Saved Bookmarks
| 1. |
If light passes near a massive object, the gravitational interation causes abending of the ray. This can be thought of as heappening due to a change in the effective refrative index of the medium given by n(r) =1 +2 GM//rc^2 where r is thedistance 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 form the original path as it graxes the object. |
Answer» Solution : `rArr` As shown in the figure , when light ray passes tangentially to the surface of central massive body of mass M and radius R , suppose it gets deviated by amount `d theta` within distacen dr. `rArr` Now, applying Snell.s law at the point where light ray is made incident on the concentric spherical suface at distance, r, FROMTHE centre of central massive body we GET , `n sin theta = (n + dn) sin ( theta + d theta)` `thereforen sin theta = (n + dn) (sin theta cos d theta + cos theta sin d theta)` `therefore n sin theta = n sin theta cos ( d theta) + n cos (d theta) + n cos theta sin (d theta) + (dn) sin theta cos (d theta) + (dn) cos theta sin (d theta)` `rArr` Since ` dtheta` is extremely small, we can take `sin (d theta ) = (d theta)` and taking`cos (theta) = ` `n sin theta =n sin theta + n cos theta (d theta)+ (dh) sin theta + (dn ) cos theta (d theta)` `therefore 0 = n cos theta( because (dn)(cos theta) (dtheta ` is negligible) `therefore - (dn) sin theta = n cos theta (d theta)` ` therefore - ((dn)/(dr)) sin theta = n cos theta ((d theta)/(dr))` `therefore - ((dn)/(dr)) tan theta = n ((d theta)/(dr))`.....(1) `rArr` From the statement , `n = 1 + (2GM)/(rc^2)` `therefore (dn)/(dr) = 0 + (2GM)/(c^2) (-(1)/(r^2)) [ because (d)/(dr)(1/r)= - (1)/(r^2)]` `therefore (dn)/(dr) = 0 + (2GM)/(r^2c^2)` ......(2) `rArr` From equation (1) and (2) , `(2GM)/(r^2c^2) tan theta = (1 + (2GM)/(rc^2))(d theta)/(dr)` `rArr` Here inside the backet on R.H.S `(2GM)/(rc^2) lt lt lt lt 1` and so neglecing it, `(2 GM)/(r^2 c^2) tan theta = (d theta)/(dr)` `therefored theta = (2GM)/(c^2) ((tan theta)/(r^2))dr ....... (3)` `rArr`From the figure `r^2= x^2 + R^2` .......(4) `therefore 2r dr = 2x dx+ 0` `therefore r dr = x dx rArr dr = (xdx)/(r)` ......(5) `rArr` From EQUATIONS (3) and (5), `d theta = (2GM)/(c^2) (tan theta)/(r^3)x dx ......(7)` `rArr` Here, `r^2 = x^2 + R^2` `therefore (r^2)^(3//2) = (x^2 + R^2)^(3//2)` `therefore r^3 = (x^2 +R^2)^(3//2)` ......(8) `rArr` From equaitons (6),(7) and (8), `d theta = (2GM)/(c^2) (R)/(x^2 + R^2)^(3//2)dx .....(9)` `rArr` Nowsuppose, `x = R tan theta phi` `therefore dx = R sec^2 phi d phi` .......(11) `rArr` Now, `(x^2 + R^2)^(3//2) = (R^2 tan^2 phi + R^(2))^(3//2)` `= (R^2 sec^2 phi )^(3//2)` ` = R^3 sec^3 phi` ........(12) `rArr` From equaitons (9),(11),(12), `d theta = (2GM)/(Rc^2) cos phi " d " phi`.......(13) `rArr` From equation (10), if ` x= - oo " then " phi = - (pi)/(2)` rad `x = + oo` then `phi = l (pi)/(2)` rad `rArr` Also when ` x = - oo, theta = 0` `x= + oo, theta - theta_0` Taking intergration on both the SIDES of eqation (13), `int_(0)^(theta0) d theta = (2GM)/(Rc^2) int_(-pi/2)^(+pi/2) cos theta " d "theta` `therefore theta_0 = (2GM)/(Rc^2){sin phi}_(-pi/2)^(+pi/2)` ` = (2GM)/(Rc^2){sin (pi/2) - sin (- pi/2)}` `thereforetheta_0 = (4GM)/(Rc^2)` `rArr` Above equation gives required result. |
|