IdPhO 2020, Експеримент — Задача 1. Crystallography

Experiment — IdPhO 2020 — Q1 — Crystallography (English (Official))

Условие

Errors – no evaluation required. A list of equipment and recommendations for its use are at the very end of the problem.

Introduction

Crystallography is when a football stadium-sized synchrotron is used to measure distances on the order of several angstroms (the length of the peptide bond between amino acids 1.31.3 Å).

The main object studied in crystallography is a crystal. It can be thought of as one periodically repeating element in space - a unit cell. (Fig. 1)

If ρ(x⃗)\rho(\vec{x}) is some real function that describes the structure of the unit cell (for example, the positions of atoms or the density of electrons) and a⃗1,a⃗2,a⃗3\vec{a}_1, \vec{a}_2, \vec{a}_3 – lattice vectors (may not be perpendicular to each other), then a complete crystal can be generated using the following equation:

ρ(x⃗+e⋅a⃗1+f⋅a⃗2+g⋅a⃗3)=ρ(x⃗),e,f,g,∈Z.(1)\rho(\vec{x} + e\cdot\vec{a}_1 + f\cdot\vec{a}_2 + g\cdot\vec{a}_3) = \rho(\vec{x}), \qquad e, f, g, \in \mathbb{Z}. \qquad (1)

In the case of 1D or 2D crystals, one (a⃗1\vec{a}_1) or two vectors (a⃗1\vec{a}_1 and a⃗2\vec{a}_2), respectively, is sufficient to create a crystal from a unit cell.

The main task of the crystallographic experiment is to determine the crystal structure (in particular, ρ(x⃗),a⃗1,a⃗2,a⃗3\rho(\vec{x}), \vec{a}_1, \vec{a}_2, \vec{a}_3).

Due to the periodicity of the crystal, one method that can be used to determine the structure is radiation diffraction.

In the crystallographic experiment (Fig. 2), an incident beam with intensity I0I_0, wavelength λ\lambda and wave vector k⃗i\vec{k}_i, (∣k⃗i∣=2π/λ|\vec{k}_i| = 2\pi/\lambda) passes through the crystal (in the case of a 2D crystal k⃗i\vec{k}_i, it is perpendicular to the crystal plane). The diffracted beam has the same wavelength λ\lambda, wave vector k⃗s\vec{k}_s, ∣k⃗s∣=∣k⃗i∣|\vec{k}_s| = |\vec{k}_i| and can be denoted by the scattering vector q⃗=k⃗s−k⃗i\vec{q} = \vec{k}_s - \vec{k}_i (Fig. 3). It is clear that q=2kisin⁡(θ/2)q = 2k_i\sin(\theta/2). The angle θ\theta between k⃗s\vec{k}_s и k⃗i\vec{k}_i can be considered much less than one, that is, q≪kiq \ll k_i.

The complex amplitude F(q⃗)F(\vec{q}) of the diffracted ray q⃗\vec{q} is called the structure factor (since it is determined by the crystal structure) and is denoted F(q⃗)F(\vec{q}). It has a module ∣F∣|F| and φ\varphi phase:

F(q⃗)=∣F∣⋅exp⁡(iφ).(2)F(\vec{q}) = |F|\cdot\exp(i\varphi). \qquad (2)

The measured intensity I(q⃗)I(\vec{q}) is the square of the modulus of the complex amplitude:

I=∣F∣2=FF∗.(3)I = |F|^2 = FF^*. \qquad (3)

Further processing of these intensities gives the densities ρ(x⃗)\rho(\vec{x}). Usually the crystals under investigation contain small inorganic (salts or complex compounds) or even large organic molecules (proteins or DNA), and the density ρ(x⃗)\rho(\vec{x}) determines the atomic positions of the compound. In this case, coherent X-rays with a wavelength of 1÷101 \div 10 Å(for example, synchrotron radiation) are used, corresponding to typical distances between atoms. If the distances are greater, a different wavelength should be chosen (for example, visible light can be used for optical gratings).

Now you will try to become a crystallographer. In Part A, using the example of a diffraction grating (1D crystal), we will study the basic laws. In Parts B, C, D you will define the lattice parameters, unit cell symmetries and then the detailed structure.

Unit lattice (a box with red lattice vectors a1, a2, a3 from origin 0) plus a motif (protein molecule) equals a unit cell; below, many unit cells stacked into a crystal block.
Figure 1: Unit cell represents whole crystal.
X-rays (orange arrow) hit a green crystal; diffracted beams go to a photographic plate; at right a diffraction pattern of spots. Labels: X-rays, crystal, diffracted beams, photographic plate, difraction pattern.
Figure 2: Scheme of a crystallographic experiment.
Incident beam |k_i| = 2π/λ (wavy line with wavelength λ) hits a Sample; the elastically scattered beam |k_s| = |k_i| leaves at angle θ to a screen with concentric rings; on the screen the vector q = k_s − k_i is drawn.
Figure 3: Scattering vector.

A. From slit to crystal

The simplest example of a crystal is an optical diffraction grating - it can be viewed as a one-dimensional crystal. Such a crystal has a slit (width bb) and an inter-slit gap as a unit cell (Fig. 4A). The lattice period a1=aa_1 = a is a one-dimensional vector that generates the crystal.

When observing Fraunhofer diffraction on a diffraction grating, the light intensity II depends on the light propagation angle θ\theta as follows

I(θ)=I0N2(sin⁡(Nπasin⁡θλ)sin⁡(πasin⁡θλ))2⋅(sin⁡(πbsin⁡θλ)(πbsin⁡θλ))2,(4)I(\theta) = \frac{I_0}{N^2}\left(\frac{\sin\left(\frac{N\pi a\sin\theta}{\lambda}\right)}{\sin\left(\frac{\pi a\sin\theta}{\lambda}\right)}\right)^2\cdot\left(\frac{\sin\left(\frac{\pi b\sin\theta}{\lambda}\right)}{\left(\frac{\pi b\sin\theta}{\lambda}\right)}\right)^2, \qquad (4)

where I0I_0 is the intensity at θ=0\theta = 0, NN is the number of illuminated slits of the diffraction grating.

To understand equation 4, you need to know that:

lim⁡x→0sin⁡xx=1.(5)\lim_{x\to 0}\frac{\sin x}{x} = 1. \qquad (5)
Two vertical x-axes labelled A and B with marks 0, b and a; in A the segment from b to a is black, in B the segment from 0 to b is black.
Figure 4: Unit cells (A and B) of a 1D crystal (diffraction grating). The non-transparent areas are shown in black.

A.1 Write the formula 4 using the scattering vector qq (q≪kiq \ll k_i). [0,3 т.]

A.2 Find scattering vector qq for the maximum numbered hh for a diffraction grating with a period aa. [0,2 т.]

A.3 Let q1q_1 be the scattering vector for the first maximum. Express qq in terms of q1q_1 for intensity maxima. How are q1q_1 and aa related? [0,2 т.]

A.4 Observe the diffraction of samples DG1-DG5. Determine experimentally q1q_1, and aa for each sample. Draw a scheme of your setup, write the quantities you measure, and write down the formulas for the calculations. [1 т.]

A.5 Conduct an experiment and determine the a/ba/b ratio for samples DG3, DG4, DG5. Explain your solution using formulas, diagrams, and pictures. It is known that b≤a/2b \le a/2. [1,5 т.]

It can be noted that the period of the crystal a1=aa_1 = a is responsible for the periodicity of the diffraction maxima, and the unit cell parameters (slit width bb) are responsible for the intensity of the maxima. This fact is used to simplify the calculation of the relative intensities of the maxima by using the structure factor F(q)F(q):

F(q)∼∫ρ(x)exp⁡(iqx)dx,(6)F(q) \sim \int \rho(x)\exp(iqx)dx, \qquad (6)

where ρ(x)\rho(x) – is transmission (the ratio of the amplitudes of the transmitted wave to the incident wave), qq – the scattering vector, which describes the position in the diffraction pattern. Integration is carried out over the entire unit cell. If we replace qq with the position of this or that maximum, it will be possible to find the relative intensities of the maxima as I=∣F∣2I = |F|^2. Here and below ρ(x)∈R\rho(x) \in \mathbb{R}.

The positions of the intensity maxima for crystal diffraction (called reflexes) can always be represented as a sum of vectors (called reciprocal lattice vectors):

1-D: q⃗=h⋅q⃗1\vec{q} = h\cdot\vec{q}_1, where h∈Zh \in \mathbb{Z};

2-D: q⃗=h⋅q⃗1+k⋅q⃗2\vec{q} = h\cdot\vec{q}_1 + k\cdot\vec{q}_2, where h,k∈Zh, k \in \mathbb{Z};

3-D: q⃗=h⋅q⃗1+k⋅q⃗2+l⋅q⃗3\vec{q} = h\cdot\vec{q}_1 + k\cdot\vec{q}_2 + l\cdot\vec{q}_3, where h,k,l∈Zh, k, l \in \mathbb{Z}.

Thus, each reflex for a 3D crystal can be denoted by three numbers (h,k,l)(h, k, l) (and (h)(h) for 1D and (h,k)(h, k) for 2D) and has its own complex amplitude F(h,k,l)F(h, k, l) and intensity I(h,k,l)I(h, k, l).

A.6 Write down ρ(x)\rho(x) for the unit cell of the diffraction grating from A1 (Fig. 4A). Use the coordinate system as shown in the figure. Suppose that the unit cell is such that the period of the lattice aa is pp times the width of the slit bb: a=pb,p∈Na = pb, \quad p \in \mathbb{N}. Calculate the structure factor FA(h)F_A(h) for this unit cell for reflex hh. Record your answer using hh and q1q_1. What maxima have intensity 0? Write the equation for hh for such maxima. [0,7 т.]

A.7 Consider another unit cell (Fig. 4B) of the diffraction grating. Calculate the structure factor FB(h)F_B(h) for this unit cell for reflex hh. What reflexes of this diffraction grating have an intensity of 0? Write the equation for hh for such reflexes. [0,7 т.]

A.8 These two diffraction gratings described above are illuminated with light of the same intensity. Find the quotients IA,h=0/IB,h=0I_{A,h=0}/I_{B,h=0} and IA,h=1/IB,h=1I_{A,h=1}/I_{B,h=1}. [0,4 т.]

B. 2D crystal

In this part you will go through the first step in defining the structure – finding the lattice parameters.

The reciprocal lattice vectors in the case of 2D

Let us consider diffraction on a two-dimensional crystal. A schematic 2D crystal of point atoms is shown in Fig. 5. Lattice vectors a⃗1\vec{a}_1 and a⃗2\vec{a}_2 has an angle α≤90∘\alpha \le 90^{\circ} between them.

If the plane of a given 2D crystal is placed perpendicular to the beam, then a periodic diffraction pattern appears on the screen behind the crystal, the position of the maxima of which is described by the expression q⃗=h⋅q⃗1+k⋅q⃗2\vec{q} = h\cdot\vec{q}_1 + k\cdot\vec{q}_2, where h,k∈Zh, k \in \mathbb{Z}.

You can get an idea of such a diffraction pattern by looking at the diffraction patterns of UC1-UC7 samples.

Oblique lattice of grey points; from one point two arrows: a1 horizontal to the right and a2 up and to the right, with angle α between them.
Figure 5: 2D crystal with vectors a⃗1\vec{a}_1 and a⃗2\vec{a}_2 as lattice vectors. The points represent the positions of the atoms in this crystal.
Four square unit cells A, B, C, D with sides a_A, a_B, a_C, a_D. Each has a black square atom of side b in the lower-left corner; C also has a larger black square b' near the upper right; D has a second black square of side b displaced half a period right and up (dotted centre lines).
Figure 6: The unit cells for crystals A-D are square. It is known that aB>aAa_B > a_A. No other size ratios are known.

B.1 Find the angle β\beta between the vectors q⃗1\vec{q}_1 and q⃗2\vec{q}_2 and their lengths q1q_1, q2q_2. Please note that these vectors must be of minimum length and the angle between them must be ≤90∘\le 90^{\circ}. Express your answer through the crystal parameters a1,a2,αa_1, a_2, \alpha (Fig. 5) [1 т.]

Samples UC1-UC4: the simplest 2D crystals

Consider four 2D crystals A, B, C, D (Fig. 6). These crystals have a square unit cell with perpendicular lattice vectors of equal length. The unit cell of each crystal contains an atom (a non-transparent square with a side bb on a transparent film). Unit cells C and D also contain displaced atoms of the second type. In crystal C, the size of an atom of the second type b′b' is larger than that of the first: b′>bb' > b. In crystal D, the second atom is the same size as the first, and is displaced half a period to the right and half a period up.

B.2 For crystals A and D, find the complex amplitude modulus ∣F(h,k)∣|F(h, k)| for the reflex (h,k)(h, k). Express your answer in terms of aa (crystal period) and bb (atom size). It is enough to indicate an expression that is true for all reflexes except for the central one (h=0h = 0, k=0k = 0) [1 т.]

B.3 Look at the diffraction patterns of samples UC1-UC4. For each UC1-UC4 sample, experimentally determine the crystal lattice period aUC1a_{UC1}, aUC2a_{UC2}, aUC3a_{UC3}, aUC4a_{UC4}. [0,6 т.]

B.4 For each UC1-UC4 sample, find the corresponding crystal structure among Fig. 6. Explain your choice using diagrams, pictures and formulas. [0,4 т.]

B.5 Determine the size of the atom bb. [0,8 т.]

Samples UC5-UC6: Still Simple 2D Crystals

B.6 Observe the diffraction patterns of samples UC5, UC6, UC7. Determine experimentally the parameters a1a_1, a2a_2 and the angle α\alpha for each sample. Explain which parameters of the diffraction pattern you are using with the help of diagrams and figures. [1,2 т.]

C. Crystal Symmetries

The second step after calculating the parameters of the unit cell is to determine the symmetry of the unit cell.

Symmetries: Understand Theoretically

The unit cells of real crystals often contain several molecules with some symmetry between them (Fig. 7). Knowing these symmetries greatly simplifies the process of defining the structure. The symmetry of the unit cell causes reciprocal symmetries and systematic absences (reflexes with zero intensity for each unit cell with this symmetry). Systematic absence is determined by special conditions for hh and kk (Fig. 8).

Typical Symmetries for Reflex Intensities:

  • mirror symmetry about some straight line. This line is called the axis of symmetry and is denoted by the equation of this line;
  • rotational symmetry of order mm (denoted as CmC_m for a specific center of rotation, m∈Nm \in \mathbb{N}). When rotated through an angle of 360∘⋅n/m360^{\circ}\cdot n/m, n∈Zn \in \mathbb{Z}, the picture transforms into itself).

Consider a diffraction image (Fig. 9) from a 2D crystal with perpendicular lattice vectors equal to each other (a1=a2a_1 = a_2). This diffraction image (only reflexes with ∣h∣,∣k∣≤2|h|, |k| \le 2) is shown in the figure.

Left: a square unit cell seen from above with four protein molecules (cyan, magenta, orange, green) arranged around the centre. Right: the same cell in perspective with a vertical red rotation axis through its centre.
Figure 7: This unit cell has the rotation symmetry C4C_4: rotation around the red axis by n⋅90∘n\cdot 90^{\circ} degrees, n∈Zn \in \mathbb{Z} gives the same cell.
Plot titled Reflections: black dots of various sizes on an h–k grid from −4 to 4; on the row k = 0 the odd-h reflexes are missing.
Figure 8: Reflexes (h=2n+1,k=0)(h=2n+1, k=0) are systematically absent. Note: Reflex (0,0)(0,0) with relatively high intensity is omitted here for clarity.
Square grid of 5 × 5 black diffraction spots of different sizes, with the largest at the centre 0; red axes q_x (right) and q_y (up) through the centre.
Figure 9: Diffraction image of 2D crystal.
Four square grids with red x and y axes, each containing a black L-shaped block: 1. initial (upper left), 2. x = 0 (mirrored to the upper right), 3. y = x (mirrored to the lower right), 4. moved (shifted near the origin).
Figure 10: The unit cells of crystals 2, 3, 4 were obtained from 1 using different symmetries.

C.1 Specify hh and kk for the point that will be the center of rotation. What orders of rotational symmetry mm are possible for a given image? Draw all possible axes of mirror symmetry in the image. Name your lines. [0,3 т.]

The axis of symmetry is – a straight line, which means it can be indicated as an equation of the straight line c1⋅qx+c2⋅qy=dc_1\cdot q_x + c_2\cdot q_y = d, where c1c_1, c2c_2, dd are some coefficients.

C.2 Specify the equation of the straight line for each axis of mirror symmetry drawn in the previous task. Do not forget to note which equation corresponds to which line. [0,2 т.]

C.3 For each rotation symmetry and axis of mirror symmetry, write down the corresponding notation (CmC_m for rotation and equation for mirror symmetry) and the equation for the intensities I(qx,qy)I(q_x, q_y), which should take place if this symmetry element is present. [0,4 т.]

C.4 Write down the equation for the intensities of the reflexes (h,k)(h, k) and (−h,−k)(-h, -k). What symmetry from question C1 corresponds to this equation? Explain your answer. [0,2 т.]

Let us consider some crystals with unit cells as in Fig. 10. Black squares represent non-transparent elements, white squares represent transparent elements. The unit cell of crystal 2 was obtained by mirror symmetry of the initial unit cell relative to the x=0x = 0 axis. The unit cell of crystal 3 was obtained by the mirror symmetry of the initial unit cell about the y=xy = x axis. Crystal 4 is obtained by shifting the original one by the vector (x1x_1, y1y_1).

C.5 Using the definition of the structure factor and symmetry find the structural factors f2(qx,qy)f_2(q_x, q_y), f3(qx,qy)f_3(q_x, q_y), f4(qx,qy)f_4(q_x, q_y) for crystals 2, 3, 4, respectively. Express your answer in terms of the structure factor F(qx,qy)=f1(qx,qy)F(q_x, q_y) = f_1(q_x, q_y) of crystal 1. [0,4 т.]

C.6 Consider an arbitrary 2D crystal (Fig. 5). Indicate what orders of mm symmetry of rotation can be in 2D crystals. Explain the answer. [0,5 т.]

Symmetries: Observe Experimentally

The answer sheet contains unit cells for the patterns PG1-PG9. Having studied the symmetry of unit cells and diffraction patterns, and reflex absences, we will find a correspondence between them.

C.7 Determine what symmetries the crystals with unit cells KK, LL, MM, NN, and PP, QQ, RR, SS, TT have. (Fig. in the answer sheet). Draw the axes of mirror symmetry, at the bottom of the picture indicate which rotational symmetries are present on it. [0,9 т.]

C.8 Observe the diffraction patterns of the samples PG 1, 2, 5, 8. These samples correspond to the unit cells KK, LL, MM, NN. Determine what symmetries the given diffraction patterns have. Find the correspondence between patterns and unit cells. [0,8 т.]

The symmetries of diffraction patterns can be calculated using the sum rule (the complex amplitude of a complex object is the sum of the complex amplitudes of the parts of that object) and knowledge of how the symmetries change the complex amplitudes of the terms. In addition, the sum rule allows you to define absences.

C.9 Observe the diffraction of the samples PG 3, 4, 6, 7, 9. These samples correspond to the unit cells PP, QQ, RR, SS, TT. Find the correspondence between samples and unit cells. Explain your solution using formulas, diagrams and pictures. [1 т.]

C.10 Observe the diffraction pattern of the UC8 sample. Could this sample be a crystal? Explain your answer. [0,3 т.]

D. You still need phases ...

When you know the parameters of the unit cell and the symmetry of the unit cell, it is time for the last step - to determine the complete structure of the crystal.

Phase problem ...

When radiation is scattered by a crystal, the complex amplitudes can be calculated using the formula (the so-called Fourier transform)

F(q⃗)∼∫ρ(x⃗)⋅exp⁡(iq⃗x⃗)dx⃗.(7)F(\vec{q}) \sim \int \rho(\vec{x})\cdot\exp(i\vec{q}\vec{x})d\vec{x}. \qquad (7)

To determine the density of a crystal using a diffraction image, it is necessary to perform the inverse Fourier transform:

ρ(x⃗)∼∫F(q⃗)⋅exp⁡(−iq⃗x⃗)dq⃗.(8)\rho(\vec{x}) \sim \int F(\vec{q})\cdot\exp(-i\vec{q}\vec{x})d\vec{q}. \qquad (8)

Since we have discrete reflexes, the above formula can be written as follows

ρ(x⃗)∼∑∣F(q⃗)∣eiφ⋅exp⁡(−iq⃗x⃗),(9)\rho(\vec{x}) \sim \sum |F(\vec{q})|e^{i\varphi}\cdot\exp(-i\vec{q}\vec{x}), \qquad (9)

where summation is performed over all reflexes. In fact, it is enough to take only the brightest reflexes, since their contribution to the sum is the largest.

The experimentally measured intensities I(q⃗)I(\vec{q}) make it possible to determine only the modulus ∣F(q⃗)∣|F(\vec{q})|. Unfortunately, the intensities do not contain information about the phases φ\varphi, so it is impossible to directly calculate ρ(x⃗)\rho(\vec{x}) from the intensities. This problem is called phase problem.

The usual way to solve the phase problem is to get some approximate phases, then calculate the density of the crystals using them, then update the phases using that density, and repeat this until you are satisfied with the resulting structure.

There are several methods for obtaining initial phases. One assumes that you already have a known crystal structure, and that the known and unknown crystals are structurally similar. In this case, you can simply use the phases calculated from the known structure with the measured intensities in the inverse Fourier transform (9) for the unknown structure.

... which you successfully solve

You have 3 different 2D crystals: MR0, MR1, MR2. The unit cell of each of them is a 4x4 square, with some squares non-transparent (ρ=0\rho = 0), and others transparent (ρ=1\rho = 1). The unit cell MR0 is known (Fig. 11). The structures MR1 (possible variants are shown in Fig. 12) and MR2 (contains 7 white squares) are unknown, but you know that they are quite similar to MR0.

Using the above approach, you will define the structures MR1 and MR2. Reflexes phases (in radians) with ∣h∣,∣k∣≤2|h|, |k| \le 2 were preliminary calculated (Fig. 13) for the MR0 structure (in the coordinate system shown in Fig. 11). Hint: the imaginary parts of the densities Imρ(x⃗)\rho(\vec{x}) vanish with good accuracy, which means that it is enough to take into account only the real part of each term (9).

k / h-2-1012
23.1420.46360-0.46363.142
10-0.46362.034-1.5710
03.142-1.57101.5713.142
-101.571-2.0340.46360
-23.1420.46360-0.46363.142

Figure 13: Reflexes phases (in radians) for the MR0 structure for the ∣h∣≤2,∣k∣≤2|h| \le 2, \quad |k| \le 2 reflexes.

4 × 4 grid of black and white squares (5 white squares) with red x and y axes whose origin 0 is at the lower-left corner.
Figure 11: MR0 structure with coordinate system. White squares – transparent, black – non-transparent.
Four 4 × 4 black-and-white unit cells labelled W, X, Y, Z, each a variant of the MR0 pattern.
Figure 12: Possible unit cells of the MR1 crystal.

D.1 The crystal (MR0 or MR2) is illuminated with light with an intensity of I0I_0. Find the intensity of the maximum at q⃗=0\vec{q} = 0. [1 т.]

D.2 Determine the structure of the unit cell of the MR1 crystal. The MR1 crystal has one of the indicated units cell (Fig. 12). Describe your solution. [2 т.]

D.3 Determine the unit cell structure of the MR2 crystal. The structure of MR2 is similar to MR0: two non-transparent squares have become transparent. [2 т.]

Equipment

Equipment list

  1. Multimeter
  2. Photodiode with magnet
  3. Battery
  4. Resistors 10 kOhm and 200 kOhm
  5. Splicing Connectors (4 pcs.)
  6. Battery holder
  7. Red laser (wavelength λ\lambda = 630 nm)
  8. Laser stand
  9. Needle
  10. Masking tape
  11. Screwdriver
  12. Ruler
  13. Large clips (2 pcs.)
  14. Tape Measure
  15. Slides ("Diffraction grating", "Unit cell", "Plane group", "Molecular replacement") with samples
  16. Film for reducing laser intensity
  17. Small clips (4 pcs.)
  18. Slide holder (needs to be assembled)
  19. Graph paper
  20. Cardboard
  21. Tablet with magnetic layer

Samples

Samples should be illuminated with a laser as follows. Position the sample so that the caption (slide title in English) is on top of the slide and reads from left to right. In this case, the laser should be in front of the slide, and the screen for observing should be behind the slide (Fig. 15).

Intensity detector

To measure intensities, use a photodiode connected to a voltmeter, resistor and battery according to the circuit shown in Fig. 17. In such a circuit, the current through the diode is proportional to the intensity of the incident light.

Pay attention to the polarity of the connection. On the battery block, the red wire is positive. By default, use a 10 kΩ resistor, in case the measured intensities are very small – use a 200 kΩ resistor.

To use the connectors (fig. 18), do the following. Lift the lever up, insert the wire into the connector, lower the lever. All three connector points are electrically connected to each other.

Laser

Mount the laser in the stand 8. Use a large clip to fix the laser on. If you need to make the laser beam narrower, then seal the laser hole with masking tape and make a hole in it with a needle. If you want to reduce the intensity of the laser – use a darkening film. Figure 19 is for your reference about that.

!WARNING! Direct exposure to the laser beam is hazardous to the eyes. Do not point the laser beam at yourself or other people. If you don't use the laser – turn it off.

!WARNING! A sharp needle is dangerous: you can get hurt! Use it carefully. When not in use, put the needle in the protective cover.

Photograph of the numbered equipment 1–21 on a table: multimeter, photodiode, battery, resistors, connectors, battery holder, green laser pointer box with lasers, laser stand, needle, masking tape, screwdriver, ruler, large clips, tape measure, slides, darkening film, small clips, slide holder, graph paper, cardboard, clipboard tablet.
Figure 14: Equipment. In the upper left corner – details of the intensity detector.
Sketch: a laser (cylinder) at lower left shines a red beam through a slide labelled IPhO; diffracted red rays continue to a rectangular screen behind the slide.
Figure 15: Place and illuminate the samples as shown.
Photograph of items 1–6: multimeter with box and leads (1), photodiode (2), battery (3), resistors (4), splicing connectors (5), battery holder (6).
Figure 16: Items of the intensity detector.
Circuit: voltmeter V in parallel with a resistor; this pair is in series with a photodiode and a battery (+ marked); the wires at the diode and at the battery are labelled red.
Figure 17: Intensity detector circuit. Pay attention to the polarity of the connection.
Photograph of a transparent splicing connector with an orange lever and a black wire inserted.
Figure 18: Scheme of splicing connector. All three connector points are electrically connected to each other.
Three photographs: laser pointer in its stand held switched on by a large binder clip; laser tip with tape and a needle; darkening film held upright in a large binder clip with a red laser spot behind.
Figure 19: Laser is kept switched on using large clip. To reduce beam size use sticky tape and a needle. Darkening film in a holder (large clip) to reduce laser intensity.

Отговори

Решение


Оригинал в Архива: IPhO 2020 Exp Q1.pdf · официални решения: IPhO 2020 Exp S1.pdf