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Consider a thin lens placed between a source (S) and an observer (O) (see figure) . Let the thicknes of the lens very as w(b) = w_0- (b^2)/(alpha), where b is the verticle distance from the pole. w_0 is constant. Using Fermat.s principle i.e. the time of transit for a ray between the source and observer is an extremum, find the condition that all paraxial rays starting from the source will converage at a point O on the axis. Find the focal length. (ii) A gravitational lens may be assumed to have a varyingwidth of the from w(b) = k_1 "ln" ((k_2)/(b)) b_("min") lt b lt b_("max") = k_1 "ln" ((k_2)/(b_"min")) b lt b_("min") Show that an observe will see an image of a point object as a right about the center of the lens with and angular radius beta=sqrt(((n-1)k_1(u)/(v))/(u+v)) |
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Answer» Solution :`rArr` (i) Time taken by light ray to travel from S to `P_1` `t_1 = (SP_1)/(C) = ((u^2+b^2)^(1//2))/(c) = ({u^2(1+b^2/u^2)}^(1/2))/(c)` `therefore t_1 = u/c(1+ b^2/u^2)^(1/2)` `rArr` Here `(b^2)/(u^2) lt lt 1` and so expanding according to binomial theorem and then retaining only first two terms, `t_1 ~~ u/c (1+(b^2)/(2u^2))` .......(1) `rArr` Similarly, time taken by light ray to travel from `P_1` to O, `t_2 = (P_1O)/(c) = ((v^2 + b^2)^(1/2))/(c) = ({v^2(1+ b^2/v^2)}^(1/2))/(c)` `therefore t_2 = v/c (1+(b^2)/(v^2))^(1/2)` `rArr` Here `(b^2)/(b^2) lt lt 1` and so expanding according to binomial theorem and then rataining only first two terms , `t_(2) = v/c (1 + (b^2)/(2v^2))`.......(2) `rArr` Now time taken by light ray to travel trough thickness W of a LENS at point `P_1` `t_(2) = v/c (1 + (b^2)/(2v^2))` `rArr` Now time taken by light ray to travel trough thickness W of a lesns at point`P_1` `t_3 = ((n-1)W)/(c)` (As PER statement `W = W_0 - (b^2)/(alpha))`......(3) `rArr` Now, total time taken by light ray to travel from S to `P_1` to O is t then, `t = t_1 + t_2+t_3` `therefore t = u/c (1 + (b^2)/(2b^2))+v/c(1+(b^2)/(2v^2)) +((n-1)W)/(c)` `therefore t = 1/c (u + (b^2)/(2u))+1/c(v+b^2/(2v))+1/c(n-1)W` `therefore t 1/c [ u+v+(b^2)/(2)(1/u +1/v)+(n-1)W]` `rArr` Here, for the sake of simplicity, if we assume `1/u + 1/v = 1/D` then ......(4) `t = 1/c[u +v + (b^2)/(2D) +(n-1)W].....(5)` `therefore t =1 /c[u+v+(b^2)/(2D)+(n-1)(w_0 - (b^2)/(alpha))]....(6)` As per statement `therefore (dt)/(db) = 1/c[0+0+1/2(2b)+(n-1)(0-1/alpha XX 2b)]` `therefore (dt)/(db) = 1/c[b/D - (2b)/(alpha)(n-1)]`........(7) `rArr` Now , according to Farmar.s principle above time t is either maximum or mimum and so its first derivative with respect to variable b should be zero. `(dt)/(db) = 0` `therefore1/c[b/c - (2b)/(alpha)(n-1)] =0` `therefore b/D = (2b)/(alpha) (n-1)` `alpha = 2 (n-1)D`.....(8) `rArr` Above equation gives required condition. Here, value of `alpha` is independent of b. Hence for the case when `b lt lt u`, all paraxial rays, incident on the lens, will be focused at point O. (ii) Now, for given gravitational lens, `W= K_1 "log"((K_2)/(b))` (As per statement ) ....(9) `rArr` From equation (5) and (9). `t = 1/c[u+v = (b)/(2D)+(n-1) K_1 "log"((K_2)/(b))]` `therefore t = 1/c[u+v+(b^2)/(2D) + (n-1)K_1{logK_2-logb}]` `therefore (dt)/(db) = 1/c[0+0 + (1)/(2D) (2b)+(n-1)K_1{0-1/b}]` `therefore (dt)/(db)=1/c[b/d-((n-1)K_1)/(b)]`........10 `rArr` Now, according to Farmat.s principle we have `(dt)/(db) = 0` ` b/D = ((n-1)K_1)/(b)` `therefore b = (n-1)K_1D` `therefore b = sqrt((n-1)K_1D)`.......(11) `rArr`Thus, those light rays which are incident on the lens at above HEIGHT from S, contribute in the formation of image. Such image is having shape of a circular ring . If its angular radius is `beta` then , `Beta = ("arc")/("radius")` `therefore beta = = (b)/(b)` `= sqrt((n-1)K_1D)/(b)` `= sqrt(((n-1)K_1)/(v^2) xx (uv)/(u+v)`) `( therefore" From equation (4) D " = (uv)/(u+v))` `thereforebeta = sqrt(((n-)K_u)/(v(u+v)))"".......(12)` `rArr` Above equation gives required result. |
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