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Question-54357

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Question Number 54357 by ajfour last updated on 02/Feb/19
Commented by ajfour last updated on 02/Feb/19
Find 𝛂 such that the ray of light being reflected the way shown, reaches C from S. For example lets take h=((5R)/2) and b=((3R)/2).
Findαsuchthattherayoflightbeingreflectedthewayshown,reachesCfromS.Forexampleletstakeh=5R2andb=3R2.
Answered by mr W last updated on 02/Feb/19
Commented by mr W last updated on 03/Feb/19
let δ=(b/R), λ=(h/R) α−β=(π/2)−θ ⇒β=α+θ−(π/2) (2b−R sin θ)(1/(tan α))+R(1−cos θ)=h ⇒tan α=((2b−R sin θ)/(h−R(1−cos θ)))=((2δ−sin θ)/(λ−1+cos θ)) (b−R tan (θ/2))(1/(sin β))=(R tan (θ/2))(1/(sin (θ+β))) (b−R tan (θ/2))(1/(sin (θ+α−(π/2))))=(R tan (θ/2))(1/(sin (2θ+α−(π/2)))) ((δ/(tan (θ/2)))−1)((cos (2θ+α))/(cos (θ+α)))=1 ((δ/(tan (θ/2)))−1)((cos 2θ−sin 2θ tan α)/(cos θ−sin θ tan α))=1 ((δ/(tan (θ/2)))−1)((cos 2θ−sin 2θ((2δ−sin θ)/(λ−1+cos θ)))/(cos θ−sin θ((2b−R sin θ)/(h−R(1−cos θ)))))=1 ⇒((δ/(tan (θ/2)))−1)(((cos 2θ)(λ−1+cos θ)−(sin 2θ)(2δ−sin θ))/((cos θ)(λ−1+cos θ)−(sin θ)(2δ−sin θ)))=1 ⇒θ=.... example: δ=(b/R)=(3/2), λ=(h/R)=(5/2) ⇒θ=80.5644°⇒α=50.4304°
letδ=bR,λ=hRα−β=π2−θ⇒β=α+θ−π2(2b−Rsinθ)1tanα+R(1−cosθ)=h⇒tanα=2b−Rsinθh−R(1−cosθ)=2δ−sinθλ−1+cosθ(b−Rtanθ2)1sinβ=(Rtanθ2)1sin(θ+β)(b−Rtanθ2)1sin(θ+α−π2)=(Rtanθ2)1sin(2θ+α−π2)(δtanθ2−1)cos(2θ+α)cos(θ+α)=1(δtanθ2−1)cos2θ−sin2θtanαcosθ−sinθtanα=1(δtanθ2−1)cos2θ−sin2θ2δ−sinθλ−1+cosθcosθ−sinθ2b−Rsinθh−R(1−cosθ)=1⇒(δtanθ2−1)(cos2θ)(λ−1+cosθ)−(sin2θ)(2δ−sinθ)(cosθ)(λ−1+cosθ)−(sinθ)(2δ−sinθ)=1⇒θ=….example:δ=bR=32,λ=hR=52⇒θ=80.5644°⇒α=50.4304°
Commented by ajfour last updated on 03/Feb/19
far too good Sir, thanks. what if i assume tangent to circle at point of reflection (with θ assumed) and take reflection of C in tangent; and then join S ′ to C ′. i think it′ll be unnecessary and equivalent to how you have solved. Dont you think so?
fartoogoodSir,thanks.whatifiassumetangenttocircleatpointofreflection(withθassumed)andtakereflectionofCintangent;andthenjoinS′toC′.ithinkit′llbeunnecessaryandequivalenttohowyouhavesolved.Dontyouthinkso?
Commented by mr W last updated on 03/Feb/19
that′s correct sir. i considered it too, but this doesn′t help so much if we try to solve with trigonometrical method. but it helps more if we try to use coordinate method.
that′scorrectsir.iconsideredittoo,butthisdoesn′thelpsomuchifwetrytosolvewithtrigonometricalmethod.butithelpsmoreifwetrytousecoordinatemethod.
Answered by ajfour last updated on 03/Feb/19
Commented by ajfour last updated on 03/Feb/19
I finally got (((h−R+Rcos θ)tan 2θ+3b−2Rsin θ)/(h−tan 2θ(2b−Rsin θ))) =((2b−Rsin θ)/(h−R+Rcos θ)) =tan 𝛂 And for h=((5R)/2) , b=((3R)/2) θ=80.5644° , α=50.43038°.
Ifinallygot(h−R+Rcosθ)tan2θ+3b−2Rsinθh−tan2θ(2b−Rsinθ)=2b−Rsinθh−R+Rcosθ=tanαAndforh=5R2,b=3R2θ=80.5644°,α=50.43038°.
Commented by mr W last updated on 03/Feb/19
only 80.5644° is the solution. light ray can not go through the circle. 90° is the solution when (h/R)=5 and (b/R)=1.5.
only80.5644°isthesolution.lightraycannotgothroughthecircle.90°isthesolutionwhenhR=5andbR=1.5.
Commented by ajfour last updated on 03/Feb/19
yes Sir, you are right.
yesSir,youareright.
Commented by mr W last updated on 03/Feb/19
B(R sin θ,−R cos θ) slope of OB: tan ϕ_(OB) =−(1/(tan θ)) slope of BS′: tan ϕ_(BS′) =((h−R+R cos θ)/(2b−R sin θ)) slope of BC: tan ϕ_(BC) =((−R+R cos θ)/(b−R sin θ)) ϕ_(BS′) −ϕ_(OB) =ϕ_(OB) −ϕ_(BC) tan (ϕ_(BS′) −ϕ_(OB) )=tan (ϕ_(OB) −ϕ_(BC) ) ((((h−R+R cos θ)/(2b−R sin θ))+(1/(tan θ)))/(1−((h−R+R cos θ)/(2b−R sin θ))×(1/(tan θ))))=((−(1/(tan θ))−((−R+R cos θ)/(b−R sin θ)))/(1−(1/(tan θ))×((−R+R cos θ)/(b−R sin θ)))) (((h−R+R cos θ)sin θ+(2b−R sin θ)cos θ)/((2b−R sin θ)sin θ−(h−R+R cos θ)cos θ))=((R(1−cos θ)sin θ−(b−R sin θ)cos θ)/((b−R sin θ)sin θ+R(1−cos θ)cos θ)) (((h−R)sin θ+2b cos θ)/(2b sin θ−(h−R)cos θ−R))=((R sin θ−b cos θ)/(b sin θ+R cos θ−R)) ⇒(((λ−1)sin θ+2δ cos θ)/(2δ sin θ−(λ−1)cos θ−1))=((sin θ−δ cos θ)/(δ sin θ+cos θ−1))
B(Rsinθ,−Rcosθ)slopeofOB:tanφOB=−1tanθslopeofBS′:tanφBS′=h−R+Rcosθ2b−RsinθslopeofBC:tanφBC=−R+Rcosθb−RsinθφBS′−φOB=φOB−φBCtan(φBS′−φOB)=tan(φOB−φBC)h−R+Rcosθ2b−Rsinθ+1tanθ1−h−R+Rcosθ2b−Rsinθ×1tanθ=−1tanθ−−R+Rcosθb−Rsinθ1−1tanθ×−R+Rcosθb−Rsinθ(h−R+Rcosθ)sinθ+(2b−Rsinθ)cosθ(2b−Rsinθ)sinθ−(h−R+Rcosθ)cosθ=R(1−cosθ)sinθ−(b−Rsinθ)cosθ(b−Rsinθ)sinθ+R(1−cosθ)cosθ(h−R)sinθ+2bcosθ2bsinθ−(h−R)cosθ−R=Rsinθ−bcosθbsinθ+Rcosθ−R⇒(λ−1)sinθ+2δcosθ2δsinθ−(λ−1)cosθ−1=sinθ−δcosθδsinθ+cosθ−1
Commented by ajfour last updated on 04/Feb/19
but if we let 𝛒^2 =(λ−1)^2 +(2δ)^2 ; φ=tan^(−1) (((λ−1)/(2δ))) 𝛍^2 =1+δ^2 and ψ=tan^(−1) ((1/δ)) ((ρcos (θ−φ))/(ρsin (θ−φ)−1)) = ((−μcos (θ+ψ))/(μsin (θ+ψ)−1)) ____________________________ ⇒ μρsin (2θ−φ+ψ) = ρcos (θ−φ)+μcos (θ+ψ) ____________________________.
butifweletρ2=(λ−1)2+(2δ)2;ϕ=tan−1(λ−12δ)μ2=1+δ2andψ=tan−1(1δ)ρcos(θ−ϕ)ρsin(θ−ϕ)−1=−μcos(θ+ψ)μsin(θ+ψ)−1____________________________⇒μρsin(2θ−ϕ+ψ)=ρcos(θ−ϕ)+μcos(θ+ψ)____________________________.
Commented by mr W last updated on 04/Feb/19
best form for result!
bestformforresult!

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