= Solution
At fixed radiation <wavelength>, differentiating <Bragg's law> gives $\cot\theta\,d\theta=-da/a$. With constant linear <thermal expansion coefficient> $\alpha_L$,
$$
\frac{\Delta a}{a}\simeq\alpha_L\Delta T=0.0018,
\qquad
\boxed{\Delta\psi=2\Delta\theta\simeq-2\alpha_L\Delta T\tan\theta.}
$$
Both scattering angles decrease as the lattice expands. Substitution gives
$$
\boxed{\Delta\psi_{111}\simeq-0.0416^\circ,\qquad
\Delta\psi_{220}\simeq-0.0702^\circ.}
$$
These are shifts of $2\theta$, not of the Bragg angle alone. As a finite-step check, using $a'=a(1+0.0018)$ gives $\psi'=2\arcsin[\sin(\psi/2)/(1.0018)]$. It produces shifts $-0.04151^\circ$ and $-0.07008^\circ$, consistent with the first-order estimate. A strictly constant differential coefficient would instead give $a'/a=e^{\alpha_L\Delta T}$; its difference from the linear length approximation is second order in this small expansion.
\Image[/past-exam-of-the-condensed-matter-physics-course-of-the-university-of-oxford/2013-diamond-diffraction-thermal-shifts.png]
{title=Calculated thermal shifts of the two lowest-angle diamond-cubic powder lines}
{description=Original illustrative line profiles centred on the stated scattering angles and their calculated shifted positions. Each line is normalized separately; the chosen common Gaussian width is not a measured instrumental width, and the panels do not predict an intensity ratio. The finite-step curves use $a'/a=1.0018$.}
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