BAAO 2025, II кръг — Задача 3. Gravitational Microlensing

a.

(i) [4]

By geometry and the small angle approximation for $\tan x$, we have:

By considering the angle $\alpha$ shown in Figure 1, we also have:

Hence:

(ii) [3]

Using the definition of parallax, we have:

Marking scheme

(a) (i) [1] Deduces $b = \theta_ED_L$
[1] Considers the angle $\alpha$ and finds one of the corresponding equations or finds another similarly useful geometric construction and a corresponding equation
[1] Makes all necessary small angle approximations for their method
[1] Makes all necessary substitutions for their method
(If none of marks 2-4 scored, award 1 mark for substitution of expression for b or δ into any correct geometric equation. Award 3 marks for any solution only missing approximations. Must be rearranged into the final form to score full marks)

(ii) [1] Finds $D_S = 11.5$kpc or $D_L = 5.1$kpc
[1] Finds all 3 $D$’s
[1] Calculates $\theta_E = 1.41\times10^{-10}$rad $(= (8.08\times10^{-9})^{\circ} = 0.0291$mas$)$
(Award one of the first 2 marks if the same error is made for both parallax calculations and $D_{LS}$ is correct for their values)

b.

(i) [6]

By small angles:

Extending the ray’s trajectory and considering the distance $y$, as shown in Figure 2, by small angle approximations:

Since we can do exactly the same construction with $\theta_-$ (using a sign convention where $\theta_-$ is negative), it satisfies the same equation. Thus the $\theta_{\pm}$ are the 2 roots of this equation, and from the quadratic formula, we have:

(ii) [4]

The displacement of the source from its unlensed position is given by the expression $\theta_+ - \varphi$. Differentiating, we have: (1)

Thus, the displacement is decreasing with $\varphi$, and so for minimal displacement we pick $\varphi = 0$, corresponding to $\theta_+ - \varphi = \theta_E$. (2)

(iii) [1]

By the Rayleigh criterion for an example wavelength of 500 nm (any wavelength corresponding to visible light is fine):

We remark that this is an absurd size for an optical telescope, so the deflection cannot be detected by conventional means (although it may be observed using interferometry).

Marking scheme

(b) (i) [1] Deduces $b = D_L\theta$
[1] Helpful geometric observation which could be used to find $\theta$ and a corresponding expression
[1] Finds any equation which would reduce to the correct quadratic upon substitutions of $b = D_L\theta$ and the formulae for $\delta$ and $\theta_E$ $D_S\theta = D_S\varphi + D_{LS}\delta$ or $\frac{D_LD_S}{D_{LS}}\varphi + D_L\delta = \frac{bD_S}{D_{LS}}$)
[1] Correctly makes any 2 of the above substitutions
[1] Obtains $\theta^2 - \varphi\theta - \theta_E^2 = 0$
[1] Deduces $\theta_- = \pm\frac{1}{2}\left(\sqrt{\varphi^2 + 4\theta_E^2} - \varphi\right)$ (Ignore sign)

(ii) (Allow full marks for all parts from (b-ii) to (c-iii) considering only $\theta_+$, $b_+$ in place of $\theta_{\pm}$, $b_{\pm}$)
[1] Considers the angle $\varepsilon = \theta_+ - \varphi$
[1] Differentiates $\varepsilon$ correctly (Award this mark for correctly differentiating $\theta_+$ if the first mark has been lost)
[1] Deduces $\varepsilon$ decreasing with $\varphi$ and hence minimised when $\varphi = 0$
[1] Deduces $\varepsilon_{\min} = \theta_E$

(iii) [1] Correct Rayleigh criterion calculation for a sensible wavelength. 500 nm gives 4300 m.

c.

(i) [3]

By small angles:

By considering Figure 3:

(ii) [3]

By differentiating:

(iii) [6]

Let $A$ be the annulus described above, and $B$ the corresponding annulus of points in $P_L$ between $b_{\pm}$ and $b_{\pm} + db_{\pm}$. Let $P$ be the luminosity of the source. Without the lensing effect:

Since $A$ is a narrow strip with length $2\pi a$ and width $da$, it has area $2\pi a\cdot da$. Hence:

When the lensing effect is present, the rays passing through $A$ which contribute to the image we are considering are exactly the ones which are incident on $B$. Thus:

Therefore, the ratio of powers on the annuli, which is the same as the ratio of brightnesses (since the power is constant over the annulus for small enough $d\varphi$), is given by:

(iv) [1]

(v) [2]

So the % increase in brightness is 3.5% (1)

Marking scheme

(c) (i) [1] Finds $b\pm$
[1] Helpful geometric observation which could be used to find $a$ and a corresponding expression (e.g. considering the length given by $\varphi D_L$ and $\frac{D_{SL}}{D_S}$ and finding either one of those expressions)
[1] Finds $a = \frac{D_LD_S}{D_{LS}}\varphi$

(ii) [1] Correctly finds $da = \frac{D_LD_S}{D_{LS}}d\varphi$ (ECF their $a$)
[1] Attempts to differentiate their $b$ and uses any one differentiation rule correctly
[1] Correctly finds $db = \frac{D_L}{2}\left(1 \pm \frac{\varphi}{\sqrt{\varphi^2 + 4\theta_E^2}}\right)d\varphi$ (ECF their b)

(iii) [1] Formula for area of annulus (in terms of $a$ or $b$)
[1] Correct brightness on either plane (allow unnecessary terms accounting for extra distance due to $a$ or $b$ as long as they are correct)
[1] Determines power on any relevant annulus (ECF the above)
[1] Determines power on the other annulus (does not need to be simplified)
[1] Finds an expression for the ratio of powers in terms of only $u$
[1] Correctly simplifies their expression

(iv) [1] Sums brightnesses from both images to obtain $A$

(v) [1] Determines $u = 2.4$
[1] Determines % increase (= 3.5%)

d.

i) [4]

Reading values from Figure 6, we see that the baseline magnitude of the source is $m = 16.35$, and the peak magnitude is $m = 15.55$. Thus: (1)

We therefore need to solve for $u$ in the equation:

This could be done with Newton-Raphson iteration, or even by by rearranging into a quadratic in $u^2$ to solve exactly, but it is easiest to use the following fixed-point iteration scheme:

Applying the scheme, with $u_0 = 1$ (any sensible value will work fine), it converges in relatively few iterations to $u = 0.527$ (1)

ii) [3]

Figure 4 shows the movement of the source relative to the lens, as it would be seen by the observer without the lensing effect. Therefore, at a time of $t_E$ from the peak brightness, the source will have traversed a distance of $\theta_E$ from the point of closest approach, and will therefore be at a value of $u = \sqrt{0.527^2 + 1^2} = 1.13$. Therefore: (1)

From the graph, this happens at $t \approx 6210$, while the peak occurs at $t \approx 6140$. Thus, $t_E = 6210 - 6140 = 70$ days. (1)

Marking scheme

(d) (i) [1] Reads $m_0 \in [16.30, 16.36]$ or $m_{\mathrm{peak}} \in [15.52, 15.58]$ from figure 6
[1] Determines $A = 2.09$ (ECF any sensible values for $m$ read from the graph)
[1] Sets up an appropriate numerical method to solve for $u$ or rearranges to a quadratic in $u^2$
[1] Finds $u = 0.527$ (ECF their A)

(ii) [1] Deduces that 1 Einstein time away from the peak, $u = \sqrt{u_0^2 + 1^2} = 1.13$
[1] Deduces $A = 1.26$ at $t_E$ away
[1] Uses the graph to see that this is $70 \pm 15$ days away from the peak