Let's Bend light

Let us consider two arcs concentric with the centre of the planet at a distance $r$ and $r+dr$ . Let the light be incident at angle $\theta$ at the surface at $r$ and leave $r+dr$ at an angle $\theta+d\theta$

As $h << R$ and the light ray initially was tangent to the surface, the marked angle can be assumed to $\theta$ . Then from the Snell's law,

$n(r) \sin (\theta)=n(r+dr) \sin (\theta+d\theta)$

Now $\frac{ n(r+dr)-n(r) }{ (r+dr)-(r) }=\frac{ dn }{ dr }$ . Which gives $n(r+dr)=n(r)+\frac{ dn }{ dr }dr$ . Replacing in the above equation and using $\sin (A+B)= \sin A \cos B+\cos B \sin A$ . We get,

$n(r)sin(\theta)=\left( n(r)+\frac{ dn }{ dr }dr \right)\left( \sin \theta \cos d\theta + \cos \theta \sin d\theta \right)$

On simplification we get,

$-\frac{ dn }{ dr } \tan\theta=n(r)\frac{ d\theta }{ dr }$ .

Now $\frac{ dn }{ dr }=-\frac{ 2GM }{ { r }^{ 2 }{ c }^{ 2 } }$ . Also, $\left( 1+\frac{ 2GM }{ r{ c }^{ 2 } } \right) \frac{ d\theta }{ dr } \approx \frac{ d\theta }{ dr }$ As the object is massive. Replacing all of this in the original equation we get

$d\theta=\frac{ 2GM }{ { c }^{ 2 } }\left( \frac{ \tan\theta }{ { r }^{ 2 } }dr \right)$ .

Now from the diagram we can see that ${ r }^{ 2 }={ (h+R) }^{ 2 }+{ x }^{ 2 }$ . And $\tan\theta=\frac{ R+h }{ x }$ . As the light ray just grazes the planet, $h<< R \rightarrow R+h \approx R$ . So ${ r }^{ 2 }={ R }^{ 2 }+{ x }^{ 2 } \rightarrow dr=\frac{ xdx }{ \sqrt{ { x }^{ 2 }+{ R }^{ 2 } } }$ and $\tan\theta=\frac{ R }{ x }$ . Replacing all of this in the original equation and integrating within proper limits we get

$\int _{ 0 }^{ { \theta }_{ 0 } }{ d\theta } =\frac{ 2GMR }{ { c }^{ 2 } }\int _{ -\infty }^{ \infty }{ \frac{ dx }{ { ( { x }^{ 2 }+{ R }^{ 2 } ) }^{ \frac{ 3 }{ 2 } } } }$

${ \theta }_{ 0 }=\boxed{ \frac{ 4GM }{ R{ c }^{ 2 } } }$