Physics 3LC
Referring to the last problem, is the image upright?
Inverted!
What is the relationship between theta mirror and the customary angles?
Theta mirror = i
A typical sheet of weighing paper is 4 inches by 4.2 inches and weighs 0.4 g. What is its absorber thickness in ?
convert inches to cm via 1in. = 2.54 cm 4 in * 2.54 = 10.16 cm 4.2 in * 2.54 = 9.91 cm absorber thickness = (g)/(cm * cm) --> (0.4g)/(10.16*10.668) = 0.00369 g/cm^2
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 for an ordinary light microscope is: x = lambda/2nsin Delta where is the wavelength of light used, and is the half viewing angle of the objective. The quantity 'n sin ' 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]. [14] F. R. Hallett, P. A. Speight, R. H. Stinson, INTRODUCTORY BIOPHYSICS (New York: John Wiley & Sons, 1977). With light of wavelength 540 nm and a lens with a numerical aperture of 0.59, what is the minimum resolvable separation of two objects?
.457um x= 540*10^-9/2*.59 = .457um
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 18 degrees. What is the index of refraction?
1.2586 n2=2sin(30 + Dmin/2)
The speed of light in a transparent medium is 1.9 x 10^8m/s. What is the index of refraction of this medium?
1.58 n(index of refraction) = 3*10^8/1.9*10^8
(1 pt) A beam of microwaves with = 0.9 mm is incident upon a 12 cm slit. At a distance of 1 m from the slit, what is the approximate width of the slit's image?
12cm
In Part 3.2.2 of the experiment, a pair of students measure Theta Mirror= 20 degrees. To find Theta Bench , they use the theoretical relationship Theta Mirror = 1/2 Theta Bench . Then they measure Theta Bench= 47 degrees. What is the discrepancy between theory and experiment for Theta Bench?
17.5%
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 4.5 mV. A signal of 19.7 mV is measured at a distance of 29.5 mm and 16.6 mV is measured at 33.5 mm. Correct the data for background and normalize the data to the maximum value. What is the normalized corrected value at 33.5 mm?
19.7- 4.5 = 15.2 mV 16. 6- 4.5 = 12.1 mV max = 15.2 12.1/15.2 = .79
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 181 mV at a distance of 4.2 cm and a value of 80 mV at a distance of 6.1 cm. Normalize the students' data to the value obtained at 4.2 cm. (Divide the signal value by 181.) Then calculate the theoretically expected (normalized) value at 6.1 cm. Normalized experimental value at 6.1 cm Theoretically expected normalized value at 6.1 cm
80/181mv = .44 = normalized experimental value v1r1^2=v2r2^2 v2=v1(r1/r2)^2 = 1(4.2/6.1)^2 = .47 = theoretical
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 9.7 m away and the image is 9.8 cm, what is the focal length? (Hint: use Lab Manual Equation 4.2: ) 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.)
9.701cm 1.01% 1/f = 1/i + 1/o
In the experiment, a meter is hooked up to a speaker to monitor the amplitude of the received sound. Suppose the background signal level is 10 mV, the signal is 94 mV with no attenuator and is 27 mV with an attenuator in place. Calculate including the background correction.
94-10 27-10 17/84 = 20%
Ultrasonic waves, like all other waves, exhibit diffraction. Recall that the minimum half angle of a beam is given by: 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. [20] Harold J. Metcalf, TOPICS IN CLASSICAL BIOPHYSICS, (Englewood Cliffs, NJ: Prentice Hall, Inc., 1980). Obstetric Ultrasounds A transducer is 21 mm in diameter and emits a wave of wavelength 0.50 mm. The beam travels a total distance of 42 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, , and recall that this is the minimum HALF angle of the beam.)
= 24.406mm delta = 1.22 *.5mm/21mm = .029 radians *180/pi = 1.66 Tan1.66 = x/420mm x= 12.203 * 2 = 24.406mm
How is the lens able to change its refractive power? A. The lens can sharpen the curvature of its front and back surfaces, increasing its focusing power. B. The lens can DECREASE its index of refraction, thus increasing its focusing power. C. The lens can INCREASE its index of refraction, thus increasing its focusing power. D. None of the above
A.
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 . 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. A red line B. A spectrum of colors C. An blue line D. A yellow line E. None of the above
A. A Red line
Which of the following is NOT an example of a type of molecular change within the cell typically leading to cellular damage? A. A hydrogen radical and a hydroxyl radical combining to form a water molecule. B. Ionization of a water molecule into a hydrogen radical and a hydroxyl radical. C. A single strand break in double stranded DNA. D. All of the above are examples of molecular change leading to cellular damage
A. A hydrogen radical and a hydroxyl radical combining to form a water molecule.
Which of the following describe the characteristics of a bundle of fiber used for image viewing? A. Coherent; aligned fiber ends B. Incoherent; aligned fiber ends C. Incoherent; no specific alignment of the fiber ends is necessary D. Coherent; no specific alignment of the fiber ends is necessary
A. Coherent; Aligned Fiber Ends
An interface between air and some other substance, like tissue or water, strongly scatters ultrasound waves. Which of the following is NOT a consequence of this? A. Images are sharper with higher ultrasound frequencies. B. Sound transducers should be carefully coupled to the skin via gel. C. Lungs have high ultrasound attenuation rates. D. All of the above are consequences.
A. Image are sharper with higher ultrasound frequencies.
What photon energies are selectively filtered (higher or lower energies) in medical imaging techniques? A. Lower B. Higher
A. Lower
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. bright spots due to constructive interference of light waves, and dark spots due to destructive interference of light waves.
Referring to the previous problem, does the absorption of the Heidbrinkions depend on the atomic charge of the absorber, Z? (Hint: 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
A. yes
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
All Facts affect visual acuity
A reflected ultrasound pulse brings back two kinds of information. The ___________ of the reflected pulse relative to that of the incident pulse gives information on the type of interface that produced the reflection. The _______________ gives information about the distance between the transducer and the reflecting interface. A. amplitude; travel time between when a pulse leaves and when it returns to the transducer; B. frequency; amplitude of the reflected pulse; C. travel time; frequency of the reflected pulse; D. None of the above
Amplitude, travel time between when a pulse leaves and when it returns to the transducer.
Where does the majority of the focusing in the human eye occur? A. Lens - vitreous humor interface B. Air - cornea interface C. Aqueous humor - lens interface D. Cornea - aqueous humor interface
B
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 go parallel to 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 diverge away from one another. C. 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. D. 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. E. None of the above
B
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. cornea; center; margin; lens B. cornea; margin; center; lens C. lens; center; margin; cornea D. lens; margin; center; cornea
B Cornea, margin, center, lens
ll 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. Beta particles; Alpha particles B. Alpha particles; Gamma rays C. Gamma rays; Beta particles D. Gamma rays; Alpha particles
B. Alpha particles; Gamma rays
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 ingestion C. By direct contact with skin D. None of the above
B. By ingestion
Which metastable radionuclides are preferable to other radionuclides for diagnostic imaging in nuclear medicine? A. Metastable radionuclides that have very low gamma radiation. B. Metastable radionuclides that have already undergone an electron-emitting transition. C. Metastable radionuclides that have a low quality factor. D. All of the above are desirable metastable radionuclides.
B. Metastable radionuclides that have already undergone an electron-emitting transition.
Which of the following does NOT help the eye compensate for chromatic aberrations? A. There is a difference in spectral sensitivity between the two types of photoreceptors. B. The fovea reflects radiation in the yellow region of the spectrum. C. The lens acts as a color filter. D. The macula lutea absorbs maximally in the violet and blue regions. E. All of the above help the eye compensate for chromatic aberrations.
B. The fovea reflects radiation in the yellow region of the spectrum.
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 , 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 . On the graph, the mass absorption coefficient is given by A. The x-intercept of the line (where the extrapolated line hits the x-axis) B. The negative of the slope of the line C. The y-intercept of the line (where the line hits the y-axis) D. The inverse of the slope of the line E. The slope of the line F. none of the above
B. The negative of the slope of the line
What is the similarity between MRI and radiation from a gas lamp? A. The radiation from both is found in the visible portion of the electromagnetic spectrum. B. They are both examples of line radiation. C. The radiation from both is created by a static magnetic field. D. They are both examples of continuous spectrum radiation.
B. They are both examples of line radiation
Why are the similarities of cesium to potassium and strontium to calcium important? A. Cesium and strontium are easily converted into potassium and calcium, respectively, INCREASING their half-life. B. They are readily assimilated into the body, INCREASING the amount of radiation damage. C. They are readily assimilated into the body, DECREASING the amount of radiation damage. D. Cesium and strontium are easily converted into potassium and calcium, respectively, DECREASING their half-life.
B. They are readily assimilated into the body, INCREASING the amount of radiation damage.
What are photons? A. electrons B. Tiny quantized packets of light. C. protons D. name of a rock group E. neutrons
B. Tiny Quantized Packets of light.
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
Blue
Where does the focusing in the eye of a fish occur? A. Water - cornea interface B. Cornea - aqueous humor interface C. the lens
C
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 : . 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]. [5] Jerry B. Marion, William F. Hornyak, GENERAL PHYSICS WITH BIOSCIENCE ESSAYS, 2nd Ed. (New York: John Wiley & Sons, 1985). What limits the resolution of the eye under optimal conditions? A. Separation between adjacent cones B. Cornea C. Diameter of the pupil D. None of the above
C. Diameter of the pupil.
In Part 7.2.2 of the experiment, a pair of students measure the signal from a cobalt source for 10 sec and find C = 25 counts. They decide to repeat the measurement, but this time read only C = 20 counts. What is wrong? A. They did not count for long enough time intervals B. Nothing is wrong; cobalt has a really short half-life C. Nothing is wrong; there is a certain amount of statistical uncertainty in the measurement D. The Geiger counter is probably broken
C. Nothing is wrong; there is a certain amount of statistical uncertainty in the measurement
Which of the following DOES NOT describe projection imaging? A. Images are formed by projecting a beam through the patient's body and casting shadows onto an appropriate receptor B. Spatial distortion is generally not a major problem C. Produces images of selected planes or slices of tissue of the patient's body D. A large volume of the patient's body can be viewed with one image E. All of the above describe projection imaging.
C. Produces Images of selected planes or slices of tissues of the patient's body. In projection imaging (radiography and fluoroscopy), images are formed by projecting a beam through the patient's body and casting shadows onto an appropriate receptor that converts the invisible beam image into a visible light image. The primary advantage of this type of image is that a large volume of the patient's body can be viewed with one image. A disadvantage is that structures and objects are often superimposed so that the image of a superfluous object interferes with the image of the structure under study. In projection imaging, spatial distortion is generally not a major problem in most clinical applications. Tomographic imaging techniques such as conventional tomography, computed tomography (CT), sonography, single photon emission tomography (SPECT), positron emission tomography (PET), and magnetic resonance imaging (MRI), produce images of selected planes or slices of tissue of the patient's body. The general advantage of a tomographic image is the increased visibility of objects within the imaged plane. One factor that contributes to this is the absence of overlying objects. The major disadvantage is that only a small slice of a patient's body can be visualized with one image. Therefore, most tomographic procedures usually require many images to survey an entire organ system or body cavity. [1]
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 filtered for a specific wavelength B. So that the light rays entering the prism are focused to a small spot C. So that the light rays entering the prism are parallel D. All of the above
C. So that the light rays entering the prism are parallel.
Why are lower energy photons filtered out in medical x-ray imaging? A. They are quickly attenuated by tissue and therefore do not penetrate the body well. B. They do not contribute to the image formation. C. They only contribute to the patient exposure to the harmful radiation. D. All of the above are reasons why these photons are filtered out.
D. ALL OF THE ABOVE
How can the CT image soft tissues with x-rays? A. It measures the x-ray penetration along each ray. B. The beam is rotated around the body section to produce views from many angles. C. Tissue densities are fed into a computer and viewed as a gray-shaded image. D. All of the above contribute to created the CT image of soft tissues with x-rays.
D. All of the above
What factors affect the biological half-life? A. Size of particle B. Mode of intake C. Water solubility of material D. All of the above affect the biological half-life
D. All of the above.
Which of the following cellular properties would make the cell more susceptible to damage by ionizing particles? A. High rate of mitosis B. Cellular division throughout a major portion of the organisms lifetime C. Lack of specialization in a cellular developmental sequence D. All of the above would make the cell more susceptible to damage
D. All of the above.
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. Positrons B. Alpha particles C. Beta particles D. Gamma rays
D. Gamma rays
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 discharge tube is surrounded by a filter. C. The gas radiates a continuous spectrum of colors. D. The gas radiates only a few discrete wavelengths. E. None of the above
D. Gas radiates only a few discrete wavelengths.
In an arrangement such as in Lab Manual Figure 10.1, the attenuators are 8 inches by 11 inches. Referring to the previous problem, compare the wavelength with the size of the attenuator. Do you expect diffraction of sound waves by the attenuators to be important? (Hint: Use the diffraction equation (Equation (6.2)) from the lab manual to compare the wavelength with the size of the attenuator. Use the wavelength that you found in the previous question. Let d=8 inches in Eq. (6.2) and find the angle θ. If the angle θ is large, then diffraction is important. It the angle is small (close to zero), then diffraction is not important.)
Difration is somewhat important.
How do radioiodine experiments diagnose hyperthyroidism/hypothyroidism? A. Administer Na, followed by external measures of -radiation intensity. B. Administer radioiodine compound and then measure the relative temperature of the thyroid with respect to a standard. C. Measure of the normal biological turnover rate of the radioactive source via excretion. D. A hypoactive thyroid may absorb up to 80\% of I, while a hyperactive thyroid may absorb as little as 15% of I. E. A hyperactive thyroid may absorb up to 80% of I, while a hypoactive thyroid may absorb as little as 15% of I. F. None of the above
E. A hyperactive thyroid may absorb up to 80% of I, while a hypoactive thyroid may absorb as little as 15% of I.
What is a diffraction grating? A. a grid of perpendicular lines. B. a series of bright and dark rings. C. a large single slit. D. bright and dark spots that appear on the screen. E. a large number of equally spaced parallel adjacent slits. F. None of the above.
E. A large number of equally spaced parallel adjacent slits.
Which of the following safety rules should you follow when working with radioactive substances? A. Return all sources to the closed, labeled container handled by your T.A. B. When counting samples, do not stand beside the source. C. Mouth pipetting, eating, drinking, and smoking are prohibited. D. Wash your hands following the laboratory E. All of the above.
E. All of the above.
What happens in the photoelectric effect? A. Electrons hit a metal and electrons are emitted. B. Light hits a metal and photons are emitted. C. Electrons hit a metal and light is emitted. D. Light hits a metal and protons are emitted. E. Light hits a metal and electrons are emitted.
E. Light hits a metal and electrons are emitted.
Given Graph Semilog paper is a convenient tool for analyzing data that changes exponentially. The scale on the y axis of the paper is logarithmic. Since the logarithm of an exponential function is a linear function, exponentially varying data appear as a straight line on semilog paper. Suppose that in Part 7.2.3 of the experiment, students created the above graph based on their data. What is the half-life in minutes?
Find point after which sample is reduced by half (ie. Time for sample to reduce from 10^3 to five tick marks above 10^2) 1 minute~ (5 ticks downward to half between 10^2 and 10^3)
Suppose that placing 0.3 inch of lead in front of a gamma source reduces the count rate from 1045 cps to 573 cps. What is um^-1 in g/cm^2 ? The density of lead is 11.4 g/cm^3 .
I = Ioe^(-umpx) I = 1045 cps; Io = 573cps; p = 11.4 g/cm^3; x = 0.3 in. --> 0.762cm um = [ln (I/Io)]/(px) = 0.06917 cm^2/g --> um^-1 = 14.4566 g/cm^2
Which of the following describe the characteristics of a bundle of fiber used for illumination? A. Incoherent; aligned fiber ends B. Coherent; aligned fiber ends C. Incoherent; no specific alignment of the fiber ends is necessary D. Coherent; no specific alignment of the fiber ends is necessary
Incoherent; no specific alignment of the fiber ends is necessary
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. Direct; It uses a reagent that changes its optical properties in response to changes in the material of interest C. Indirect; It uses a reagent that changes its optical properties in response to changes in the material of interest D. Indirect; It senses changes the specific optical properties of the material of interest
Indirect; It uses a reagent that changes its optical properties in response to
Which of the following is true for the reflection coefficient for a perfect mirror? A. The ratio of reflected intensity to initial intensity is zero B. It / Io = 1 C. Ir / Io = 1 D. Ir / Io = It / Io E. All of the above
Ir / Io = 1
Depending on the application, a wide variety of particles and waves are used to probe the body. The basic requirements for a suitable probe beam are that it can penetrate the body (which excludes ordinary light), that is interacts with the sample to produce useful information, and that harmful, unwanted interactions with the body be minimal. One of the remarkable results of modern physics is that all particles have wave properties and all waves have particle properties. For example, "photons" and "light waves" are both electromagnetic radiation. Conceptually, all imaging techniques consist of some form of radiation interacting with matter. There are three things that can happen when particles/waves strike a sample (Lab Manual Fig. 1.4). Some of the particles (a fraction of the wave energy) are reflected. If the wave is characterized by an initial intensity I, the ratio of the reflected intensity I to incident intensity is called the reflection coefficient r. Some of the particles penetrate the sample. The ratio of transmitted intensity I to initial intensity, I/ I, is called the transmission coefficient t. Finally, some energy can be absorbed into the sample. Because energy is conserved, the sum of the reflected energy, the transmitted energy, and the absorbed energy equals the energy in the initial wave. In medical imaging, the reflections and absorption of various radiations by structures in the body are exploited to infer the properties of internal body structures. In the 3LC Lab, we study the interaction of some of these radiations with matter. Which of the following is NOT a requirement for a suitable probe beam? Which of the following is NOT a requirement for a suitable probe beam? A. It interacts with the sample to produce useful information B. It can penetrate the body C. Harmful, unwanted interactions with the body are minimal D. It interacts with a large number of cells E. All of the above are requirements for a suitable probe beam
It interacts with a large number of cells.
How does the flux of light from an collimated source depend on the distance r from the source? A. r2 B. It is a constant independent of distance. C. r -2 D. r 1 E. r -2 F. none of the above
It is a constant independent of distance.
In Part 4.2.5 of the experiment, the expected magnification of the microscope is given by Lab Manual Equation 4.3: . 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 16 cm. The distance between the objective lens and the real, inverted image is 41 cm. The focal length of the eyepiece is 6 cm. When viewing the ruled screen (as described in Part 4.2.5), you observe 3.5 magnified, millimeter divisions filling the 82 mm width of the screen. What eye-to-object distance is consistent with this data?
L = m*O1*f2/i (82/3.5)(16)(6)/41 = 54.87cm
Which statement is correct? A. A desk lamp is a isotropic source. A laser is an isotropic source. B. A desk lamp is an isotropic source. A laser is a collimated source. C. A desk lamp is a collimated source. A laser is a collimated source. D. A desk lamp is a collimated source. A laser is an isotropic source. E. none of the above
Lamp - Isotropic Laser Collimated
Which of the following DOES NOT describe tomographic imaging? A. A large volume of the patient's body can be viewed with one image B. Produces images of selected planes or slices of tissue of the patient's body C. Includes PET and MRI scans D. Increased visibility of objects within the imaged plane due to absence of overlying objects E. All of the above describes tomographic imaging
Large volumes of patients body viewed with one image.
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
Laser intense low divergence, energy focused small spot
What is the mass per unit area of an 7 inch by 10.5 inch lead sheet that weighs 208 gm.
Mass of the sheet is m =208 g Area of the sheet is A=7inch *10.5 inch = 73.5 inch^2 = 73.5 *(2.54 cm)^2 =474.19cm^2 Then mass per unit area is m/A = 208g / 474.1cm^2 =0.4386 g/cm^2
For an unknown sample of the experiment, students measure 1660 counts when they first receive their sample and 1309 counts 5 minutes later. Calculate the half-life of their sample. min ?
N = N0e^-lambda(t) lambda = ln(N/N0)/t Lambda = ln(1309/1660)/5min t1/2 = ln2/lambda A: 14.589 (NOTE That TIME CANNOT BE NEGATIVE)
This refers to the previous question in which a wave propagates in air and is transmitted through a piece of paper. Does increasing the frequency increase the amplitude of a sound wave transmitted through the piece of paper?
No
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. Outer Epithelium D. Stroma of Cornea E. None of the above
Pigmented Epithelium
An ordinary dental X-ray uses which imaging technique(s)? A. Sonography B. Projection C. Tomography D. None of the above
Projection
In Eq. (10.1) in the lab manual, the constant is 41.1 gm-cmsec in air. A piece of paper has a mass per unit area of approximately . If the frequency is 4.8 kHz, what does theory predict for the ratio of the transmitted amplitude to the incident amplitude of the sound wave? (A sound wave is a pressure wave.)
Pt/Pi= Rad (1/1 + (pi*γ*sigma/pv)^2)) Using the formula let: γ = 4800 Hz σ = 7*10^-3 gm/cm^2 ρv = 41.1 p_t/p_i = 0.5288
Referring to the last two problems, is the image real or imaginary?
REAL
At a distance of 9 cm from a presumably isotropic, radioactive source, a pair of students measure 65 cps (cps = counts per second). On average, how many counts per second do you expect at a distance of 29 cm? (Note that the average number of counts per second need not be an integer.)
Set-up: cps2/cps1 = r1^2/r2^2 cps2 = cps1 x r1^2 / r2^2 cps2 = 65 x 9^2 / 29^2 A: 6.26
The length of a fiberscope is mostly limited by what? A. Spectral Capability - what wavelengths can be conveyed or transmitted B. Bending Radius C. Jacket Thickness D. Spectral Capacity - how much of the transmitted wavelengths are conveyed
Spectral Capacity - how much of the transmitted wavelengths are conveyed
What factors DO NOT make fiber optics well suited to deliver the laser power? A. Light Weight B. Stiffness C. High Transmission D. Small Size E. All of the above factors make fiber optics well suited to deliver laser power
Stiffness
As an example, mercury-203 is a radioisotope that is important for kidney function studies and renal imaging. The physical half-life, , is 46.6 days and biological half-life, , of is 10.3 days. What is the effective half life of mercury-203? days
Teff = (TbTp)/(Tb+Tp) Teff = (10.3 x 46.6)/(10.3 + 46.6) A: 8.4355 days
To resolve two point sources, what distribution of cones must occur where the image strikes the retina? A. There must be an overexcited cone between two cones that produce the full response. B. Two adjacent cones must both produce the full response. C. There must be a cone producing less than the full response between two cones that produce the full response. D. None of the above
There must be a cone producing less than the full response between two cones that produce full response.
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 Theta Bench and between the laser and the normal to the mirror Theta Mirror are measured (Lab Manual Fig. 3.7). What is the relationship between theta Bench and the customary angles?
Theta Bench = i + r
In order to form a reflection interface for sound waves, what must be true of the materials on either side of the interface? A. They must differ in their indices of refraction. B. They must be separated by a certain distance. C. They must differ in their acoustic impedance. D. They must be at different temperatures. E. None of the above
They must be dif in acoustic impedance
Some wave energy can penetrate and pass through a sample. This is a description of what type of interaction with matter? A. Transmission B. Absorption C. Reflection D. Incidence
Transmission
In the photometer, light strikes a metal and electrons are emitted. The moving electrons constitute a current that flows through a resistor . The digital multimeter (DMM) can be set to read the voltage drop across the resistor . The DMM has a high input impedance (resistance) that is in parallel with . Let's denote the input impedance . Note that . Note that even with no input, e.g., the photometer turned off, the DMM set on 200 mV will output a signal because of ambient electromagnetic waves, e.g., radio waves in the air. These radio waves have an oscillating electric field that pushes electrons in the cable connected to the DMM. These moving electrons constitute a small current. Recall Ohm's law: . For the DMM, we can write . So if is large, even a tiny current can produce a finite voltage . This is why your DMM will read a voltage even if the input cable is not connected to anything. The cable acts like an antenna. What should the digital multimeter be set to read? A. power B. impedance C. current D. voltage E. resistance
Voltage
The photometer is a device that converts light to voltage which is read out by the digital multimeter (DMM). This is due to the photoelectric effect. (Einstein won his Nobel Prize for his explanation of the photoelectric effect.) In the photoelectric effect, light hits a piece of metal. Quantum mechanics tells us that light can be thought of as tiny packets of energy called photons. The photons are absorbed by electrons and increase the energy of the electrons. Some of the electrons become so energetic that they escape from the metal. In other words electrons are emitted from the metal, producing a current that is amplified by the photometer. The current goes through a resistor in the photometer. Ohm's law (V=IR) tells us that current I going through a resistor R is associated with a voltage drop V. The DMM measures the voltage drop across the resistor. Saturation occurs when the metal emits the maximum flux of electrons that it can produce. (Or it may be the maximum amplification of the photometer.)
What does a photometer do? B. It converts light into electric current. A. It emits light. B. It converts light into electric current. C. It measures the frequency of light. D. It emits sound.
When a wave strikes a new medium, three things can happen: reflection, transmission, and absorption. Consider an echo. Do you think reflection is an important process when sound waves produce an echo?
YES
The densities of cardboard, aluminum, and lead are 0.6 , 2.7 , and 11.4 g/cm^3 , respectively. Suppose that you are studying the range of a (nonexistent) elementary particle, the Heidbrinkion, and that it takes 54 cm of cardboard, or 40 cm of aluminum, or 15 cm of lead to stop half of the Heibrinkions emitted from a source. Calculate the absorber thickness for each material.
absorber thickness = (density) * (thickness) a.t. of cardboard = (0.6 g/cm^3)(54 cm) = 32.4 g/cm^2 a.t. of Al = 2.7 * 40 = 108 g/cm^2 a.t. of Pb = 11.4 * 15 = 171 g/cm^2
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: where is the wavelength of light visualized, d is the diameter of the aperture, and is the angular separation. The wavelength of light in a material is smaller than the wavelength in vacuum by the ratio = / n, where n is the index of refraction. So the formula can be generalized to: 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: = 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.) Limitations of the human eye 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
alpha = 1.22 λm/d a= 1.22* (500*10^-9)/1.33*(2*10^-3) = 2.29*10^-4 a=d/l l = .025 D= alpha*l = 2.29*10^-4 (.035m) = 5.73 um
Suppose you have measured the diffraction pattern of a grating with d = 0.11 mm and have found that the spots were separated by s = 1.5 cm. Now you want to determine d for an unknown grating. With the unknown grating, the spots are separated by s = 3.5 cm. What is d? mm
d2= d1*s1/s2 =.11mm*1.5cm / 3.5cm = .0471428571mm
Assume you have a convex lens with f = 12 cm. If the object is placed 22 cm from the lens, how far is the image from the lens? What is the lateral magnification?
di = fdo / do - f = (12)(22) / 22-12 = 26.4 m = -di/do -1.2 cm
The index of refraction of the core of a typical fiber optic is ncore = 1.46; the cladding has nclad = 1.42. Calculate the critical angles for the total internal reflection icrit and acrit . icrit = degrees, alpha crit= degrees
icrit = 76.656 alpha crit = 19.85 Sin(core) (1.46) = sin90 *1.42 Clad is 90 for this. which = 1 Alpha crit is 90- answer.
Suppose you want to take a chest X-ray with an X-ray source that has a divergence of 1 . If the film is 1 meters from the (point) source, how big is the spot size at the film in centimeters?
r = 1m divergence = 1 * pi/180 = .017 degrees 1 * .017 *100cm/m = 1.7cm Or use: Tan = theta/2 = D/2 / d D = d(tantheta) = 1.7cm
How does the flux of light from an isotropic source depend on the distance r from the source? A. r 2 B. r -2 C. r -1 D. r 1 E. r 1/2 F. none of the above
r-2
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 6.2.5 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 = 1.6 cm, L = 11 m, and = 7 x m, what is the width of the hair?
w= λ*L/s .48125mm
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 11 m away, the spots are 4.5 cm apart, and the lines of the grating are separated by 0.3 mm, what is wavelength ? nm
λ = s* l /L s = spot separation = 3 cm l = distance btw lines = 0.3mm L = distance to sreen = 11 m λ= 1227
The velocity of sound in air at STP (Standard Temperature and Pressure) is approximately 344 m/s. If the frequency of the sound wave is = 5.3 kHz, what is the wavelength in cm?
λ = v / f = 344 / [5.3 *10^3] = .0649056 m *100cm/m = = 6.49056 cm
