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IPhO 2013, Експеримент — Задача 2.7. The effect of the optical vessel (large cuvette) on the solar cell current

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Всички материали от IPhO 2013

Solar cells — E2 · 7 юли 2013 г.

Warnings and notices

WARNING: Do not use the multimeter as an ohmmeter on the solar cells since the measuring current can damage them. When changing the measuring range on the multimeters, please turn the dial with caution. It can be unstable and may break. Check whether there is a number under the decimal point when measuring – if the dial is not fully in place, the multimeter will not measure, even if there are digits in the display.

Notice: Do not change the voltage on the power supply. It must be 12 V throughout the experiment. (The power supply for the light source should be connected to the outlet (230 V ~) at your table.)

Notice: Uncertainty considerations are only expected when explicitly mentioned.

Notice: All measured and calculated values must be given in SI units.

Notice: For all measurements of currents and voltages in this experiment, the LED-light source is supposed to be on.

Условие

Mount the light source in the box and place solar cell A in the single holder with the circular aperture immediately in front, so that there is approximately 50 mm between the solar cell and the light source. Place the empty optical vessel immediately in front of the circular aperture as shown in Fig. 2.8.

Photograph of the black box from above: the LED light source on the left shining through the transparent optical vessel placed in front of the holder with the circular aperture and solar cell; the variable resistor shaft protrudes from the right wall. — Figure 2.8 Experimental set-up for question 2.7.

2.7a Measure the current $I$ , now as a function of the height, $h$ , of water in the vessel, see Fig. 2.8. Make a table of the measurements and draw a graph. [1 т.]

2.7b Explain with only sketches and symbols why the graph looks the way it does. [1 т.]

Mount the light source in the box and place solar cell A in the single holder so that the distance between the solar cell and the light source is maximal. Place the circular aperture immediately in front of the solar cell.

2.7c For this set-up do the following:

Measure the distance $r_1$ between the light source and the solar cell and the current $I_1$ .
Place the empty vessel immediately in front of the circular aperture and measure the current $I_2$ .
Fill up the vessel with water, almost to the top, and measure the current $I_3$ . [0,6 т.]

2.7d Use your measurements from 2.7c to find a value for the refractive index $n_{\mathrm{w}}$ for water. Illustrate your method with suitable sketches and equations. You may include additional measurements. [1,6 т.]

Отговори

Решение

2.7a

h	I
mm	mA
2	2.54
22	2.55
28	2.56
34	2.57
38	2.42
42	2.21
45	2.13
46	2.08
48	2.15
49	2.54
50	2.97
52	3.36
53	3.61
57	3.96
59	3.99
63	3.89
67	3.6
69	3.49
72	3.47

2.7b Exemple drawings for position A, B, C and D on previous graph:

Sketch A: the LED light source on the left, the vessel with a low water level, the aperture and the solar cell connected to a milliammeter mA; red rays from the diodes pass above the water through the aperture to the cell. — A

Sketch B: the water surface is just below the aperture; a red ray is refracted into the water, reflected at the surface and refracted out so that it misses the aperture. — B

2.7c NOTE: The exemplar measurements are from a different lamp than in 2.1. For a solution to 2.7d using the distance graph it is necessary to refer to the graph below.

Hand-drawn graph on millimetre paper of r² (mm², ticks 0, 1000, 5000, 10000) against 1/I (mA⁻¹, ticks 0.1, 0.5, 1.0, 1.5) with measured points on a straight line.

$r_1 = 103.5\ \mathrm{mm}$ ; $I_1 = 0.81\ \mathrm{mA}$ ; $I_2 = 0.705\ \mathrm{mA}$ ; $I_3 = 0.85\ \mathrm{mA}$

\frac{1}{I_3}\cdot\frac{I_2}{I_1} = 1.024\ \mathrm{mA}^{-1}\sim r_c^2 = 8800\ \mathrm{mm}^2\sim r_c = 93.8\ \mathrm{mm}

2.7d

h = (b-\Delta r)\tan\theta_1 = b\tan\theta_2 \Rightarrow \frac{b}{b-\Delta r} = \frac{\tan\theta_1}{\tan\theta_2} \approx \frac{\sin\theta_1}{\sin\theta_2} = n,\ \mathrm{da}\ \ \theta_2 < \theta_1 \ll 1.

n_w \approx \frac{b}{b-\Delta r} = \frac{b}{b-(r_1-r_c)} = \frac{26.0\ \mathrm{mm}}{26.0\ \mathrm{mm} - (103.5-93.8)\mathrm{mm}} = 1.6

NOTE: Better results may be obtained. The uncertainty is rather large in this method because of the subtraction of two large numbers for $\Delta r$

A different method is to determine the shift by actually moving the set-up and perhaps making an interpolation in directly measured data.

Chart of I/mA (0 to 4.5) against h/mm (0 to 80) with the measured points: flat near 2.55 mA (region A), a dip to about 2.1 mA (region B) between two green vertical lines marking the aperture (about 42 and 52 mm), a peak near 4 mA (region C) and a decline to about 3.5 mA (region D).

Sketch C: the water surface is just above the aperture; a red ray from a low diode is refracted into the water, totally reflected at the surface and exits through the aperture onto the solar cell. — C

Sketch D: the vessel is full of water; red rays from the diodes are refracted at both walls and pass through the aperture to the solar cell, with dashed lines showing the unrefracted directions. — D

Geometry sketch: a ray inclined at angle θ1 enters a water layer of width b between two vertical lines, travels at the smaller angle θ2 inside and leaves again at θ1, rising a height h; the dashed line marks the apparent source shifted by Δr from the left wall; labels b, h, θ1, θ2, Δr and r0. — 2.7d

Оригинал в Архива: exp2.pdf · официални решения: exp2_sol.pdf