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  • 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

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\frac{GM}{rc^{2}}

Where r is the distance of the point of consideration from the center 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.

 

Answers (1)

Let us consider two spherical surfaces of radius r and r + dr. Let the light be incident at an angle \theta at surface r and r + dr at an angle \theta+d\theta.

From Snell's Law,

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

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

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

For small angle, \sin d\theta\approx d\theta\; and \; \cos d\theta\approx 1

Ignoring the product of differentials

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

or we have, -\frac{dn}{dr}\tan \theta=n(r)\frac{d\theta}{dr}

\frac{2GM}{r^{2}c^{2}}\tan \theta=\left ( 1+\frac{2GM}{rc^{2}} \right )\frac{d\theta}{dr}\approx \frac{d\theta}{dr}

\int_{0}^{\theta_{0}}d\theta=\frac{2GM}{c^{2}}\int_{-\infty }^{\infty }\frac{\tan \theta dr}{r^{2}}

Now, r^{2}=x^{2}+R^{2}\; and \; \tan \theta=\frac{R}{x}

2rdr=2xdx

Now substitution for integrals, we have

\int_{0}^{\theta_{0}}d\theta=\frac{2GM}{c^{2}}\int_{-\infty }^{\infty }\frac{R}{x}\frac{xdx}{\left ( x^{2}+R^{2} \right )^{\frac{3}{2}}}

Put     x=R\; \tan \phi

dx=R\; sec^{2} \phi \; d\phi

\therefore \theta_{0}=\frac{2GMR}{c^{2}}\int_{-\pi /2}^{\pi /2}\frac{R\; \sec ^{2}\phi d\; \phi }{R^{3}\sec ^{3}\phi }

      \theta_{0}=\frac{2GM}{Rc^{2}}\int_{-\pi /2}^{\pi /2}\cos \phi d\; \phi =\frac{4GM}{Rc^{2}}

\Rightarrow \theta_{0}=\frac{4GM}{Rc^{2}}. This is the required proof.

 

Posted by

infoexpert23

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