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PHYSICS 3LC FINAL (2,3,4,5,6,8,10)
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, the - Answer - A. bright spots due to constructive interference of light waves, and dark spots due to destructive interference of light waves. 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. - Answer - E. A large number of equally spaced parallel adjacent slits. 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 - Answer - λ = s* l /L s = spot separation = 3 cm l = distance btw lines = 0.3mm L = distance to sreen = 11 m λ= 1227 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. cm. What is d?
mm - Answer - d2= d1s1/s =.11mm1.5cm / 3.5cm = .0471428571mm (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? - Answer - 12cm 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? - Answer - w= λL/s .48125mm 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. where 0.025 is the - Answer - alpha = 1.22 λm/d a= 1.22 (50010^-9)/1.33(210^-3) = 2.2910^- 4 a=d/l l =. D= alphal = 2.2910^-4 (.035m) = 5.73 um 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,
x= 54010^-9/2.59 = .457um 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 - Answer - C. So that the light rays entering the prism are parallel. 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 - Answer - Blue 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? - Answer - 1. n2=2sin(30 + Dmin/2) 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 - Answer - A. A Red line 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 - Answer - D. Gas radiates only a few discrete wavelengths.
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. - Answer - B. The fovea reflects radiation in the yellow region of the spectrum. What photon energies are selectively filtered (higher or lower energies) in medical imaging techniques? A. Lower B. Higher - Answer - A. Lower 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. - Answer - 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. - Answer - D. All of the above 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. - Answer - B. They are both examples of line radiation 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
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 - Answer - B 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 - Answer - B Where does the focusing in the eye of a fish occur? A. Water - cornea interface B. Cornea - aqueous humor interface C. the lens - Answer - C 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 - Answer - B Cornea, margin, center, lens 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 - Answer - A. The speed of light in a transparent medium is 1.9 x 10^8m/s. What is the index of refraction of this medium? - Answer - 1. n(index of refraction) = 310^8/1.910^
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? - Answer - Theta Bench = i + r 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 - Answer - A. Coherent; Aligned Fiber Ends 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? - Answer - 17.5% 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 - Answer - icrit = 76. alpha crit = 19. Sin(core) (1.46) = sin90 *1. Clad is 90 for this. which = 1 Alpha crit is 90- answer. What is the relationship between theta mirror and the customary angles? - Answer - Theta mirror = i 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 - Answer - Incoherent; no specific alignment of the fiber ends is necessary
C. r - 2 D. r 1 E. r - 2 F. none of the above - Answer - It is a constant independent of distance. 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 - Answer - Lamp - Isotropic Laser Collimated 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 - Answer - 80/181mv = .44 = normalized experimental value v1r1^2=v2r2^ v2=v1(r1/r2)^2 = 1(4.2/6.1)^2 = .47 = theoretical 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? - Answer - 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 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? - Answer - 19.7- 4.5 = 15.2 mV
- 6- 4.5 = 12.1 mV max = 15. 12.1/15.2 =. 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 - Answer - Laser intense low divergence, energy focused small spot 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 - Answer - Pigmented Epithelium 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 - Answer - All Facts affect visual acuity 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.
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 - Answer - B. The negative of the slope of the line 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 - Answer - B. Alpha particles; Gamma rays 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. - Answer - B. Metastable radionuclides that have already undergone an electron-emitting transition. 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 - Answer - E. A hyperactive thyroid may absorb up to 80% of I, while a hypoactive thyroid may absorb as little as 15% of I.
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 - Answer - D. Gamma rays 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? - Answer - λ = v / f = 344 / [5.3 10^3] = .0649056 m 100cm/m = = 6.49056 cm 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.) - Answer - Difration is somewhat important. 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.) - Answer - Pt/Pi= Rad (1/1 + (piγsigma/pv)^2)) Using the formula let: γ = 4800 Hz σ = 7*10^-3 gm/cm^ ρv = 41. p_t/p_i = 0. 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? - Answer - No
B. frequency; amplitude of the reflected pulse; C. travel time; frequency of the reflected pulse; D. None of the above - Answer - Amplitude, travel time between when a pulse leaves and when it returns to the transducer.
