Solution

According to General Relativity, the deflection angle of a light ray passing a mass with an impact parameter $b$ is:

$$\alpha = \frac{4GM_{enc}}{c^2 b}$$

Let's consider paraxial rays ($b \ll R$) to determine the focal length. For a ray passing close to the center, the path length inside the spherical galaxy is approximately equal to its diameter ($L \approx 2R$).

By symmetry, the gravitational effect on the photon is exerted only by the matter enclosed within a cylinder of radius $b$, whose axis coincides with the line of sight. The volume of this cylinder is approximately:

$$V_{cyl} \approx \pi b^2 \cdot 2R = 2\pi R b^2$$

Since the density of matter in the galaxy is uniform ($\rho = \frac{M}{\frac{4}{3}\pi R^3}$), the enclosed mass within the cylinder is:

$$M_{enc} = \rho V_{cyl} = \frac{M}{\frac{4}{3}\pi R^3} \cdot 2\pi R b^2 = \frac{3M}{2R^2} b^2$$

Substituting this mass into the deflection angle formula, we get:

$$\alpha(b) = \frac{4G}{c^2 b} \left( \frac{3M}{2R^2} b^2 \right) = \frac{6GM}{c^2 R^2} b$$

From geometrical optics, a thin lens focuses paraxial rays if the deflection angle is proportional to the impact parameter: $\alpha = \frac{b}{F}$. Thus, the central part of the galaxy acts as an ideal lens with a focal length $F$:

$$\frac{1}{F} = \frac{6GM}{c^2 R^2} \quad \implies \quad F = \frac{c^2 R^2}{6GM}$$

Now, we estimate the numerical values in SI units:

• $R \approx 20000 \text{ pc} \approx 6 \cdot 10^{20} \text{ m}$
• $M \approx 3 \cdot 10^{11} M_\odot \approx 6 \cdot 10^{41} \text{ kg}$
• $c \approx 3 \cdot 10^8 \text{ m/s}$
• $G \approx 6.67 \cdot 10^{-11} \text{ m}^3/(\text{kg} \cdot \text{s}^2)$

Calculating the focal length in meters:

$$F = \frac{(3 \cdot 10^8)^2 \cdot (6 \cdot 10^{20})^2}{6 \cdot 6.67 \cdot 10^{-11} \cdot 6 \cdot 10^{41}} \approx 1.35 \cdot 10^{26} \text{ m}$$

Converting the distance back to parsecs ($1 \text{ pc} \approx 3 \cdot 10^{16} \text{ m}$):

$$F \approx \frac{1.35 \cdot 10^{26}}{3 \cdot 10^{16}} = 4.5 \cdot 10^9 \text{ pc}$$