Physics 3LC Final review

a) Pigmented Epithelium

a) Pigmented Epithelium

READING 2

Pulsed ruby lasers have been used to repair retinal tears; the photocoagulation process takes about 1.0 milliseconds. With argon lasers, which are also used, photocoagulation takes about 100 milliseconds. After the iris is dilated, a laser focal spot ~ 100 micrometers in diameter is imaged on the loose tissue. The actual weld made by the laser occurs on the pigmented epithelium (the layer furthest to the back of the retina) because the first layers of the retina are transparent at optical frequencies. The mechanism of photocoagulation is essentially thermal, and a small scar is left at the laser focal spot. The green light of the argon laser is preferred over the red light of pulsed ruby lasers because green light is more heavily absorbed in the pigmented tissue and somewhat sharper foci are possible, with 50 micrometers being the nominal limit. Since the fovea of the retina is approximately 1000 micrometers across, the decreased size of the scar left by the argon laser can be significant in treatment of the patient's problem [3].

Where does the actual 'weld' take place on the retina, based upon the physiology of the eye?

a) Pigmented Epithelium

b) Sphincter or Constrictor Pupillae

c) Stroma of Cornea

d) Outer Epithelium

e) None of the above

d) A desk lamp is an isotropic source. A laser is a collimated source.

Which statement is correct?

a) A desk lamp is a isotropic source. A laser is an isotropic source.

b) A desk lamp is a collimated source. A laser is an isotropic source.

c) A desk lamp is a collimated source. A laser is a collimated source.

d) A desk lamp is an isotropic source. A laser is a collimated source.

e) None of the above

d) It is a constant independent of distance.

How does the flux of light from an collimated source depend on the distance r from the source?

 

a) r^-1

b) r¹

c) r^-2

d) It is a constant independent of distance.

d) r²

e) None of the above

d) All of the factors above affect visual acuity

READING 3

The light sensing retina consists of a collection of sensors called rods and cones, which are packed closely together on the retinal surface. The human eye contains about 6 x 1066 x 106 cones and 120 x 106120 x 106 rods. The rods and cones are not distributed evenly over the retinal surface; the fovea is a region of the retina that contains only cones, which are tightly packed. These cones are ineffective at low levels of illumination, but are responsible for all the visual information concerning color.

Our ability to see is described in terms of how well we can distinguish two closely spaced point sources, which is referred to as the acuity of the eye. Maximum acuity is possible only for light that strikes the fovea, where the density of receptors is exceptionally large. Highest acuity is therefore obtained when viewing objects on the visual axis within a very narrow cone. If the image falls upon the retina outside the fovea, the acuity is considerably decreased because of the lower density of receptors. Secondly, acuity is influenced by the level of illumination; if light is intense, the low-sensitivity cones are effective. If the level of light is low, the cones fail, and since there are no rods in the fovea, the acuity in this region falls.

Without this occurring, the two point sources would not be resolved - they would appear as only one point source to the viewer.

What factors DO NOT affect visual acuity?

a) Region of retina image falls upon

b) Density of receptors

c) Level of illumination

d) All of the factors above affect visual acuity

a) Because the laser beam is intense and has very low divergence its energy can be focused onto a small spot.

INTRODUCTION: MEDICAL LASERS AND ACUITY

Three types of lasers used medically are: the argon laser, the carbon dioxide CO2 laser and the yttrium-aluminum-garnet (YAG) laser. Each laser, due to its physical characteristics has found its way into different medical applications. The argon laser is a lower powered laser. It is absorbed quickly by hemoglobin and is very effective in stopping bleeding where many small blood vessels are clustered. It coagulates bleeding ulcers, destroys vascular tumors of the bladder and cleans up lesions like the portwine birthmark stain. The CO2 laser is high powered. When teamed with an operating microscope, it is an excellent high-precision surgical knife because it only cuts 0.1 mm for every pass made by the surgeon. The CO2 laser is outstanding in neurosurgical work on vulnerable areas such as the brain and spinal cord. While annihilating the cell it touches, it leaves neighboring cells untouched - a plus when vaporizing malignant brain tumors. It is also used to treat cervical cancer, remove nodes from the larynx and open up blocked fallopian tubes. The YAG laser can penetrate very deeply within tissue. Because the YAG passes through clear liquid, it can be used in fluid-filled cavities such as the eyeball. Since it can stop the bleeding in large vessels, it was first used on patients hemorrhaging from stomach ulcers. While it can vaporize large tumors, it cannot be used for precision surgery. Recent reports have this laser even destroying early bladder cancer [2].

READING 1

A cauterizing instrument such as a focused laser beam could have wide application in general surgical procedures. The organ or tissue to be excised can be cut free at the same time that the vascular input is sealed so as to minimize subsequent bleeding. (Major organs have been removed using laser surgery, and relatively few complications are reported [3].) The laser light is absorbed in a small volume of tissue in which the energy is quickly converted to heat. Direct contact of tissue by the laser is not required, and the depth and confinement of the laser instrumentation can be controlled. Because the laser beam is intense and has very low divergence its energy can be focused onto a small spot. When the temperature exceeds 100 degrees Celsius at the tissue, vaporization of tissue water can occur. As the laser power and doses are increased, greater amounts of steam are produced without much further increase in temperature. If sufficient steam is generated rapidly within the tissue, physical separation or cutting occurs. Tissues bordering the cut are heated sufficiently by dissipated thermal energy to effect cautery [4]. Thus, the effect is one of intense heating, with perhaps 1000 J being delivered to the lesion [3].

Why is damage to neighboring tissue minimized?

a) Because the laser beam is intense and has very low divergence its energy can be focused onto a small spot.

b) Because the laser is operated at an extremely low power level.

c) Because direct contact of the tissue by the laser confines the damage to those cells.

d) All of the above

a) There must be a cone producing less than the full response between two cones that produce the full response.

To resolve two point sources, what distribution of cones must occur where the image strikes the retina?

a) There must be a cone producing less than the full response between two cones that produce the full response.

b) There must be an overexcited cone between two cones that produce the full response.

c) Two adjacent cones must both produce the full response.

d) None of the above

d) r^-2 

How does the flux of light from an isotropic source depend on the distance r from the source?

a) r^1/2 

b) r¹ 

c) r^-1 

d) r^-2 

e) r²

f) None of the above

Normalized experimental value at 6.2 cm

—> 0.443
Theoretically expected normalized value at 6.2 cm

—> 0.416

A common technique in analysis of scientific data is normalization. The purpose of normalizing data is to eliminate irrelevant constants that can obscure the salient features of the data. The goal of this experiment is to test the hypothesis that the flux of light decreases as the square of the distance from the source. In this case, the absolute value of the voltage measured by the photometer is irrelevant; only the relative value conveys useful information. Suppose that in Part 2.2.2 of the experiment, students obtain a signal value of 185 mV at a distance of 4 cm and a value of 82 mV at a distance of 6.2 cm. Normalize the students' data to the value obtained at 4 cm. (Divide the signal value by 185.) Then calculate the theoretically expected (normalized) value at 6.2 cm.

0.686

A bothersome feature of many physical measurements is the presence of a background signal (commonly called "noise"). In Part 2.2.4 of the experiment, some light that reflects off the apparatus or from neighboring stations strikes the photometer even when the direct beam is blocked. In addition, due to electronic drifts, the photometer does not generally read 0.0 mV even in a dark room. It is necessary, therefore, to subtract off this background level from the data to obtain a valid measurement. Suppose the measured background level is 5.1 mV. A signal of 20.7 mV is measured at a distance of 29 mm and 15.8 mV is measured at 32.5 mm. Correct the data for background and normalize the data to the maximum value. What is the normalized corrected value at 32.5 mm?

8.73 cm

Suppose you want to take a chest X-ray with an X-ray source that has a divergence of 5^o . If the film is 1 meters from the (point) source, how big is the spot size at the film in centimeters?

_______________________________cm

a) Spectral Capacity - how much of the transmitted wavelengths are conveyed

READING 2

The length of the fiberscope is limited by the spectral capability (what wavelengths can be conveyed or transmitted) and spectral capacity (how much of those wavelengths are conveyed). Most optics can transmit visible light and some infrared radiation, but will transmit ultraviolet rays with only poor efficiency. The length of the fiber is limited mostly by spectral capacity; as the length of the lightguide increases, the percentage of input light transmitted decreases. This is due to absorption within the fiber and coupling losses at the exit and entry points [6]. There is a limit to the bending radius of the fiberscope; as the thickness increases, the flexibility decreases because the minimum bending adius becomes larger. However, the jacket that covers the fiber optics and protects it from chemical attack prevents this limit from being exceeded. Magnification is restricted to lower orders (x20) because the outline of the fibers can become superimposed upon the image.

The length of a fiberscope is mostly limited by what?

a) Spectral Capacity - how much of the transmitted wavelengths are conveyed

b) Jacket Thickness

c) Bending Radius 

d) Spectral Capability - what wavelengths can be conveyed or transmitted

d) Coherent; aligned fiber ends

READING 1

Because of the multiple reflections that occur within a single optical fiber, an image cannot be passed through a single fiber. But an image can be broken down into a series of fine dots of various shades of light and dark and each portion is sent through a single fiber [8]. The basic requirement to transfer an image is that the fibers must maintain identical relative positions on the input and output faces of the bundle [9]. This is called a coherent fiber bundle, and in this way, the variation of light intensity used to create the image is maintained. Thus, it is necessary to use a coherent bundle of fibers with aligned fiber ends in order to transmit the image through the angioscope. Since light does not require such an organization of fibers in order to illuminate the visualized area, incoherent bundles can be used. Incoherent bundles do not maintain identical relative positions on the input and output faces, and thus are not used for image transfer.

Which of the following describe the characteristics of a bundle of fiber used for image viewing?

a) Coherent; no specific alignment of the fiber ends is necessary

b) Incoherent; aligned fiber ends

c) Incoherent; no specific alignment of the fiber ends is necessary

d) Coherent; aligned fiber ends

c) Incoherent; no specific alignment of the fiber ends is necessary

Which of the following describe the characteristics of a bundle of fiber used for illumination?

a) Incoherent; aligned fiber ends

b) Coherent; no specific alignment of the fiber ends is necessary

c) Incoherent; no specific alignment of the fiber ends is necessary

d) Coherent; aligned fiber ends

d) 𝜃_bench = i+r

In Part 3.2.2 of the experiment, the angles are defined differently than in Lab Manual Fig. 3.1. In the experiment, the angles between the laser and the optical bench 𝜃_bench and between the laser and the normal to the mirror _𝜃mirror are measured (Lab Manual Fig. 3.7).

What is the relationship between 𝜃_bench and the customary angles? 

a) 𝜃_bench = r-i

b) 𝜃_bench = i-r

c) 𝜃_bench = r

d) 𝜃_bench = i+r

e) None of the above

d) Indirect; It uses a reagent that changes its optical properties in response to changes in the material of interest

READING 4

Another medical application of fiber optics is physical and chemical measurements. A typical miniature fiber optic pH sensor has been applied in opthamology. The operation of this sensor is based upon a color change of a pH-sensitive agent, which is contained within a small chamber with ion permeable walls. The subminiature chamber is attached to a standard optical fiber 150 micrometers in diameter. The measurements are performed by comparing colored light optical pulses to white light optical pulses. This miniature, elastic medical pH sensor measures pH changes in real time and in vivo. This technology has been further applied to perform calcium and other ion concentration measurements in inaccessible places [11].

Oxygen saturation can also be determined utilizing optical fiber sensors. Oxygen saturation refers to the amount of oxygen carried by the hemoglobin in red blood cells relative to the maximum carrying capacity (typical arterial blood is 95venous blood may be 75in the cardiovascular or cardiopulmonary system. Optical measurement of oxygen saturation depends on the difference in the absorption spectra of oxyhemoglobin and hemoglobin. The absorption coefficients of hemoglobin and oxyhemoglobin (absorption coefficient is a characteristic of the absorbant; the larger the coefficient, the better the absorption capacity of the medium) are sufficiently low at wavelengths above 620 nm that sufficient light transmission through whole blood can occur over distances that are compatible with the geometric requirements of practical fiber-optic catheters. At 650 nm, the signal is measured where there is a useful difference in absorbance between oxyhemoglobin and hemoglobin.

Optical sensors fall into two categories: those that sense changes in the specific optical properties of the material of interest (direct sensors) and those that use a reagent that changes its optical properties in response to changes in the material of interest (indirect sensors). The pH sensor is an indirect sensor because it utilizes a pH sensitive agent, which changes its optical properties in response to the pH. On the other hand, the oxygen saturation sensor is direct; it measures the difference in optical properties between the two materials of interest. Other fiber optic sensors include those which detect changes in flow, pressure, temperature, and other chemicals within the body.

Which type of sensor is the pH sensor? Why?

a) Direct; It senses changes in the specific optical properties of the material of interest

b) Indirect; It senses changes the specific optical properties of the material of interest

c) Direct; It uses a reagent that changes its optical properties in response to changes in the material of interest

d) Indirect; It uses a reagent that changes its optical properties in response to changes in the material of interest

a) Stiffness

READING 3

Laser coronary angioplasty has received considerable interest recently as a potential adjunct to coronary bypass surgery and transarterial treatment by balloon angioplasty. Angioplasty is a technique that removes interarterial plaque (buildup), thereby unclogging the artery for blood flow. The mechanism by which the laser removes the arterial plaque is by vaporization or thermal decomposition. Presently, this technique has been used in unclogging a femoral artery; however, the eventual aim is to perform laser angioplasty in the coronary artery; further development in the ability to view, steer, and aim inside small arteries is required [10].

A key factor for the effectiveness of a laser in medical applications is the laser delivery system. High transmission, light weight, small size, and flexibility make fiber optics well suited to deliver the laser power. The best optical fibers with regard to transmission, flexibility, and nontoxicity are the silica based fibers, used in conjunction with the Er:YAG laser.

What factors DO NOT make fiber optics well suited to deliver the laser power?

a) Stiffness

b) Light Weight

c) Small Size

d) High Transmission

e) All of the above factors make fiber optics well suited to deliver laser power

30.6

In Part 3.2.2 of the experiment, a pair of students measure 𝜃_mirror = 18 degrees. To find 𝜃_bench, they use the theoretical relationship 𝜃_mirror = 1/2 𝜃_bench. Then they measure 𝜃_bench = 47 degrees. What is the discrepancy between theory and experiment for 𝜃_bench?  Report your answer as a percent difference, do not type in "%". Round your answer to one digit after decimal point.30.6

2.3

The speed of light in a transparent medium is 1.3 x 10⁸ m/s. What is the index of refraction of this medium? Round your answer to one digit after decimal point.

i_crit = 73.5 degrees,

alpha_crit= 24.5 degrees

The index of refraction of the core of a typical fiber optic is n_core = 1.46; the cladding has n_clad = 1.4. Calculate the critical angles for the total internal reflection i_crit and 𝛼_crit . Round your answers to one digit after decimal point.

b) cornea; margin; center; lens

READING 3

One type of aberration that occurs in the eye is SPHERICAL ABERRATION (Lab Manual Figure 4.6). This is found on all lenses bound by spherical surfaces (the cornea). The marginal portions of the lens bring rays to a shorter focus than the central region. The image of a point is therefore not a point, but a small, blurred circle. The cornea is flatter at its margin than at its center, and the lens is denser in the center and hence refracts light more strongly at its core than at its outer layers [15].

For an image to come into focus, the light received by the eye must be bent so that the rays converge at the fovea. The closer the object being viewed, the more the light must be bent if the object is to be seen clearly. The cornea, vitreous humor, and aqueous humor have a fixed index of refraction, but the lens has the ability to adapt. The lens can sharpen the curvature of its front and back surfaces, increasing its focusing power. When the eye is focused upon infinity (for human beings, this begins about 20 feet away), ciliary muscles that encircle the equator of the lens relax, and thus expand. This causes the zonules (which connect the ciliary muscles to the lens) to be pulled taut, flattening out the front and back of the lens and increasing the diameter of the equator. In this state, the ability of the lens to bend light is at a minimum; at this point the combined refractivities of the components of the eye allow it to focus on a distant object. When the eye focuses on a closer object, ciliary muscles contract, relieving stress on the zoncules and the lens undergoes an elastic recovery. The lens becomes thicker, front to back, its surfaces become more sharply curved and the diameter of the equator shrinks. This gives the lens the added refractive power needed for focusing upon closer objects.

How has the eye accomplished correction for spherical aberrations? Fill in the blanks. The____________

is flatter at its___________________ than at its_____________ ,and the_________________ is denser in the center and hence refracts more strongly at its core than at its outer layers.

a) lens; center; margin; cornea

b) cornea; margin; center; lens

c) lens; margin; center; cornea

d) cornea; center; margin; lens

c) the lens

Where does the focusing in the eye of a fish occur?

a) Cornea - aqueous humor interface

b) Water - cornea interface

c) the lens

d) Air - cornea interface

READING 1

Consider the indices of refraction of the various materials of the eye.

The eye accomplishes focusing by bending light which is also known as refraction. Refraction of light is greatest at the interface of materials with widely differing indices of refraction. At the aqueous humor-lens or lens-vitreous humor interface, the indices of refraction differ by 0.07. However at the air-cornea interface, the indices differ by 0.38. Therefore, the lens is only of secondary importance for image formation for animals that live in air.

Under water, the human eye cannot focus sharply because the air-cornea interface is replaced by a water-cornea interface. For creatures that live underwater, the eye cannot rely on the corneal surface for focusing. Without the cornea, all of the focusing must be done by the lens, which has to be of much higher power than the human eye. There are several modifications that take place in the fish eye. First, since there is no focusing done by the cornea, its shape doesn’t matter; the eye can be bulbous or flat. The lens takes on a different shape - in humans, it is an ellipse (seen from the side), but in the fish it is spherical. Furthermore, the index of refraction for the fish lens is very high, typically 1.65.

Where does the majority of the focusing in the human eye occur? 

a) Cornea- aqueous humor interface 

b) Lens - vitreous humor interface

c) Aqueous humor - lens interface

d) Air - cornea interface

a) A convex lens has both sides rounded outward causing the light rays to converge towards one another. A concave lens has both sides rounded inward, causing the light rays to diverge away from one another.

Which of the following statements is correct?

a) A convex lens has both sides rounded outward causing the light rays to converge towards one another. A concave lens has both sides rounded inward, causing the light rays to diverge away from one another.

b) A convex lens has both sides rounded outward causing the light rays to converge towards one another. A concave lens has both sides rounded inward, causing the light rays to go parallel to one another.

c) A convex lens has both sides rounded outward causing the light rays to diverge away from one another. A concave lens has both sides rounded inward, causing the light rays to converge towards one another.

d) A convex lens has both sides rounded inward causing the light rays to converge towards one another. A concave lens has both sides rounded outward, causing the light rays to diverge away from one another.

e) None of the above

a) The lens can INCREASE its index of refraction, thus increasing its focusing power.

How is the lens able to change its refractive power?

a) The lens can INCREASE its index of refraction, thus increasing its focusing power.

b) The lens can be made less sharp for the curvature of its front surface, increasing its focusing power.

c) The lens can DECREASE its index of refraction, thus increasing its focusing power.

d) None of the above

Distance from lens to image = 22.5 cm

Lateral Magnification = -1.25 times

Assume you have a convex lens with f = 10 cm. If the object is placed 18 cm from the lens, how far is the image from the lens?  What is the lateral magnification?

b) Inverted

Referring to last problem, is the image upright, or Inverted?

a) Upright

b) Inverted

c) Impossible to determine

a) Real

Referring to the last two problems, is the image real or imaginary?

a) Real

b) Imaginary

c) Impossible to determine

Focal Length = 9.7 cm (keep 1 digit after 1 decimal point)

Discrepancy = 1% (round answer to integer number)

In Part 4.2.2, you will determine the focal length of a convex lens by focusing on an object across the room. If the object is 10 m away and the image is 9.8 cm, what is the focal length? (Hint: use Lab Manual Equation 4.2: 1 / o+ 1 / i =  1 / f, convert m into cm) Suppose one estimated the focal length by assuming f = i.  What is the discrepancy between this approximate value and the true value? (Hint: When the difference between 2 numbers is much smaller than the original numbers, round-off error becomes important. So you may need to keep more digits than usual in calculating the discrepancy.)

154 cm

In Part 4.2.5 of the experiment, the expected magnification of the microscope is given by Lab Manual Equation 4.3: m = i₁L / O₁f₂. Refer also to Fig. 4.4 for a definition of the components and distances used in Eq. 4.3. Suppose you obtain the following data. The distance between the object and the objective lens is 15 cm. The distance between the objective lens and the real, inverted image is 38 cm. The focal length of the eyepiece is 10 cm. When viewing the ruled screen (as described in Part 4.2.5), you observe 2 magnified, millimeter divisions filling the 78 mm width of the screen. What eye-to-object distance is consistent with this data? Round up answer to integer.

b) Blue

Blue light is bent more by a prism than orange light. Does blue light or orange light have a larger index of refraction in glass?

a) Orange

b) Blue

a) Lower

READING 2

An x-ray beam is made up of photons with different energies. As the beam passes through many materials, photons of certain energies penetrate better than others. This selective attenuation of photons, according to their energy, is referred to as FILTRA TION. Low energy photons are filtered out by diagnostic x-ray equipment, usually by placing several millimeters of aluminum in front of the beam before it enters the patient. These pho tons are filtered out because they do not contribute to the image formation (they are quickly attenuated by tissue and therefore do not penetrate the body well to form the image); they only contribute to the patient’s exposure to the harmful radiation [1].

What photon energies are selectively filtered (higher or lower energies) in medical imaging techniques?

a) Lower

b) Higher

c) So that the light rays entering the prism are parallel

Why is it necessary to collimate the light source before using the prism to disperse the light?

a) So that the light rays entering the prism are focused to a small spot

b) So that the light rays entering the prism are filtered for a specific wavelength

c) So that the light rays entering the prism are parallel

d) All of the above

b) They are both examples of line radiation.

READING 4

MRI (Lab Manual Figure 5.5) stands for Magnetic Resonance Imaging, a medical imaging technique that gives very clear pictures. MRI exploits the fact that certain materials, when placed in a magnetic field, absorb and reradiate electromagnetic radiation that has a specific frequency. Like the radiation from a gas lamp, it is an example of line radiation. For MRI, the radiation is associated with the nucleus rather than the atom; this is why its original name is Nuclear Magnetic Resonance (NMR). The radiation is typically in the form of radio signals. Since the frequency of the emission is proportional to the magnetic field, spatial information is obtained by placing the subject in a spatially varying magnetic field; the frequency of the radio emission then becomes proportional to position in the subject. The intensity of the emission is proportional to the density of resonant nuclei. The magnetic resonance image is actually an array of pixels (bins or picture elements) showing the intensity of the radio signals originating from each tissue volume element within the body section [18, 19]. Paul Lauterbur and Peter Mansfield won the 2003 Nobel Prize in Physiology and Medicine for their development of MRI.

For more information on the Nobel Prizes

What is the similarity between MRI and radiation from a gas lamp?

a) The radiation from both is created by a static magnetic field.

b) They are both examples of line radiation.

c) They are both examples of continuous spectrum radiation.

d) The radiation from both is found in the visible portion of the electromagnetic spectrum.

d) All of the above contribute to created the CT image of soft tissues with x-rays.

READING 3

CT (computed tomography), like conventional projection x-ray imaging, uses x-rays. However, CT produces a clearer picture of soft tissue. The CT generates its image from multiple views. A thin, fan shaped x-ray beam is projected through the edges of the body slice being imaged. The detectors do not see a complete image of the body slice, only a profile from one direction. The profile data are measurements of the x-ray penetration along each ray. In order to produce enough information to create a full image, the x-ray beam is rotated, or scanned, around the body section to produce views from many angles. Then, tissue density values for each view are fed into a computer, where they are organized into a two-dimensional matrix. The matrix is displayed on the system monitor, and appears as a gray-shaded image. The image represents a cross-sectional slice of human anatomy, typically a thickness of 1 cm [17].

How can the CT image soft tissues with x-rays?

a) It measures the x-ray penetration along each ray.

b) Tissue densities are fed into a computer and viewed as a gray-shaded image.

c) The beam is rotated around the body section to produce views from many angles.

d) All of the above contribute to created the CT image of soft tissues with x-rays.

c) A red line

Filter manufacturers sell ’interference filters’ that only transmit light within a narrow range of wavelengths. For example, an interference filter exists for use with a HeNe laser that strongly attenuates light at wavelengths other than the red laser wavelength (𝜆 = 632nm). Suppose that in Part 5.2.4 of the experiment you held one of these filters between the prism and the telescope. What would you see?

a) An blue line 

b) A spectrum of colors

c) A red line

d) A yellow line

e) None of the above

d) All of the above are reasons why these photons are filtered out.

Why are lower energy photons filtered out in medical x-ray imaging? 

a) They only contribute to the patient exposure to the harmful radiation.

b) They do not contribute to the image formation.

c) They are quickly attenuated by tissue and therefore do not penetrate the body well.

d) All of the above are reasons why these photons are filtered out.

b) The fovea reflects radiation in the yellow region of the spectrum.

READING 1

All lenses made of a single material refract rays of shorter wavelength more strongly than those of longer wavelength, and so bring blue light to a shorter focus than red, a phenomenon called CHROMATIC ABERATION (Lab Manual Figure 5.4). The result is that the image of a point of white light is not a white point but a blue circle fringed with color. The error is actually moderate between the red end of the spectrum and the blue-green, but it increases rapidly at shorter wavelengths - the blue-violet and ultraviolet. The first device which helps the eye compensate for this type of aberration is the lens, which acts as a color filter. The lens passes the visible spectrum, but cuts off sharply at the violet end, where chromatic aberration is worst. Secondly, there is a difference in spectral sensitivity between the two types of photoreceptors. Rods are maximally sensitive in the blue-green, whereas cones have their maximum in the green. Thus, as one moves from dim light to bright light, the pattern of maximum sensitivity of the eyes moves away from the end of the spectrum where aberration is at its worst. The third device is found in the fovea. A region around it, called the macula lutea, is colored with a yellow pigment, xanthophyll. The pigment absorbs maximally in the violet and blue regions of the spectrum, just where absorption by the lens falls to low values, and hence remove further parts of the spectrum where chromatic aberration is worst [15].

Which of the following does NOT help the eye compensate for chromatic aberrations?

a) The lens acts as a color filter.

b) The fovea reflects radiation in the yellow region of the spectrum.

c) The macula lutea absorbs maximally in the violet and blue regions.

d) There is a difference in spectral sensitivity between the two types of photoreceptors.

e) All of the above help the eye compensate for chromatic aberrations

c) The gas radiates only a few discrete wavelengths.

Why does a gas discharge tube (e.g., a neon light) have a certain color?

a) The tungsten filament in the tube glows.

b) The gas radiates a continuous spectrum of colors.

c) The gas radiates only a few discrete wavelengths.

d) The discharge tube is surrounded by a filter.

e) None of the above

1.4

In Part 5.2.3 of the experiment, you will measure the index of refraction of yellow light using Lab Manual Equation 5.2. Suppose the minimum angle of deviation is 29 degrees. What is the index of refraction?

a) bright spots due to constructive interference of light waves, and dark spots due to destructive interference of light waves.

READING

When light passes through a slit and hits a screen, a pattern of bright and dark spots appear on the screen if the wavelength of the light is comparable to the size of the slit. To understand this pattern, we can use Huygens’ principle and imagine the wavefront that hits the slit as consisting of a series of point sources of light. Each point source is like a tiny light bulb radiating light in isotropically every direction. Suppose the slit has a right edge and a left edge. We can think of the piece of the wavefront that hits the right edge as a point source radiating light that hits the screen. Similarly the left edge also can be associated with a point source. The light from each point source is a wave that has maxima (crests) and minima (troughs). When the waves from the two edges of the slit hit a point on the screen, they may interfere either constructively or destructively. If they interfere constructively, then the maxima arrive at the same time and the waves are in phase. So a bright spot results. If the two waves interfere destructively, then the maximum from one wave arrives at the same time as the minimum from the other wave, and they cancel each other out. In this case the waves are out of phase and a dark spot results.

When light passes through a slit whose size is comparable to the wavelength of the light, the screen shows which of the following

a) bright spots due to constructive interference of light waves, and dark spots due to destructive interference of light waves.

b) bright and dark spots due to destructive interference of light waves.

c) bright spots due to destructive interference of light waves, and dark spots due to constructive interference of light waves.

d) nothing.

e) bright and dark spots due to constructive interference of light waves.

f) None of the above.

a) Diameter of the pupil

READING 2

The resolving power of the eye depends on the distribution of light sensing elements on the retina as well as diffraction effects. The average diameter of a foveal cone is about 2.0 micrometers. Assuming cone-cone contact, let’s determine the angular separation of the cones from the lens (given that distance between the cones and the lens is 23 mm), and thus the limit of resolution of the eye due to cone density (at its maximum). Clearly, if two point sources are to be resolved, the images cannot evoke equal responses from adjacent cones. Instead, there must be an unexcited cone between the two cones that produce the full response. Therefore, the separation of two excited cones with one unexcited cone between them is 4 micrometers. The angular separation of these cones, measured from the eye lens is : 𝛼 = 4.0x10^-6 / 23x10^-3 = 1.8x10^-4rad. Since the angle of resolution for diffraction-limited resolution governed by the diameter of the pupil is a larger angle, the diameter of the pupil under optimal condition limits eye resolution. However, the two values still coincide very closely [5].

What limits the resolution of the eye under optimal conditions?

a) Diameter of the pupil

b) Separation between adjacent cones

c) Cornea

d) None of the above

e) a large number of equally spaced parallel adjacent slits.

What is a diffraction grating?

a) bright and dark spots that appear on the screen.

b) a large single slit.

c) a grid of perpendicular lines.

d) a series of bright and dark rings.

e) a large number of equally spaced parallel adjacent slits.

f) None of the above.

23.16

READING 3

Ultrasonic waves, like all other waves, exhibit diffraction. Recall that the minimum half angle of a beam is given by:

𝛼 = 1.22𝜆_m /d

The wavelengths used in ultrasound imaging techniques are generally in the range from 0.3 mm to 0.75 mm [longer waves (lower frequencies) penetrate more deeply but provide less resolution of detail than the short waves]. In order for an ultrasonic beam to spread by less than a centimeter after it has traveled across a human torso (20 cm), the source of the waves must be about 10 or 20 times larger than the wavelength [20]. The source of the waves is called a transducer.

A transducer is 19.5 mm in diameter and emits a wave of wavelength 0.50 mm. The beam travels a total distance of 37 cm through the patient and back to the transducer. How much has the beam spread after returning to the transducer? (Hint: Start with the formula, 𝛼 = 1.22𝜆_m /d, and recall that this is the minimum HALF angle of the beam.)

0.013 mm

Suppose you have measured the diffraction pattern of a grating with d = 0.05 mm and have found that the spots were separated by s = 1.1 cm. Now you want to determine d for an unknown grating. With the unknown grating, the spots are separated by s = 4.3 cm. What is the spacing between lines d in the unknown grating?

_________ mm

13 cm

A beam 14 cm in diameter (a) consisting of microwaves with wavelength 𝜆  = 0.9 mm is incident upon a d=13 cm slit. At a distance of 2 m from the slit, what is the approximate width of the slit’s image?

_____________ cm

0.15 mm

Babinet’s principle of complementary screens. A surprising result of wave theory is that the diffraction pattern of a dark screen with a hole cut in it (of arbitrary shape) is identical to the pattern of an object with the same shape as the hole. (Lab Manual Fig. 6.3).

An application of this principle is that a line mounted on transparent slide casts the same diffraction pattern as a dark film with a slot of equal size cut in it. In Part 6.3.4 of the experiment, you will exploit this principle to measure the width of a hair. If the distance between the first spot and the central minimum is s = 2.4 cm, L = 9 m, and 𝜆  = 4 x 10^-7 m, what is the width of the hair?

_________ mm

5.73 micrometers

READING 1

Two point sources can be resolved by an optical system if the corresponding diffraction patterns are sufficiently small or sufficiently separated. By definition, the ”minimum resolvable separation” is when the maximum of the diffraction pattern of one source falls on the first minimum of the diffraction pattern of the other. For a circular aperture, this distance in angular measure (radians) is given by:

𝛼 =  1. 22𝜆_m / d

where 𝜆_m is the wavelength of light visualized, d is the diameter of the aperture, and  𝛼  is the angular separation. The wavelength of light in a material 𝜆_m is smaller than the wavelength in vacuum 𝜆_o by the ratio 𝜆_m = 𝜆_o / n, where n is the index of refraction. So the formula can be generalized to:

𝛼 =  1. 22𝜆_o / ( n * d)

This formula determines the diffraction-limited resolving power of an aperture with diameter d. This angular separation can be converted to an approximate linear separation, D, at the retina using:

𝛼  = D / 0.025

where 0.025 is the corneal-retinal distance in meters. (Note: The lab manual has a typo and should have D in this last equation. D is not the same as d.)

What is the linear separation D at the retina? The minimum diameter of the pupil is about 2 mm, the eye is most sensitive to wavelengths of about 500 nm (air), the index of refraction of the aqueous humor is n = 1.33, and the distance between the cornea and retina is about 2.5 cm.

____________ micrometers

(This is a very small distance - approximately 1/100 the size of a period.)

0.52 micrometers

READING 4

The resolution of a microscope is diffraction limited, just as the eye, but gains a significant advantage because an object can be placed very close to its objective lens. The smallest resolvable linear separation xm for an ordinary light microscope is:

X_m = 𝜆/( 2nsinΘ)

where 𝜆  is the wavelength of light used, and Θ  is the half viewing angle of the objective. The quantity ’n sin Θ’ is called the numerical aperture of the objective, where n is the index of refraction. The minimum resolvable separation of two objects can be reduced by placing a liquid with a large index of refraction between the object and the objective, as in an oil immersion microscope [14].

With light of wavelength 490 nm and a lens with a numerical aperture of 0.47, what is the minimum resolvable separation of two objects?

____________ micrometers

1,167 nm

In Part 6.2.2, you will determine the wavelength of the laser by shining the laser beam on a ”diffraction grating”, a set of regularly spaced lines. Suppose the pattern is displayed on a screen a distance L from the grating and the spots are separated by s. If the screen is 9 m away, the spots are 3.5 cm apart, and the lines of the grating are separated by 0.3 mm, what is 𝜆 ?

___________ nm

a) Yes

The densities of cardboard, aluminum, and lead are 0.6 g/cm³, 2.7 g/cm³, and 11.4 g/cm³, respectively. Suppose that you are studying the range of a (nonexistent) elementary particle, the Heidbrinkion, and that it takes 50 cm of cardboard, 40 cm of aluminum and 15 cm of lead to stop half of the Heidbrinkions emitted from a source. Does the absorption of the Heidbrinkions depend on the atomic charge of the absorber, Z? (Hint: 𝜇_m is a constant if there is no Z dependence.) Cardboard is mostly hydrocarbons (Z=6), aluminum has Z=13, and lead has a Z of 82.

a) Yes

b) No

b) A hyperactive thyroid may absorb up to 80% of the tagged iodine; a hypoactive gland may absorb as little as 15% of ¹³¹I.

READING 3

The thyroid needs iodine to function properly. (This is why iodine is added to table salt.) The radioisotope most commonly used in thyroid studies is ¹³¹ I. Tracer doses of Na¹³¹ I in the range of 5 to 50 microcuries are given orally to the patient, either in a solution or in a capsule. An identical dose is set aside to be used as a standard. An external radioassay of the thyroid is done 24 hours later. Scintiscans of an abnormal thyroid may reveal areas of greater or lesser radioactivity than the rest of the gland. Some of these areas are called nodules, and they can be felt by palpitation. The scintiscan will reveal whether a nodule is ”hot” (very active, accumulates radioiodine) or ”cold” (inactive). Cold nodules may be either centers of neoplastic malignancy or adenomas, cysts or hemorrhagic areas. Hot nodules are almost always benign [21]. Radioiodine experiments can also determine the activity of the thyroid, which is used to diagnose hyperthyroidism or hypothyroidism.

The diagnosis of hyperthyroidism/hypothyroidism is a similar procedure (administration of compounds of ¹³¹ I, followed by external measures of 𝛽 -radiation intensity in the thyroid). A hyperactive thyroid may absorb up to 80% of the tagged iodine; a hypoactive gland may absorb as little as 15% before normal biochemical turnover elsewhere in the body reduces concentration via excretion. Mapping of the thyroid by ¹³¹ I scintography is common practice. Both the outline of the organ and its turnover rate can be obtained from maps made at different time intervals after administration. The maximum activity of the emission is a direct measure of the uptake of iodine by the thyroid.

How do radioiodine experiments diagnose hyperthyroidism/ hypothyroidism?

a) Measure of the normal biological turnover rate of the radioactive source via excretion.

b) A hyperactive thyroid may absorb up to 80% of the tagged iodine; a hypoactive gland may absorb as little as 15% of ¹³¹I.

c) A hypoactive thyroid may absorb up to 80% of ¹³¹I, while a hyperactive thyroid may absorb as little as 15% of ¹³¹I.

d) Administer radioiodine compound and then measure the relative temperature of the thyroid with respect to a standard.

e) Administer ¹³¹Na, followed by external measures of 𝛽-radiation intensity.

f) None of the above

c) Alpha particles; Gamma rays

READING 1

The three types of radiation (𝛼, 𝛽, 𝛾)  have different effects upon the biological system. This relative effect is called the QUALITY FACTOR (or relative biological damage caused by the radiation) [21].

Alpha particles are doubly charged, slow moving particles, which allows them to interact effectively with atomic electrons and produce a high degree of ionization. This is because the electric field of a slowly moving particle can act on an atomic electron for a greater length of time, thus increasing the probability of driving an electron out and causing ionization. The 𝛾 ray, on the other hand, has no electric change (it consists of a photon) and is rapidly moving. Furthermore, all the ionization accomplished by 𝛾 rays is produced by secondary electrons. Thus, the effect of  𝛾 rays is not as severe as 𝛼 particles. Note that this has an inverse relationship to the penetrating power of the radiations: g  rays can penetrate the body, b  particles penetrate a few millimeters of tissue, and a  particles will penetrate only to a depth of about 40 micrometers. Neutrons are relatively damaging because they penetrate deep into tissue and produce high energy protons, which behave like alphas [5].

Fill in the blanks. Lower energy particles can act on an atomic electron for a greater length of time, thus increasing the probability of driving an electron out and causing ionization. Thus___________ are more damaging than ____________________.

a) Gamma rays; Alpha particles

b) Gamma rays; Beta particles

c) Alpha particles; Gamma rays

d) Beta particles; Alpha particles

c) By ingestion

An alpha cannot penetrate your skin. Once inside the body, however, an alpha source is the most hazardous radiation since it causes the most ionizing events per unit length. There is really only one way a sealed alpha source can hurt you. What is it?

a) If it comes from a uranium source

b) By direct contact with skin

c) By ingestion

d) None of the above

d) The negative of the slope of the line

Suppose that in Part 8.2.4 a pair of students obtain data measuring the range of gammas in lead using thicknesses of 0, 3, 6, and 9 cm. For 0 cm thickness, they measure 2742 counts; for 3 cm thickness, they measure 638 counts; for the 6 cm thickness, they record 155 counts; and for 9 cm thickness, they record 26 counts. Next, they plot their results on semilog paper. Along the x-axis they plot 𝜌x, where x is the thickness. Along the y-axis they plot the number of counts. Because they use semilog paper, this is the same as if they did a linear plot with the y-axis being the logarithm of the number of counts. Then they fit a line through the data, and use the graph to determine the mass absorption coefficient 𝜇_m . On the graph, the mass absorption coefficient is given by

a) The y-intercept of the line (where the line hits the y-axis)

b) The inverse of the slope of the line

c) The slope of the line

d) The negative of the slope of the line

e) The x-intercept of the line (where the extrapolated line hits the x-axis)

f) none of the above

a) Metastable radionuclides that have already undergone an electron-emitting transition. 

READING 2

In medical imaging techniques, many different radioisotopes are utilized: I¹³¹ for thyroid studies, Fe⁵⁹ for blood studies, etc. However, in recent years, the desirability of Tc^99m (m stands for metastable) and other metastable isotopes for use in nuclear medicine has increased. Metastable states are of particular interest in nuclear medicine because they make possible the separation of electron and photon radiation. In a diagnostic procedure it is undesirable to have electron radiation in the body because it contributes to radiation dosage but not to image formation. By using a nuclide that has already undergone an electron-emitting transition and is in a metastable state, it is possible to have a radioactive material that emits only gamma radiation [1].

Which metastable radionuclides are preferable to other radionuclides for diagnostic imaging in nuclear medicine?

a) Metastable radionuclides that have already undergone an electron-emitting transition. 

b) Metastable radionuclides that have a low quality factor.

c) Metastable radionuclides that have very low gamma radiation.

d) All of the above are desirable metastable radionuclides.

c) Gamma rays

READING 4

In positron emission tomography (PET), the distribution of a positron source within the body is reconstructed from the profile of gamma rays, the product of positron annihilation. Short lived radioisotopes such as ¹¹ C, ¹³ N, ¹⁵ O, and ¹⁸ F (as a substitute for hydrogen) replace natural isotopes of C, N and O in various organic compounds. These isotopes all decay by the emission of positrons (antimatter electrons) that combine with electrons to annihilate and produce two 511 keV 𝛾  rays that are emitted 180 apart. These gamma rays easily penetrate body tissues and can be externally detected. The only radioisotopes of these elements that can be detected outside the body are positron emitters. The use of these specific radioisotopes leads to an attractive use of the PET technique. Positron emission tomography is an analytical imaging technique that provides a way of making in vivo measurements of the anatomical distribution and rates of specific biochemical reactions. ¹¹ C, ¹³ N, and ¹⁵ O are used to label biologically active substrates without disrupting their biochemical properties. Radiation detectors record the emission of 𝛾  rays from the tissue distribution of positron activity. Data collected are used to form a tomographic image of the cross-sectional distribution of tissue concentration of radioactivity according to the principles of computed tomography. This provides quantitative, non-invasive measurements in humans of biochemical activity [27].

Fill in the blanks. In the PET technique, detectors record the emission of_____________________ from active substrates to form a tomographic image of the cross-sectional distribution of tissue concentration.

a) Alpha particles

b) Beta particles

c) Gamma rays

d) Positrons

0.0034 g / cm²

A typical sheet of weighing paper is 4.5 inches by 4.3 inches and weighs 0.43 g. What is its absorber thickness in g / cm² ? 

______________g / cm²

116.1 g / cm² for aluminum

The densities of cardboard, aluminum, and lead are 0.6 g / cm³, 2.7 g / cm3 , and 11.4 g /cm³ , respectively. Suppose that you are studying the range of a (nonexistent) elementary particle, the Heidbrinkion, and that it takes 53 cm of cardboard, or 43 cm of aluminum, or 12 cm of lead to stop half of the Heidbrinkions emitted from a source.

Calculate the absorber thickness for aluminum.

_____________g / cm² for aluminum

136.8 g / cm² for lead

The densities of cardboard, aluminum, and lead are 0.6 g / cm³, 2.7 g / cm3 , and 11.4 g /cm³ , respectively. Suppose that you are studying the range of a (nonexistent) elementary particle, the Heidbrinkion, and that it takes 53 cm of cardboard, or 43 cm of aluminum, or 12 cm of lead to stop half of the Heidbrinkions emitted from a source.

Calculate the absorber thickness for lead.

_____________g / cm² for lead

b) Projection

An ordinary dental X-ray uses which imaging technique(s)?

a) Sonography

b) Projection

c) Tomography

d) None of the above

a) Tiny quantized packets of light.

What are photons?

a) Tiny quantized packets of light.

b) neutrons

c) name of a rock group

d) protons

e) electrons

c) Light hits a metal and electrons are emitted. 

What happens in the photoelectric effect?

a) Light hits a metal and photons are emitted.

b) Light hits a metal and protons are emitted.

c) Light hits a metal and electrons are emitted. 

d) Electrons hit a metal and electrons are emitted.

e) Electrons hit a metal and light is emitted.