IPhO 2019, Експеримент — Задача 1.2. Part B: The parameters of a diffraction grating

• B.1 λ/d = 0.4015 (m = 1), λ/d = 0.4012 (m = 2)

• B.2 0.4

  λ/d = 0.400 for both m = 3 and m = 4.

Section B – parameters of a diffraction grating

B.1: We will mark on the table a point Q, at a distance of about $H = 70cm$ from the screen – the wall of the experimental chamber - and at an equal distance from the chamber’s side walls. Using the given measuring tape we will mark on the screen two points $P_1$ and $P_2$, at an equal distance of about 100cm from the left and form the right of the marked point Q. On the screen, we will mark a point $P$, placed in the middle of the interval $P_1P_2$. Then, we will aim a laser to go through the points QP. This beam will be perpendicular to the wall that will be used as a screen.

Standard method:

We will place the grating such that the beam passes through it. By gently rotating the grating we will make sure that diffraction ordered 1 and $-1$ as well as 2 and $-2$ will appear in symmetrically around the zero order point. Note that the position of the zero order on the screen does not depend on the angle $\alpha$. In this situation it ok to assume that the incident angle of the beam on the grating is $\alpha = 0$.

As in the sketch, we will measure $H, L_1$ and $L_2$ and use the relation $d\sin\theta_m = m\lambda$.

The measured values are $2L_1 = 53.3\mathrm{cm}$, $2L_2 = 163.5\mathrm{cm}$ and $H = 60.8\mathrm{cm}$.

For the first order we get $\frac{\lambda}{d} = 0.4015$. For the second order we get $\frac{\lambda}{d} = 0.4012$.

B.2: A second method

Getting higher orders is not possible at an incident angle of $\alpha = 0$. Thus we will change $\alpha$ and as a result the angle $\theta_m$ will change. There is an angle in which $\theta_m$ is minimal. By differentiating the relation $d\left(\sin\alpha + \sin(\theta_m - \alpha)\right) = m\lambda$ by $\alpha$ we get that at the minimum ($\frac{d\theta_m}{d\alpha} = 0$) one gets $\cos\alpha - \cos(\theta_m - \alpha) = 0 \Rightarrow \alpha = \frac{\theta_m}{2}$. From this we get $2d\sin(\frac{\theta_m}{2}) = m\lambda$.

Note that there is no need to measure the angle $\alpha$, but rather to identify, by changing $\alpha$, the minimum of $\theta_m$.

Using this method it is possible to measure also ordered $m = 1$ and $m = 2$. For $m = 2$ and $m = -2$ we can verify that the beam is perpendicular to the screen by making sure the distance of these two ordered from the zero order is identical.

For $m = 3$, we will change $\alpha$ to get $\theta_{3min}$ and measure the distances $L$ and $h_3$.

The measured values, as shown in the sketch below, are $H = 67.0\mathrm{cm}$, $L = 100.2\mathrm{cm}$, $h_3 = 37.8\mathrm{cm}$.

We get $\tan\theta_{3\min} = \frac{L}{H - h3} = \frac{100.2}{67.0 - 37.8} = 3.432$ and hence $\theta_{3\min} = 73.75^{\circ}$

Therefore: $\frac{\lambda}{d} = \frac{2}{3}\sin\frac{\theta_{3\min}}{2} = \frac{2}{3}\sin\frac{73.75^{\circ}}{2} = 0.400$

For $m = 4$ we will change $\alpha$ to get $\theta_{4\min}$ and measure the distance $h_4$.

The measured values are $H = 67.0\mathrm{cm}$, $L = 100.2\mathrm{cm}$, $h_4 = 96.3\mathrm{cm}$.

From the sketch we get $\tan(\theta_{4\min} - 90^{\circ}) = \frac{h_4 - H}{L} = \frac{96.3 - 67.0}{100.2} = 0.2924$

Hence $\theta_{4\min} = 106.3^{\circ}$, therefore $\frac{\lambda}{d} = \frac{2}{4}\sin\frac{\theta_{4\min}}{2} = \frac{1}{2}\sin\frac{106.3^{\circ}}{2} = 0.400$