Physics in Modern Medicine Fall 2010 Homework #3 Chapter 3 Lasers in Medicine Questions Q3.1 Absorption in melanin increases with decreasing wavelength, and has a maximum, according to figure 3.23 in the UV, but will absorb the green light of an Argon laser (488nm) as well as the blue/green (514) of the Argon. The absorption spectrum will dictate which laser is appropriate for treatment. Q3.5 Since the chief pigment in the endometrial tissue is blood, a good choice of laser would be one that is in the blue portion of the spectrum and one that would be inappropriate would be a laser whose output is in the red portion of the spectrum. An appropriate choice for the laser would be something in the 400 600nm range, say an Argon laser or maybe a copper vapor. An inappropriate choice for the laser would be lasers whose wavelength is in the 600nm and larger range, say He-Ne, Krypton, ruby and any IR laser. Q3.9 Eximer lasers, which are ultraviolet lasers, are used in photorefractive surgery because the cornea is transparent to any visible light, such as that in the argon laser. The cornea is a good absorber of UV radiation. The retina on the other hand has many blood vessels and to target these in a surgery one would want the argon laser since it operates in the blue-green portion of the EM spectrum and would be preferentially absorbed by the red blood cells here. Problems P3.5 a) To achieve a power density of 80 milliwatts/cm 2 with a 1 watt CW laser, you would need a spot size given by (Eq. 3.7): I = P/(πr 2 ) Where I = power density in watts/cm 2, P = power in watts, and r = spot radius in cm. Solving for the spot radius, r, we get:
So a spot with a radius of 2 cm would be appropriate. b) The exposure time would be given by: T = F / I = (100 J/cm 2 )/( 80 milliwatts/cm 2 ) = (100 J/cm 2 )/( 80 x 10 3 watts/cm 2 ) = (100 J/cm 2 )/( 80 x 10 3 J/s /cm 2 ) = (100/0.080) sec (1 min /60 sec) = 21 min This is within reasonable bounds for a therapeutic procedure. c) The goals would be to maximize absorption for the photosensitizer, in this case phthalocyanine, and at the same time to minimize hemoglobin absorption, which is not specific to tumors. Since phthalocyanine has an absorption peak and hemoglobin absorbs relatively weakly at the region between 600 to 720 nm, the red gold vapor laser (630 nm), Krypton laser (676.4 nm), and dye laser operating in this wavelength range would be reasonable choices. The heliumneon laser has a good match, but in fact the powers achievable tend to be too low for this use. The ruby laser is a similarly good wavelength match, but the pulsed operation is not desirable in this application. d) The reason lasers are a good choice involves both their spatial coherence, necessary because of the need for relatively high power densities over a small region of tissue, and their temporal coherence, which is important to guarantee that the power density delivered is concentrated at the absorption peak of the photosensitizer, while avoiding high levels of absorption by other body tissues, such as blood. The very similar absorption properties of hemoglobin and many photosensitizers especially make the latter criterion highly important. Chapter 4 Diagnostic Ultrasound Questions Q4.3 Low frequency foghorns have long wavelengths. These long wavelengths, when they disperse through space, can envelope the ship so that anyone on the ship can hear the sound. If high frequency sound were used, with a
shorter wavelength, then you would only target perhaps, a small area of the ship and the ship may not be aware of the danger if no one hears the sound. Q4.4 Increasing the frequency may improve your resolution, but the penetration depth of the sound into the sample would go to zero. The half intensity depth is inversely related to the frequency. As the frequency approaches infinity, the penetration depth approaches zero. Q4.5 Miscalibration of the US scanner will produce distortions in the US image. To correct for miscalibrations the sonographer or technician would use test samples, or phantoms for calibration. Other effects that will cause distortions are refraction of the sound waves as the waves pass from one material into another and reverberations. Refraction produces echoes that are displaced from the expected path. Refraction can produce double images and there s not much to fix this problem. Reverberations are multiple high-intensity echoes from two strongly reflecting interfaces. Acoustic shadowing is another problem. US scanners compensate for the fact that the US pulse intensity decreases with increasing depth into the sample by artificially increasing the brightness. Acoustic shadows are created if a strong absorbing or very reflective surface is encountered by the US wave and the scanner software over compensates. Problems P4.1 a) We will use the average speed of sound in soft tissue here, v s = 1540 m/s, and the equations given in Chapter 4 relating the wavelength to frequency for a given medium, and hence speed of sound: Such a frequency is actually used in ultrasound imaging, and it would be adequate to resolve fairly small features of the heart or a large abdominal organ, for example. b) For this second calculation, we use the value of the speed of sound in air, instead: 344 m/s: Note that the wavelength associated with a particular frequency depends strongly upon the medium in which the sound wave travels, because of the very different values of the speed of sound.
P4.2 a) We wish to have an ultrasound wavelength of 0.50 mm in soft tissues, for which the speed of sound is v s = 1540 m/s. This requires us to solve Eq. 4.1 for the frequency: This is indeed in the range of frequencies used for abdominal and obstetrical ultrasound imaging. b) Now we are asked to solve for the speed of sound, given the frequency and wavelength of a wave in the material as 2.0 MHz and 1.75 mm. Using Eq. 4.1, we have: Although it was not required for the answer to this problem, you may have noted that 3500 m/s is the speed of sound in bone (Table 4.2). P4.3 a) We can apply Eq. 4.8 to translate the time it takes to send and receive echoes into distance measurements. The time for the pulse to travel to and reach the interface is given as 20 ms = 20 10 3 s, so this corresponds to a distance of: Here we ve used the book s estimate of 1540 m/s as an average travel time for ultrasound within the body. Even in such a short amount of time (about onesixth of a second), ultrasound waves travel a very great distance. b) If the interfaces are ΔD = 5.0 cm apart, we can again use the equation from a) and the value of the speed of sound to estimate how much time would elapse between the echoes. We first need to note that the extra distance traveled by the echo from the second interface (i.e., the one farthest from the transducer) will be 10.0 cm (0.100 m), because the second pulse must travel to the interface (5.0 cm) and travel an extra 5.0 cm on its return trip. We can then compute the extra time for the second echo pulse to reach the transducer from: t = 2 ΔD/ vs = 2 (5.0 cm) / Δ = 0.100 m /1540 m/s = 6.5 10 5 s