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NCERT Exemplar · Q30

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/rc2n(r) = 1 + 2\, GM/rc^2 where rr is the distance of the point of consideration from the centre of the mass of the massive body, GG is the universal gravitational constant, MM the mass of the body and cc 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.

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The gravitational bending of light can be treated as refraction through a thin prism of continuously varying refractive index. For a ray grazing a spherical mass, the total deviation is δ=4GMRc2\delta = \frac{4GM}{R c^2}, where RR is the object's radius.

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

n(r)=1+2GMrc2n(r) = 1 + \frac{2GM}{rc^2}

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

Tip

The key insight: treat the curved path as a series of infinitesimal refractions. For a grazing ray, the total bending angle is twice the Newtonian prediction — a famous result from general relativity.

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 RR. Let the ray approach from infinity, pass tangent to the surface at closest approach r=Rr = 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 xx be the coordinate along the original (undeviated) direction, with x=0x=0 at the point of closest approach. The distance from the centre at any point is r=R2+x2r = \sqrt{R^2 + x^2}.

2. Relate bending to the refractive index gradient

For a medium with n(r)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 ∇n\nabla n is given by

dθds=1ndndrsin⁡ϕ\frac{d\theta}{ds} = \frac{1}{n} \frac{dn}{dr} \sin\phi

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

For a ray passing at distance rr from centre, the local bending rate is

dδdx=1n(r)dndrRr\frac{d\delta}{dx} = \frac{1}{n(r)} \frac{dn}{dr} \frac{R}{r}

where δ\delta is the cumulative deviation angle from the original straight path.

3. Compute the gradient

From n(r)=1+2GMrc2n(r) = 1 + \frac{2GM}{rc^2}, we get

dndr=−2GMr2c2\frac{dn}{dr} = -\frac{2GM}{r^2 c^2}

The negative sign means nn decreases as rr 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 δ\delta is the integral of all infinitesimal bendings along the path. Since n≈1n \approx 1 (the correction is tiny — for the Sun, 2GM/Rc2≈4×10−62GM/Rc^2 \approx 4 \times 10^{-6}), we can approximate n≈1n \approx 1 in the denominator.

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

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