Important points

• Cameras create a inverted, real imageon film or CCD.

• Eyes use biological fluids to create alens that focuses light onto the retina.

• Magnifiers change the size of imagesusing lenses; so do microscopes (2lenses) and telescopes.

Important equations

• Angular magnification of a magnifier

M = θ/

θ′

= (25 cm)/

s

= (1/ f − 1/s′)× 25 cm

• Microscope

M = m1 M2

=(25 cm) s′1

f1 f2

PHYS 202 Notes, Week 10Greg Christian

March 28 & 30, 2016Last updated: 03/31/2016 at 12:28:48

This week we learn about optical instruments, interference, and thinfilms.

Optical Instruments

Cameras

A camera is a device that is able to capture images and store themeither on film or in a digital medium. Figure 1 shows a diagram of howa camera operates. The light coming off the object to be photographedpasses through a lens, creating a real and inverted image on the film (fora traditional camera) or CCD array (for a digital camera). The imageis then un-flipped by developing the film or with digital processingsoftware, to create what you see when looking at photos.

In designing a camera, there are a couple of things which can bealtered to affect the image that forms. One is the intensity of lightreaching the film or CCD array, which can be adjusted by changingthe size of the aperture surrounding the lens. The size of this apertureis usually given in terms of something called the f -number,

f − number =Focal length

Aperture diameter=

fD

, (1)

where f is the focal length and D is the aperture diameter.

Figure 1: Operation of a camera.

As you probably know, cameras can take images of objects at manydifferent distances. This is accomplished by moving the lens in or out.This in turn adjusts the image distance. The goal is to place the lenswhere the image distance matches the recording medium. Then the

phys 202 notes, week 10 2

camera is “in focus”. Most modern cameras do this automatically, butolder and/or high-end cameras still rely on manual focusing by thephotographer.

Eyes

Figure 2: Operation of an eye.

Figure 2 shows the operation of the human eye. Basically, the eyeworks by forming a lens made up of biological material, focusing in-coming light onto the retina where rods and cones sense the image andtransmit it to the brain. The parts of the eye forming the lens are (inorder):

• The aqueous humor, index of refraction n = 1.336.• The crystalline lens, index of refraction n = 1.437.• The vitreous humor, index of refraction n = 1.336.

In order to form a sharp image, the image point of the eye needs tobe exactly on the retina surface. Given that the retina is in a fixed loca-tion, the image distance, s′ is always fixed. In order to accommodatedifferent object distances, s, the eye changes its effective focal length. Itdoes that by contracting or relaxing a muscle called the ciliary muscle,which has the effect of changing the radius of curvature, and thus thefocal length of the eye.

Contracting the ciliary muscle serves to decrease the radius of curva-ture, and thus focal lngth, of the eye. This allows up-close objects to bein focus. Then the ciliary muscle is relaxed, the focal length increases,creating focus on far-away objects. This process is called accomodation.

As people age, the ciliary muscle loses its ability to work, decreasingthe range in which the focal length can be changed. This is why even-tually nearly everyone starts to have difficulty with seeing up-close

phys 202 notes, week 10 3

Figure 3: Myopia and hyperopia.

objects without reading glasses. The term near point describes the clos-est distance upon which the eye can focus. For the average person thisdecreases from about 7 cm at age 10 to 200 cm at age 60.

The eye also needs to accommodate different intensities of light. Itdoes this by dilating or contracting the pupils, to let more or less lightin.

Vision problems

Many people have eyes which are either too long or too short. Thiscauses light to be focused either in front of or behind the retina. Thefirst case is called myopia, or nearsightedness, because it causes imagesof far-away objects to appear out of focus. The second is called hyper-opia, or farsightedness, because it causes close-up images to be out offocus. Drawings of these are shown in Figure 3.

Each of these vision problems can be fixed using lenses. In the caseof hyperiopia, we can use a converging lens to focus the image ontothe retina (Figure 4), thus creating a clear image. In the case of myopia,we use a diverging lens to achieve the same result (Figure 5).

Figure 4: Hyperopia correction.

Magnifiers

Another optical instrument is a magnifying glass, or simply a magnifier.This instrument uses a converging lens to form a virtual image that islarger and farther away than the original. Magnifiers are typicallyanalyzed in terms of the angular size, or the total angle subtended bythe object, as shown in the left of Figure 6.

When dealing with magnifiers, we are concerned with the angularmagnification, or the change in angular size of the object. This is equal

phys 202 notes, week 10 4

Figure 5: Myopia correction.

to

M =θ′

θ. (2)

For a magnifier with focal length f , we can calculate the angularmagnification as

M =(25 cm)

s(3)

=

(1f− 1

s′

)× 25 cm, (4)

where

• s is the object distance; and• We assume that the near point of the eye is 25 cm.

Figure 6: Operation of a magnifier.

Microscopes

To look at very small objects, a simple magnifier isn’t enough. Herewe need a microscope. The operation of a microscope is shown in

phys 202 notes, week 10 5

Figure 7: Operation of a microscope.

Figure 7.A microscope consists of two lenses:

1. The objective lens, which has focal length f1.2. The eyepiece, which has focal length f2.

It works in the following way:

1. The objective lens creates a real, inverted image inside the focal pointF2 of the eyepiece.

2. The eyepiece uses the image I as its object and creates an enlarged,virtual image I′ which is still inverted.

The most important quantity for a microscope is the overall angularmagnification. This is given by the following equation,

M = m1M2 (5)

=(25 cm) s′1

f1 f2, (6)

where

• m1 is the lateral magnification of the objective lens (focal length f1).• M2 is the angular magnification of the eyepiece (focal length f2).

Telescopes

A telescope allows viewing of objects that are large, but very far away.One type of telescope is called the astronomical telescope, and it has avery similar design to a microscope, namely a combination of objectivelens and eyepiece. An example is shown in Figure 8.

Figure 8: Operation of a astronomicaltelescope.

Because the object to be viewed is so far away, the first image isformed nearly at the second focal point of the objective lens ( f1). Thisimage is then the object for the eyepiece. Ideally, we want this to forma final image at infinity, which means that the first image needs to belocated at the focal point of the eyepiece ( f2). This means that the totallength of the telescope will be equal to the sum of the focal lengths,f1 + f2.

phys 202 notes, week 10 6

The overall angular magnification for a telescope can be workedout from the geometry shown in Figure 8. The result is that it’s simplyproportional the the ratio of focal lengths (with a negative sign),

M = − f1

f2. (7)

Note that due to the negative sign, the telescope forms an invertedimage. This can be corrected using a device called prism binoculars,shown in Figure 9. These are placed between the objective and theeyepiece and serve to flip the image (making it upright) by way offour internal reflections.

Figure 9: Prism binoculars, to make theimage formed by a telescope upright.

For practical reasons, it’s often more feasible to construct telescopesout of reflecting mirrors rather than lenses. These devices are calledreflecting telescopes. Some schemes for constructing reflecting telescopesare shown in Figure 10.

Figure 10: Examples of reflecting tele-scopes.In one scheme, (part a in the figure), there is a “cage” inside the tele-

scope which receives the reflected light, typically in a camera. This de-sign is most practical for very large telescopes because the cage blocksincoming light rays.

phys 202 notes, week 10 7

Important points

• Constructive interference occurswhen waves are in phase.

• Destructive interference occurs whenwaves are 1/2 cycle out of phase.

• Double slits and thin films can bothform interference patterns.

Important equations

• Interference

r2 − r1 = mλ (constructive)

r2 − r1 = (m + 1/2) λ (destructive)

• Double-slit experiment

d sin θ = mλ (bright)

d sin θ = (m + 1/2) λ (dark)

ym = Rmλ/

d (bright)

ym = R (m + 1/2) λ/

d (dark)

• Thin film interference

2t = mλ (constructive)

2t = (m + 1/2) λ (destructive)

λ = λ0/nfilm

Figure 11: Example of in- and out-of-phase waves.

Other designs (parts b and c in the figure) use a secondary mirror toreflect light either along the direction of incoming light (which requiresa hole in the primary mirror), or out the side of the telescope (part c).

Optical Interference

So far, we’ve been working with the ray model of light, which treatslight as something traveling along a straight line path. Now we’ll moveinto talking about some phenomena which rely on the wave nature oflight.Recall that light is a form of electromagnetic wave, consisting ofoscillating

#»E and

#»B fields.

Interference is the phenomena where two or more waves overlap inspace. Related to this is something called the Principle of Superposi-tion:

When two or more waves overlap, the resulting displacement at anypoint and any time is found by adding the instantaneous displacementsthat would be produced at that place and time if the individual waveswere each present alone.

In other words, you can basically just sum up the individual displace-ments to get the total result. In the case of light, what we mean by “dis-placement” is the magnitude of the electric or magnetic field formingthe wave.

One concept that we need to introduce is that of phase. Phase ba-sically tells whether or not the peaks and troughs making up a waveare lined up or not. If two waves are “in phase”, the displacementsare in step with each other. If they are “one cycle out of phase”, thedisplacements are exactly counter-aligned: the peak of one wave linesup with the trough of the other. Examples are shown in Figure 11.

Depending on the phase of two waves, they can interfere either con-structively or destructively. Constructive interference (or “reinforce-ment”) happens when the waves are in phase, resulting in an increasein amplitude. The condition for constructive interference to happenis that the path difference from the two sources must line up with aninteger number of wavelengths:

r2 − r1 = mλ (m = 0, ± 1, ± 2, ± 3, . . .), (8)

with

• The resulting amplitude is the sum of the amplitudes of wave 1 andwave 2;

• r1 and r2 are the distances from the wave source to the point ofoverlap; and

• Each of the waves has wavelength λ.

phys 202 notes, week 10 8

Figure 12: Wave interference.

1 Often referred to as “Young’s Double-Slit Experiment”.

An example of this is shown in part (b) of Figure 12.The opposite phenomena, destructive interference (or “cancella-

tion”), happens when the waves are exactly a half cycle out of phase.This results in a reduced amplitude. For destructive interference tohappen, the path differences align with a half integer multiple of thewavelength,

r2 − r1 = (m + 1/2) λ (m = 0, ± 1, ± 2, ± 3, . . .), (9)

with the following being true:

• If the two waves have the same amplitude, they completely canceleach other.

• r1 and r2 are the same as in Eq. (8).

An example of this is shown in part (c) of Figure 12.

Double-Slit Interference

One of the first experiments showing interference of light waves wasthe double-slit experiment performed in 1800 by Thomas Young.1 Agraphical representation of Young’s experiment is shown in Figure 13.The experiment consists of the following steps:

1. Monochromatic light passes through a single slit S0, creating a cylin-drical wave front on the other side (as a consequence of Huygen’sprinciple).

2. The wave front travels to two separate slits S1 and S2 which areat different vertical distances from S0. This creates two separatecylindrical wave fronts on the other side.

3. These two wavefronts travel to a screen where they are observed.

phys 202 notes, week 10 9

Figure 13: Thomas Young’s double-slitexperiment.

Figure 14: Fringes deom the double slitexperiment.

The result, observed on the screen in the last step, is a series ofbright and dark bands. The bright bands occur when the light fromS1 and S2 traveled some integer multiple of the wavelength, satisfyingEq. (8) and giving constructive interference. Likewise, the dark bandsoccur when the distances satisfy Eq. (9) (half integer multiple of λ) andgive destructive interference.

The path lengths in the bottom of Figure 13 can be analyzed to fig-ure out the angles for which constructive and destructive interferenceoccur. From part (c) in the figure, the difference in path lengths isgiven by

r2 − r1 = d sin θ, (10)

where d is the distance from the slits to the screen. Combining withEq. (8), we can figure out the angles for which constructive interfer-ence occurs. These are given by

d sin θ = mλ (m = 0, ± 1, ± 2, ± 3, . . .). (11)

Similarly, we can work out the location of destructive interferencefrom Eqns. (10) and (9):

d sin θ = (m + 1/2) λ (m = 0, ± 1, ± 2, ± 3, . . .). (12)

phys 202 notes, week 10 10

Figure 15: Thin film interference.

The bands that form on the screen are often called fringes. A pictureof what these look like is shown in Figure 14. Sometimes it’s usefulto have an equation for the vertical positions of these fringes, ym. Thiscan be worked out from Eq. (11) and the geometry of the problem,

ym = Rmλ

d(m = 0, ± 1, ± 2, ± 3, . . .), (13)

where R is the distance from the slits to the screen and d is the distancebetween the slits.

Note that Eq. (13) gives the position of the bright fringes. Similarly,the position of the dark fringes is given by

ym = R(m + 1/2) λ

d(m = 0, ± 1, ± 2, ± 3, . . .), (14)

Thin Film Interference

Another way to create interference patterns is to pass light through athin film of refracting material, as shown in the top of Figure 15. Whenthis happens, part of the light is reflected directly from the top of thefilm to the observer’s eye, while the rest is refracted; reflected at thebottom of the film; and then refracted to the observer. This processcreates a path difference resulting in interference.

Since the path length is fixed, we get either constructive or destruc-tive interference depending on the wavelength of the light. The for-mation of constructive or destructive interference is described by thefollowing equations:

Constructive

2t = mλ (m = 0, ± 1, ± 2, ± 3, . . .) (15)

Destructive

2t = (m + 1/2) λ (m = 0, ± 1, ± 2, ± 3, . . .), (16)

where t is the thickness of the film. Note that the wavelength λ inEqns. (15) and (16) is the wavelength in the film medium. This isrelated to the incoming wavelength λ0 by

λ = λ0/nfilm. (17)

This is why rainbow-like patterns form on a thin film, such as theoil slick shown in Figure 15(b). In this case, the incoming light is madeup of many different colors (each a different wavelength). Some ofthe colors have the right wavelength to give constructive interference;others give destructive interference. Those colors interfering construc-tively are the ones that we see as the bright rainbow spots.

In the example above, we discussed thin film interference of lightof all colors. Figure 16 shows another situation where thin film inter-ference can occur on monochromatic (single color/wavelength) light. It

phys 202 notes, week 10 11

Figure 16: Thin film reflection ofmonochromatic light due to a gap be-tween refracting materials.

involved stacking two refracting materials close together with an an-gled gap. The thickness of the gap at any point is labeled t. The seriesof events leading to interference patterns is:

1. Some of the incoming light is reflected back to the observer by thetop surface.

2. The rest travels through thickness t before reflecting off the bottomsurface.

3. In the second reflection, the light undergoes a half-cycle phaseshift. This happens whenever light reflects off a surface with nb >

na.4. The reflected light travels back to the observer, with path length

difference 2t and phase shift of 1/2.5. Interference patterns occur, but the patterns are the opposite of what

we would expect if there wasn’t a phase change. In other words,Eqns. (15) and (16) are interchanged in terms of whether they referto constructive or destructive interference.

Note that in general, whether Eqns. (15) and (16) apply to construc-tive or destructive interference depends on whether or not the first andsecond reflections induce a phase shift. To summarize,

• Neither reflection induces a phase shift

– Eq. (15) constructive– Eq. (16) destructive

• Only one reflection induces a phase shift

– Eq. (16) constructive– Eq. (15) destructive

• Both reflections induce a phase shift

– Eq. (15) constructive– Eq. (16) destructive.