Finding the Best Focus of a Concave Mirror

Radius=r
 

Using Figure 1, the extra phases, with respect to the x position of the axial surface of the mirror (at the vertical line) and due to the presence of the curved mirror are.

δ ϕ inc =−k( r− r 2 − y 2 ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOabaeqabaGaeqiTdq Maeqy1dy2aaSbaaSqaaiaadMgacaWGUbGaam4yaaqabaGccqGH9aqp cqGHsislcaWGRbWaaeWaaeaacaWGYbGaeyOeI0YaaOaaaeaacaWGYb WaaWbaaSqabeaacaaIYaaaaOGaeyOeI0IaamyEamaaCaaaleqabaGa aGOmaaaaaeqaaaGccaGLOaGaayzkaaaabaaaaaa@4786@  

(0.1)    

δ ϕ exit =k ( f−( r− r 2 − y 2 ) ) 2 + y 2 −kf MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabes7aKjabew 9aMnaaBaaaleaacaWGLbGaamiEaiaadMgacaWG0baabeaakiabg2da 9iaadUgadaGcaaqaamaabmaabaGaamOzaiabgkHiTmaabmaabaGaam OCaiabgkHiTmaakaaabaGaamOCamaaCaaaleqabaGaaGOmaaaakiab gkHiTiaadMhadaahaaWcbeqaaiaaikdaaaaabeaaaOGaayjkaiaawM caaaGaayjkaiaawMcaamaaCaaaleqabaGaaGOmaaaakiabgUcaRiaa dMhadaahaaWcbeqaaiaaikdaaaaabeaakiabgkHiTiaadUgacaWGMb aaaa@5196@  

(0.2)

We want to minimize the variation due to y so we can take the derivative of the sum of these phase increments with respect to y and solve for the f value that does that.  I have done that in Mathematica but it turns out to need numerical solution so we will be satisfied with showing the results of an approximation.

It turns out that, for y/r<<1, f=r/2 is very close to the solution as may be shown by expanding the square root terms in Taylor series.

δ ϕ inc =−k( r− r 2 − y 2 )≈−kr[ 1−( 1− y 2 2 r 2 ) ]=−k y 2 2r MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabes7aKjabew 9aMnaaBaaaleaacaWGPbGaamOBaiaadogaaeqaaOGaeyypa0JaeyOe I0Iaam4AamaabmaabaGaamOCaiabgkHiTmaakaaabaGaamOCamaaCa aaleqabaGaaGOmaaaakiabgkHiTiaadMhadaahaaWcbeqaaiaaikda aaaabeaaaOGaayjkaiaawMcaaiabgIKi7kabgkHiTiaadUgacaWGYb WaamWaaeaacaaIXaGaeyOeI0YaaeWaaeaacaaIXaGaeyOeI0YaaSaa aeaacaWG5bWaaWbaaSqabeaacaaIYaaaaaGcbaGaaGOmaiaadkhada ahaaWcbeqaaiaaikdaaaaaaaGccaGLOaGaayzkaaaacaGLBbGaayzx aaGaeyypa0JaeyOeI0Iaam4AamaalaaabaGaamyEamaaCaaaleqaba GaaGOmaaaaaOqaaiaaikdacaWGYbaaaaaa@5E0D@  

(0.3)

δ ϕ exit =k ( f− y 2 2r ) 2 + y 2 ≈k f 2 −2f y 2 2r + y 2 −kf MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabes7aKjabew 9aMnaaBaaaleaacaWGLbGaamiEaiaadMgacaWG0baabeaakiabg2da 9iaadUgadaGcaaqaamaabmaabaGaamOzaiabgkHiTmaalaaabaGaam yEamaaCaaaleqabaGaaGOmaaaaaOqaaiaaikdacaWGYbaaaaGaayjk aiaawMcaamaaCaaaleqabaGaaGOmaaaakiabgUcaRiaadMhadaahaa WcbeqaaiaaikdaaaaabeaakiabgIKi7kaadUgadaGcaaqaaiaadAga daahaaWcbeqaaiaaikdaaaGccqGHsislcaaIYaGaamOzamaalaaaba GaamyEamaaCaaaleqabaGaaGOmaaaaaOqaaiaaikdacaWGYbaaaiab gUcaRiaadMhadaahaaWcbeqaaiaaikdaaaaabeaakiabgkHiTiaadU gacaWGMbaaaa@5AAF@  

(0.4)

where we dropped y4 terms.

If now we substitute in f=r/2 we obtain:

δ ϕ exit ≈kf( 1−2 y 2 2fr + y 2 f 2 −1 )= kr 2 ( 1−2 y 2 r 2 +4 y 2 r 2 −1 ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabes7aKjabew 9aMnaaBaaaleaacaWGLbGaamiEaiaadMgacaWG0baabeaakiabgIKi 7kaadUgacaWGMbWaaeWaaeaadaGcaaqaaiaaigdacqGHsislcaaIYa WaaSaaaeaacaWG5bWaaWbaaSqabeaacaaIYaaaaaGcbaGaaGOmaiaa dAgacaWGYbaaaiabgUcaRmaalaaabaGaamyEamaaCaaaleqabaGaaG OmaaaaaOqaaiaadAgadaahaaWcbeqaaiaaikdaaaaaaaqabaGccqGH sislcaaIXaaacaGLOaGaayzkaaGaeyypa0ZaaSaaaeaacaWGRbGaam OCaaqaaiaaikdaaaWaaeWaaeaadaGcaaqaaiaaigdacqGHsislcaaI YaWaaSaaaeaacaWG5bWaaWbaaSqabeaacaaIYaaaaaGcbaGaamOCam aaCaaaleqabaGaaGOmaaaaaaGccqGHRaWkcaaI0aWaaSaaaeaacaWG 5bWaaWbaaSqabeaacaaIYaaaaaGcbaGaamOCamaaCaaaleqabaGaaG OmaaaaaaaabeaakiabgkHiTiaaigdaaiaawIcacaGLPaaaaaa@62C1@  

(0.5)

and finally using a Taylor Series expansion for the square root we obtain.

δ ϕ exit ≈ kr 2 ( 1+2 y 2 r 2 −1 )≈ kr 2 y 2 r 2 =k y 2 2r =−δ ϕ inc MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr 0=vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabes7aKjabew 9aMnaaBaaaleaacaWGLbGaamiEaiaadMgacaWG0baabeaakiabgIKi 7oaalaaabaGaam4AaiaadkhaaeaacaaIYaaaamaabmaabaWaaOaaae aacaaIXaGaey4kaSIaaGOmamaalaaabaGaamyEamaaCaaaleqabaGa aGOmaaaaaOqaaiaadkhadaahaaWcbeqaaiaaikdaaaaaaaqabaGccq GHsislcaaIXaaacaGLOaGaayzkaaGaeyisIS7aaSaaaeaacaWGRbGa amOCaaqaaiaaikdaaaWaaSaaaeaacaWG5bWaaWbaaSqabeaacaaIYa aaaaGcbaGaamOCamaaCaaaleqabaGaaGOmaaaaaaGccqGH9aqpcaWG RbWaaSaaaeaacaWG5bWaaWbaaSqabeaacaaIYaaaaaGcbaGaaGOmai aadkhaaaGaeyypa0JaeyOeI0IaeqiTdqMaeqy1dy2aaSbaaSqaaiaa dMgacaWGUbGaam4yaaqabaaaaa@6193@  

(0.6)

                                                                                                                                               

Thus for y<<r and f=r/2 the total phase shift due to the y offset becomes approximately zero and independent of y.