If light passes near a massive object, the gravitational interaction causes a bending of the ray. This can be thought of as happening due to a change in the effective refractive index of the medium given by `n( r) = 1+ 2 GM//rc^2`
where `r` is the distance of the point consideration from the centre of the mass of the massive body, `G` is the universal gravitational constant, `M` the mass of the body and `c` the speed of light in vacuum. Considering a spherical object, find the deviation of the ray from the original path as it grazes the object.

Explanation

Let us consider two planes at r and r+dr . Let the light be incident at an angle `theta` at the plane at r and leave r +dr at an angle `theta +d theta`. Then from Snell's law

`" " n(r)sin theta =n (r+dr) sin (theta+d theta)`
`rArr" "n(r)sin theta,(n(r)+(dn)/(dr)dr)(sin theta cos d theta +cos theta sin d theta )(n(r)+(dn)/(dr)dr) (n(r)+(dn)/(dr)dr)(sin theta + cos theta d theta )`
Ignoring the product of differntials
`" "n(r) sin theta, n(r) sin theta +(dn)/(dr)dr sin theta+n(r) cos theta d theta `
or we have, `" "-(dn)/(dr)tan theta =n(r) (d theta)/(dr)`
`" " (2GM)/(r^(2)c^(2))tan theta=(1+(2GM)/(rc^(2)))(d theta)/(dr)~~(d theta)/(dr)`
`" " int_(0)^(theta_(0))d theta =(2GM)/(c^(2))int_(-oo)^ (oo)(tan thetadr)/(r^(2))`
Now substitution for integrals, we have
Now, `" " r^(2)=x^(2)+R^(2)and than theta =(R)/(x)`
`" " 2dr=2xdx`
`" " int_(-0)^(theta_(0))d theta=(2GM)/(c^(2))int_(-oo)^(oo)(R)/(x)(xdx)/((x^(2)+R^(2))^((3)/(2)))`
Put `" " x=R tanphi`
`" " dx=R sec^(2)phidphi`
`:. " " theta_(0)=(2GMR)/(c^(2))int_(-pi//2)^(pi//2)(Rsec^(2)phid phi)/(R^(3)sec^(3)phi)`
`" " =(2GM)/(Rc^(2))int_(-pi//2)^(pi//2)cosphid phi=(4GM)/(Rc^(2))`