Q.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 refrative index of the medium given by $n(r) = 1 + 2\, GM/rc^2$ where $r$ is the distance of the point of 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.

Concept: Refraction at Spherical Surface — The gravitational field acts like a medium with a radially varying refractive index, bending the light ray.

Reasoning:

1. For a ray grazing a spherical object of radius $R$, the refractive index varies only with $r$. The bending angle $\delta$ is given by integrating the gradient of $n$ along the path:

where $x$ is along the radial direction and $y$ along the straight-line path. …

The Physics at a Glance

Einstein's general relativity predicts that light bends when passing near a massive object. Remarkably, this effect can be understood using a classical analogy: gravity creates a gradient in the effective refractive index of space. The problem gives us this index as

where $r$ is the distance from the centre of mass $M$. As light travels through a medium with a varying refractive index, it bends toward regions of higher $n$ — just like a mirage on a hot road. Here, $n$ increases as $r$ decreases, so light bends toward the massive object.

Step-by-Step Solution

1. Set up the geometry

Consider a ray of light that just grazes the surface of a spherical object of radius $R$. Let the ray approach from infinity, pass tangent to the surface at closest approach $r = R$, and recede to infinity. By symmetry, the bending is symmetric about the point of closest approach.

We'll work in a plane containing the ray and the centre of the object. Let $x$ be the coordinate along the original (undeviated) direction, with $x=0$ at the point of closest approach. The distance from the centre at any point is $r = \sqrt{R^2 + x^2}$.

2. Relate bending to the refractive index gradient

For a medium with $n(r)$ varying slowly, the ray bends according to Snell's law applied locally. A standard result from geometrical optics: the curvature of a ray in a medium with gradient $\nabla n$ is given by

where $\theta$ is the angle the ray makes with some reference, $s$ is the path length, and $\phi$ is the angle between the ray direction and the gradient direction. For our radial gradient, $\phi$ is the angle between the ray and the radial line.

3. Compute the gradient

From $n(r) = 1 + \frac{2GM}{rc^2}$, we get

The negative sign means $n$ decreases as $r$ increases — the gradient points inward, so light bends toward the object.

4. Set up the integral for total deviation

For a grazing ray, the total deviation $\delta$ is the integral of all infinitesimal bendings along the path. Since $n \approx 1$ (the correction is tiny — for the Sun, $2GM/Rc^2 \approx 4 \times 10^{-6}$), we can approximate $n \approx 1$ in the denominator.

The geometry gives $\sin\phi = R/r$ (the component of the gradient perpendicular to the ray). So …

Method: Total Deviation From a Radially-Varying Refractive Index (Grazing-Ray Integration)

Use this method for problems where a ray passes near (grazing) a spherically symmetric region with a refractive index that depends only on distance $r$ from the centre, and you need the total accumulated deviation over the whole path — not just a local rate, as for a short horizontal segment.

Steps

Step 1: Set up coordinates along and perpendicular to the undeviated path

Let $x$ be the coordinate along the ray's original straight-line direction, with $x=0$ at the point of closest approach (here, at the surface, distance $R$ from the centre). The distance from the centre at any point along the path is then $r = \sqrt{R^2+x^2}$.

Step 2: Write the local bending rate using the radial gradient of $n$

Differentiate the given $n(r)$ to get $dn/dr$. The component of this gradient perpendicular to the ray (the part that actually bends it) involves the geometric factor $\sin\phi = R/r$ (from the same right triangle used in Step 1), giving a local bending rate

Since the index correction is tiny, approximate $n\approx 1$ in this expression.

Step 3: Substitute $r=\sqrt{R^2+x^2}$ and set up the full integral

Express $\dfrac{d\delta}{dx}$ purely in terms of $x$ (and constants $R$, and whatever parameters define $n(r)$), then integrate over the entire path, $x$ from $-\infty$ to $\infty$ (using symmetry to write it as twice the integral from $0$ to $\infty$ where convenient).

Step 4: Evaluate the integral with a standard substitution …