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1、IPhO2002 I- 1 THEORETICAL COMPETITION Tuesday, July 23 rd, 2002 Solution I: Ground-Penetrating Radar 1. Speed of radar signal in the material vm: constant z-constant ttz-=+(0.2 pts) = m v 2/ 1 2/1 22 2 1)1( 2 1 + = m v (0.4 pts) 1 ) 11( 2 1 2/1 = + =mv(0.4 pts) IPhO2002 I- 2 2. The maximum depth of
2、detection (skin depth, ) of an object in the ground is inversely proportional to the attenuation constant: (0.5 pts) (0.3 pts) (0.2 pts) 1/21/21/ 2 21/ 2 2 2 22 22 22 1111 1 1 . 11 11 22 22 2 a = +- +- 2/1 2 = . Numerically () r 31. 5 = m, where is in mS/m. (0.5 pts) For a medium with conductivity o
3、f 1.0 mS/m and relative permittivity of 9, the skin depth () 5.31 9 15.93 m 1.0 =(0.3 pts) + (0.2 pts) IPhO2002 I- 3 3. Lateral resolution: Antenna (1.0 pts) r =0.5 m, d =4 m: 1 22 14 2216 =+, 2 3240+-=(0.5 pts) The wavelength is =0.125 m. (0.3 pts) + (0.2 pts) The propagation speed of the signal in
4、 medium is rroororo m v 1111 = 1and 1 where,m/ns 3.0 r= oorrr m c c v m/s10m/ns1.0 8 =mv(0.5 pts) The minimum frequency need to distinguish the two rods as two separate objects is v f= min (0.5 pts) MHz800Hz10 125.0 9 3. 0 9 min=xf(0.3 pts) + (0.20 pts) 4 d + r d 222 () 4 rdd +=+ 1 22 216 d r =+ rod
5、 rod IPhO2002 I- 4 4. Path of EM waves for some positions on the ground surface The traveltime as function of x is 2 22 2 t v dx=+, (1.0 pts) 22 44 ( ) dx t x v + =(1.0 pts) 122 2 ( ) 0.3 r t xdx =+ For x =0 (1.0 pts) 100 = 2 (3/0.3) d d = 5 m (0.5 pts) 1, 1 T R T R d 1 2 3 x Buried rod( 2, 2) T R Antenna Positions Scanning direction Graph of traveltime , t(x) Antenna Positions x