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Unformatted text preview: 2005-2006 Physics Olympiad Preparation Program – University of Toronto – Solutions. Set 4: Optics and Waves Problem 1 A bat, moving at 5.00 m/s, is chasing a flying insect. If the bat emits a 40.0 kHz chirp and receives back an echo at 40.4 kHz, at what speed is the insect moving toward or away from the bat? Take the speed of sound in air to be v = 340 m/s. Solution We should use the Doppler shift formula for the moving source andthe moving observer: fvvvvfSO-+=′(1.1) where fis the original frequency of sound produced by a source; f'is a frequency of sound received by the observer; vis a speed of sound in the medium (air in our problem); vOis the speed of the observer in the motionless frame of reference; and vSis the speed of the source in respect to the motionless frame of reference. In expression (1.1), the signs for the values substituted for vOand vSdepend on the direction of the velocity. A positive value is used for motion of the observer or the source toward the other, and a negative sign for motion of one away from the other. In our problem, it is convenient to separate two processes: 1) the bat is a source of sound, and the insect is the observer; and 2) the insect reflects the sound wave and becomes a source, while the bat now is the observer. For the first part of solution, the frequency of sound detected by the insect is given by: fvvvvfbatinsins-+=Let us decide that the bat is chasing the insect, and the sign for its velocity must be always positive. The sign of the velocity of the insect is unknown. Thus, kHzvfinsins4000.5340340-+=For the echo frequency, we can obtain the following: .4000.534034034000.53404.40-+-+=-+=insinsinsinsbatvvfvvvvSolving the equation, we can find that vins= - 3.31 m/s. It means that the insect is flying away from the bat! This is, of course, natural for the insect in such circumstances! Therefore, the bat is gaining on its prey at (5.00 – 3.31) m/s = 1.69 m/s. Problem 2 A surface of a glass plate is covered with a thin layer of water. Light with the wavelength λ = 0.680 μm strikes the surface at the angle 30.0owith respect to the normal to the surface. Due to the evaporating of the water layer, the intensity of the reflected light changes periodically. The time interval between the appearances of the maximum of intensity is equal to 15.0 min. Indices of refraction of glass and water are respectively 1.50 and 1.34....
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This note was uploaded on 03/24/2010 for the course PHYSICS 2202 taught by Professor Mihalisin during the Spring '09 term at Temple.
- Spring '09