# 国际物理林匹克竞赛试题b

Solution to Experimental Question 2
Section 1
i. A typical geometric layout is as shown below. (a) Maximum distance from ruler to screen is advised to increase the spread of the dir

action pattern. (b) Note that the grating (ruler) lines are horizontal, so that diraction is in the vertical direction.

70 mm RULER

LASER β

SCREEN

1400 mm FRINGES

ii. Vis a vis the diraction phenomenon, β =

y 1400 mm

The angle β is measured using either a protractor (not recommended) or by measuring the value of the fringe separation on the screen, y, for a given order N . If the separation between 20 orders is measured, then N = ±10 (N = 0 is central zero order). The values of y should be tabulated for N = 10. If students choose other orders, this is also acceptable. N 2y mm y mm ±10 39.0 19.5 ±10 38.5 19.25 ±10 39.5 19.75 ±10 41.0 20.5 ±10 37.5 18.75 ±10 38.0 19.0 ±10 39.0 19.5 ±10 38.0 19.0 ±10 37.0 18.5 ±10 37.5 18.75

Mean Value = (19.25 ± 1.25) mm i.e. Mean "spot" distance = 19.25 mm for order N = 10. From observation of the ruler itself, the grating period, h = (0.50 ± 0.02) mm. Thus in the relation Nλ = ±h sin β

N = 10 h = 0.5 mm y sin β β = = 0.01375 1400 mm 10λ = 0.006875 mm λ Since β is small, δh δy δλ + 10% λ h y i.e. measured λ = (690 ± 70) nm The accepted value is 680 nm so that the departure from accepted value equals 1.5%. = 0.0006875 mm

Section 2
This section tests the student's ability to make semi-quantitative measurements and the use of judgement in making observations. i. Using the T = 50% transmission disc, students should note that the transmission through the tank is greater than this value. Using a linear approximation, 75% could well be estimated. Using the hint about the eye's logarithmic response, the transmission through the tank could be estimated to be as high as 85%. Any gure for transmission between 75% and 85% is acceptable. ii. Calculation of the transmission through the tank, using T =1R=1 n1 n 2 n1 + n 2
2

for each of the four surfaces of the tank, and assuming n = 1.59 for the perspex, results in a total transmission Ttotal = 80.80%

Section 3
With water in the tank, surfaces 2 and 3 become perspex/water interfaces instead of perspex/air interfacs, as in (ii). The resultant value is Ttotal = 88.5%

Section 4
TANK
~30 mm y x ~550 mm

SCREEN

LASER
TRANSMISSION FILTER IN/OUT

Possible conguration for section 4 (and sections 2 and 3) With pure water in the tank only, we see from Section 3 that the transmission T is TWater 88%

The aim here is to determine the beam divergence (scatter) and transmission as a function of milk concentration. Looking down on the tank, one sees
35 mm ~30 mm 25 mm 2θ ' 2x'

LASER

BEAM DIAMETER 2x = 2.00 mm

i. The entrance beam diameter is 2.00 mm. The following is an example of the calculations expected: With 0.5 mL milk added to the 50 mL water, we nd Scatterer concentration = Scattering angle 2x = 0.073 30 Transmission estimated with the assistance of the neutral density lters 2x = 2.2 mm ; 2θ = Ttotal = 0.7 . Hence Tmilk = Note that Tmilk = Ttotal Twater 0.7 = 0.79 0.88 and Twater = 0.88 (1) 0.5 = 1% = 0.01 50

If students miss the relationship (1), deduct one mark. ii. & iii. One thus obtains the following table of results. 2θ can be determined as shown above, OR by looking down onto the tank and using the protractor to measure the value of 2θ . It is important to note that even in the presence of scattering, there is still a direct beam being transmitted. It is much stronger than the scattered radiation intensity, and some skill will be required in measuring the scattering angle 2θ using either method. Making the correct observations requires observational judgement on the part of the student. Typical results are as Milk volume (mL) % Concentration 2x 2θ (Degrees) Tmilk follows: 0 0.5 0 1 2.00 2.2 0 4 1.0 0.79 1.0 2 6.2 12 0.45 1.5 3 9.4 18 0.22 2.0 4 12 23 0.15 2.5 5 28 0.12 3.0 3.5 6 7 Protractor 36 41 0.08 0.06 4.0 8 48 0.05

iii. From the graphed results in Figure 1, one obtains an approximately linear relationship between milk concentration, C, and scattering angle, 2θ (= φ) of the form φ = 6C . iv. Assuming the given relation I = I0 ez = Tmilk I0 where z is the distance into the tank containing milk/water. We have Tmilk = ez Thus ln Tmilk = z , and = constant × C Hence ln Tmilk = αzC. Since z is a constant in this experiment, line. Typical data for such a plot are as % Concentration 0 1 2 Tmilk 1.0 0.79 0.45 ln Tmilk 0 -0.24 -0.8 a plot of ln Tmilk as a function of C should yield a straight follows: 3 4 5 6 7 8 0.22 0.15 0.12 0.08 0.06 0.05 -1.51 -1.90 -2.12 -2.53 -2.81 -3.00

An approximately linear relationship is obtained, as shown in Figure 2, between ln T milk and C, the concentration viz. ln Tmilk 0.4C = z Thus we can write Tmilk = e0.4C = ez For the tank used, z = 25 mm and thus 0.4C = 25 or = 0.016C whence α = 0.016 mm1 %1

By extrapolation of the graph of ln Tmilk versus concentration C, one nds that for a scatterer concentration of 10% = 0.160 mm1 .

50

40

30

φ
20 10 0 0

2

4 6 % Concentration

8

10

Figure 1: Sample plot

0

-1

ln Tmilk

-2

-3

-4 0

2

4 6 % Concentration

8

10

Figure 2: Sample plot

Detailed Mark Allocation
Section 1 A clear diagram illustrating geometry used with appropriate allocations Optimal geometry used - as per model solution (laser close to ruler) Multiple measurements made to ascertain errors involved Correctly tabulated results Sources of error including suggestion of ruler variation (suggested by non-ideal diraction pattern) Calculation of uncertainty Final result Allocated as per: ±10% (612, 748 nm) ±20% (544, 816 nm) ± anything worse Section 2 For evidence of practical determination of transmission rather than simply "back calculating". Practical range 70 90% For correct calculation of transmission (no more than 3 signicant gures stated) Section 3 Correct calculation with no more than 3 signicant gures stated and an indication that the measurement was performed Section 4 Illustrative diagram including viewing geometry used, i.e. horizontal/vertical For recognizing the dierence between scattered light and the straight-through beam For taking the Twater into account when calculating Tmilk Correctly calculated and tabulated results of Tmilk with results within 20% of model solution Using a graphical technique for determining the relationship between scatter angle and milk concentration Using a graphical technique to extrapolate Tmilk to 10% concentration Final result for Allocated as ±40% [2], ±60% [1], anything worse [0] A reasonable attempt to consider uncertainties TOTAL [1] [1] [1] [1] [1] [1] [2] [2] [1] [0]

[1] [1]

[1] [1] [1] [1] [1] [1] [1] [2] [1] 20

x ? y U R? X 1 , 3 , 5 历届国际物理奥林匹克竞赛试题与解答 第4届(1970 年于苏联的莫斯科) 【题 1】如图 4.1(a)、 b) ( ,在质量 M=1kg 的...

1-8届国际物理奥林匹克竞赛试题与解答
x ? y U R? X 1 , 3 , 5 历届国际物理奥林匹克竞赛试题与解答 第4届(1970 年于苏联的莫斯科) 【题 1】如图 4.1(a)、 b) ( ,在质量 M=1kg 的...