Actually the contribution of the book of Straumann, "General Relativity and relativistic Astrophysics" is an exercise. So at the end you will have to compute the deviation on your own.

The deviation of electromagnetic wave by the Corona of the sun is caused by the electron plasma with an electron density $n_e(r)$:

$$n_e(r) = \frac{A}{(r/R_{sun})^6} + \frac{B}{(r/R_{sun})^2}$$

with $A= 10^8\text{electrons}/cm^3$ and $B= 10^6\text{electrons}/cm^3$.

The dispersion law for transverse waves in a plasma is given by the formula:

$$\omega^2 = c^2 k^2 + \omega^2_p$$

where $\omega_p$ is the plasma angular frequency given by a "scaled" electron density:

$$ \omega_p^2 = \frac{4\pi n_e e^2}{m_e}$$

We can compute the refraction index from the dispersion law:

$$n =\frac{kc}{\omega} =\sqrt{ 1 - (\frac{\omega_p}{\omega})^2}$$

If $s$ denotes the path length coordinate of the light ray the refraction index fulfills the following differential equation for varying refraction index:

$$\frac{d}{ds}\left( n \frac{d\mathbf{x}}{ds}\right) =\nabla n $$

As the deviation angle is rather small, we can integrate the differential equation along the undisturbed trajectory with the coordinates $-\infty \leq x\leq\infty$ and distance $y=b$. At the integration borders $\pm \infty$ the refraction index $n=1$:

$$\frac{d\mathbf{x}}{ds}|_\infty- \frac{d\mathbf{x}}{ds}|_{-\infty} \cong \int_{-\infty}^{\infty} \nabla n( x\mathbf{e_x} + b\mathbf{e_y}) dx$$

The integral is actually an integral over $ds$ that is replaced by an integral over $dx$ because of the assumption that we integrate along the undisturbed trajectory.

The gradient of the refraction index as a function of r can be expressed as $\mathbf{\nabla} n(r) = n'(r)\frac{\mathbf{x}}{r}$. The scattering angle $\delta_\omega$ can now be found by multiplication of the lhs of the last equation with the unit vector $\mathbf{e_y}$:

$$\delta_\omega \cong \int_{-\infty}^{\infty} \nabla n(x\mathbf{e_x} + b\mathbf{e_y})\cdot \mathbf{e_y} dx = \int_{-\infty}^{\infty} n'(\sqrt{x^2 +b^2})\frac{y}{\sqrt{x^2+b^2}} dx = \int_{-\infty}^{\infty} n'(\sqrt{x^2 +b^2})\frac{b}{\sqrt{x^2+b^2}}dx$$

So actually for every angular frequency of the electromagnetic wave one has to compute the refraction index $n =\sqrt{ 1 - (\frac{\omega_p}{\omega})^2}$ and the plasma frequency is given by the electron density given at the beginning of the post. The evaluation of the integral requires to put everything together and to do the integral frequency by frequency. Actually "Straumann" recommends to compute the integral with the saddle point method. The result at the end of the exercise is not given. But I guess, it was already done and the result will be probably much smaller than the famous angle deviation given by GR: $\delta = 1.75''$.

Actually the contribution of the book of Straumann, "General Relativity and relativistic Astrophysics" is an exercise. So at the end you will have to compute the deviation on your own.

The deviation of electromagnetic wave by the Corona of the sun is caused by the electron plasma with an electron density $n_e(r)$:

$$n_e(r) = \frac{A}{(r/R_{sun})^6} + \frac{B}{(r/R_{sun})^2}$$

with $A= 10^8\text{electrons}/cm^3$ and $B= 10^6\text{electrons}/cm^3$.

The dispersion law for transverse waves in a plasma is given by the formula:

$$\omega^2 = c^2 k^2 + \omega^2_p$$

where $\omega_p$ is the plasma angular frequency given by a "scaled" electron density (SI-units):

$$ \omega_p^2 = \frac{n_e e^2}{m_e \epsilon_0}$$

We can compute the refraction index from the dispersion law:

$$n =\frac{kc}{\omega} =\sqrt{ 1 - (\frac{\omega_p}{\omega})^2}$$

If $s$ denotes the path length coordinate of the light ray the refraction index fulfills the following differential equation for varying refraction index:

$$\frac{d}{ds}\left( n \frac{d\mathbf{x}}{ds}\right) =\nabla n $$

As the deviation angle is rather small, we can integrate the differential equation along the undisturbed trajectory with the coordinates $-\infty \leq x\leq\infty$ and distance $y=b$. At the integration borders $\pm \infty$ the refraction index $n=1$:

$$\frac{d\mathbf{x}}{ds}|_\infty- \frac{d\mathbf{x}}{ds}|_{-\infty} \cong \int_{-\infty}^{\infty} \nabla n( x\mathbf{e_x} + b\mathbf{e_y}) dx$$

The integral is actually an integral over $ds$ that is replaced by an integral over $dx$ because of the assumption that we integrate along the undisturbed trajectory.

The gradient of the refraction index as a function of r can be expressed as $\mathbf{\nabla} n(r) = n'(r)\frac{\mathbf{x}}{r}$. The scattering angle $\delta_\omega$ can now be found by multiplication of the lhs of the last equation with the unit vector $\mathbf{e_y}$:

$$\delta_\omega \cong \int_{-\infty}^{\infty} \nabla n(x\mathbf{e_x} + b\mathbf{e_y})\cdot \mathbf{e_y} dx = \int_{-\infty}^{\infty} n'(\sqrt{x^2 +b^2})\frac{y}{\sqrt{x^2+b^2}} dx = \int_{-\infty}^{\infty} n'(\sqrt{x^2 +b^2})\frac{b}{\sqrt{x^2+b^2}}dx$$

So actually for every angular frequency of the electromagnetic wave one has to compute the refraction index $n =\sqrt{ 1 - (\frac{\omega_p}{\omega})^2}$ and the plasma frequency is given by the electron density given at the beginning of the post. The evaluation of the integral requires to put everything together and to do the integral frequency by frequency. Actually "Straumann" recommends to compute the integral with the saddle point method. The result at the end of the exercise is not given. But I guess, it was already done and the result will be probably much smaller than the famous angle deviation given by GR: $\delta = 1.75''$.