are both far enough from the doorway for Fraunhofer diffraction to apply. You can ignore effects of reflections. 

Express your answer in radians. 


For diffraction patterns in light there are a number of dark fringes in which waves of light completely cancel each other. For sound 

waves this cancellation corresponds to silent spots at certain angles. 

In this problem, we would like to know the angle at which the sound is completely canceled out. In order to determine this angle, we 

first need to find the wavelength of the sound wave. This can be determined from the speed of sound and the frequency. You can then 

determine the angle to the first destructive diffraction point from the standard Fraunhofer diffraction equation. 

Hint 2. The equation for "dark" fringes 

For single-slit Fraunhofer diffraction, the equation that relates the angle between subsequent dark (in this case silent) fringes and the 

centerline to the wavelength of the incoming sound wave is 

where is the wavelength of the sound, is the width of the diffracting slit, and is the angle between the th dark fringe and the 

centerline of the image (see the figure). 

Hint 3. Find the wavelength of the sound wave 

Calculate the wavelength of the sound wave. 

Express your answer in meters. 

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,


Hint 1. Relation between wavelength and velocity 

Recall that wavelength and velocity of a wave are related by the equation , where is the velocity of the wave, is the 

wavelength, and is the frequency. 

ANSWER: 

ANSWER: 

Problem 35.44 

Description: A very thin sheet of brass contains two thin parallel slits. When a laser beam shines on these slits at normal incidence and room 

temperature (20.0 degree(s) C), the first interference dark fringes occur at pm32.5 degree(s) from the original... 

A very thin sheet of brass contains two thin parallel slits. When a laser beam shines on these slits at normal incidence and room temperature (20.0 

), the first interference dark fringes occur at 32.5 from the original direction of the laser beam when viewed from some distance. 

Part A 

Part B 

If this sheet is now slowly heated up to 135 , by how many degrees do these dark fringes change position? Coefficient of linear expansion 

for brass .Ignore any effects that might occur due to change in the thickness of the slits. (Hint: Since thermal expansion 

normally produces very small changes in length, you can use differentials to find the change in the angle.) 

Express your answer using two significant figures. 

ANSWER: 

Do they move closer together or get farther apart? 

ANSWER: 

= = 0.269 

= 8.4×10 −2 

= 0.230 radians 

dark fringes get farther apart 

dark fringes move closer together 

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