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A plane light wave with wavelength lambda = 0.60 mu m falls normally on a sufficiently large glass plate having a round recess on the oppiste side (Fig.) For the observation point P that recess correcponds to the first one and and a half Fresnel zones. Find the depth h of the recess at which the intensity of light at the point P is (a) maximum, (b) minimum, (c ) equal to the intensity of incident light. |
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Answer» Solution :We could require the comtribution to the amplitude of a wave at a point from half a Fresnel zone. For this we proceed DIRECTLY from the Fresnel Huyghens principle. The complex amplitude is written as `E = int K (varphi) (a_(0))/(r ) e^(-ikr) dS` Here `K(varphi)` is a factor which depends on the angle `varphi` between a normal `overset rarr(n)` to the area `dS` and the direction from `dS` to the point `P` and `r` is the distance from the element `dS` to `P`. We see that for the first Fresnal zone  (using `r = B + (rho^(2))/(2b)` (for `sqrt(rho^(2)b^(2))))` `E = (a_(0))/(b) int_(0)^(sqrt(b lambda)) e^(-ikb - ik rho^(2)//2b) 2pi rho d rho (K(varphi) ~~1)` For the first Fresnel zone `r = b+ lambda//2` so `r^(2) = b^(2) + b lambda rho` and `rho^(2) = b lambda` Thus `E = (a_(0))/(b)e^(-ikb) 2pi int_(0)^((b lambda)/(2)) e^(-(kx)/(b)) dx` `= (a_(0))/(b) 2pi e^(ikb) (e^(iklambda//2) - 1)/(-ik//b)` `= (a_(0))/(k) 2piie^(-ikb) (-2) = - (4PI)/(k)ia_(0)e^(-ikb) = A_(1)` For the next half `E = (a_(0))/(b)e^(-ikb) 2pi int_((b lambda)/(2))^((3b lambda)/(4)) e^(-ikx//b) d xx` `= (a_(0))/(k) 2pi ie^(-ikb) (e^(-i)(3klambda)/(4) - e^(-iklambda//2))` `= (a_(0))/(k) 2pi ie^(-ikb) (+1+i) =- (A_(1)(1+i))/(2)` If we calaclate the contribution of the full `2^(nd)` Fresnel zone we will get `-A_(1)`. If we take account of the factors `K(varphi)` and `(1)/(r )` which decrease monotonically we expect the contribution to change to `-A_(2)`. Thus we write for the combtribution of the half zones in the `2^(nd)` Fresnel zone as `-(A_(2)(1 + i))/(2)` and `-(A_(2) (1 -i))/(2)` The part lying in the RECESS has an extra phase difference equal to `-del =- (2pi)/(lambda) (n - 1)h`. Thus the full amplitude is (note that the correct from is `e^(-ikr)`) `(A_(1)-(A_(2))/(2) (1+i))e^(+idel) - (A_(2))/(2) (1-i) + A_(3) - A_(4) +........` `~~((A_(1))/(2)(1-i))e^(+idel) - (A_(2))/(2) (1-i) + (A_(3))/(2)` `~~ ((A_(1))/(2)(1-i))e^(+i del) + i(A_(1))/(2)` (as `A_(2)~~A_(3)~~A_(1))` and `A_(3) - A_(4) + A_(5)......= (A_(3))/(2)`. The corresponding intensity is `I = (A_(1)^(2))/(4) [(1-i)e^(+i del) + (i)/(e)][(1 + i)e^(-i del) -i]` `I_(0)[3 -2 cos del + 2sin del] = I_(0) [3 xx 2sqrt(2) sin (del-(pi)/(4))]` `(a)` For maximum intensity `sin (del - (pi)/(4)) = +1` or `del - (pi)/(2) = 2kpi + (pi)/(2), k = 0, 1, 2,....` `del = 2kpi + (3pi)/(4) = (2pi)/(lambda) (n - 1) h` so `h = (lambda)/(n -1) (k + (3)/(8))` `(b)` For minimum intensity `sin(delta - (pi)/(4)) = 1` `del -(pi)/(4) = 2kpi+(3pi)/(2)` or `del = 2k pi + (7pi)/(4)` so `h = (lambda)/(n - 1) (k + (7pi)/(8))` (c ) `{:(For I=I_(0)cos delta=0),(sin delta=-1):}}` or `{:{(sin delta=0),(cos delta=+1):}` Thus `delta = 2k pi + (3pi)/(2) h = (k lambda)/(n - 1)` or `delta = 2kpi + (3pi)/(2), h (lambda)/(n - 1) (k + (3pi)/(4))` |
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