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Question Number 54357 by ajfour last updated on 02/Feb/19

Commented by ajfour last updated on 02/Feb/19

$$Find\mathit{\alpha }suchthattherayoflight$$$$beingreflectedthewayshown,$$$$reachesCfromS.$$$$Forexampleletstake\mathit{h}=\frac{5R}{2}$$$$and\mathit{b}=\frac{3R}{2}.$$

Answered by mr W last updated on 02/Feb/19

Commented by mr W last updated on 03/Feb/19

$$let\delta =\frac{b}{R},\lambda =\frac{h}{R}$$$$\alpha -\beta =\frac{\pi }{2}-\theta$$$$\Rightarrow \beta =\alpha +\theta -\frac{\pi }{2}$$$$(2b-R\sin \theta )\frac{1}{\tan \alpha }+R(1-\cos \theta )=h$$$$\Rightarrow \tan \alpha =\frac{2b-R\sin \theta }{h-R(1-\cos \theta )}=\frac{2\delta -\sin \theta }{\lambda -1+\cos \theta }$$$$(b-R\tan \frac{\theta }{2})\frac{1}{\sin \beta }=\left(R\tan \frac{\theta }{2}\right)\frac{1}{\sin (\theta +\beta )}$$$$(b-R\tan \frac{\theta }{2})\frac{1}{\sin (\theta +\alpha -\frac{\pi }{2})}=\left(R\tan \frac{\theta }{2}\right)\frac{1}{\sin (2\theta +\alpha -\frac{\pi }{2})}$$$$(\frac{\delta }{\tan \frac{\theta }{2}}-1)\frac{\cos (2\theta +\alpha )}{\cos (\theta +\alpha )}=1$$$$(\frac{\delta }{\tan \frac{\theta }{2}}-1)\frac{\cos 2\theta -\sin 2\theta \tan \alpha }{\cos \theta -\sin \theta \tan \alpha }=1$$$$(\frac{\delta }{\tan \frac{\theta }{2}}-1)\frac{\cos 2\theta -\sin 2\theta \frac{2\delta -\sin \theta }{\lambda -1+\cos \theta }}{\cos \theta -\sin \theta \frac{2b-R\sin \theta }{h-R(1-\cos \theta )}}=1$$$$\Rightarrow (\frac{\delta }{\tan \frac{\theta }{2}}-1)\frac{\left(\cos 2\theta \right)(\lambda -1+\cos \theta )-\left(\sin 2\theta \right)(2\delta -\sin \theta )}{\left(\cos \theta \right)(\lambda -1+\cos \theta )-\left(\sin \theta \right)(2\delta -\sin \theta )}=1$$$$\Rightarrow \theta =....$$$$example:$$$$\delta =\frac{b}{R}=\frac{3}{2},\lambda =\frac{h}{R}=\frac{5}{2}$$$$\Rightarrow \theta =80.5644°\Rightarrow \alpha =50.4304°$$

Commented by ajfour last updated on 03/Feb/19

$$fartoogoodSir,thanks.$$$$whatifiassumetangenttocircle$$$$atpointofreflection\left(with\theta assumed\right)$$$$andtakereflectionofCintangent;$$$$andthenjoinS{}^{\prime }toC{}^{\prime }.ithink{it}^{\prime }ll$$$$beunnecessaryandequivalentto$$$$howyouhavesolved.$$$$Dontyouthinkso?$$

Commented by mr W last updated on 03/Feb/19

$${that}^{\prime }scorrectsir.$$$$iconsideredittoo,butthis{doesn}^{\prime }t$$$$helpsomuchifwetrytosolvewith$$$$trigonometricalmethod.butithelps$$$$moreifwetrytousecoordinate$$$$method.$$

Answered by ajfour last updated on 03/Feb/19

Commented by ajfour last updated on 03/Feb/19

$$Ifinallygot$$$$\frac{(h-R+R\cos \theta )\tan 2\theta +3b-2R\sin \theta }{h-\tan 2\theta (2b-R\sin \theta )}$$$$=\frac{2b-R\sin \theta }{h-R+R\cos \theta }=\tan \mathit{\alpha }$$$$Andforh=\frac{5R}{2},b=\frac{3R}{2}$$$$\theta =80.5644°,\alpha =50.43038°.$$

Commented by mr W last updated on 03/Feb/19

$$only80.5644°isthesolution.lightray$$$$cannotgothroughthecircle.90°is$$$$thesolutionwhen\frac{h}{R}=5and\frac{b}{R}=1.5.$$

Commented by ajfour last updated on 03/Feb/19

$$yesSir,youareright.$$

Commented by mr W last updated on 03/Feb/19

$$B(R\sin \theta ,-R\cos \theta )$$$$slopeofOB:$$$$\tan {\phi }_{OB}=-\frac{1}{\tan \theta }$$$$slopeof{BS}^{\prime }:$$$$\tan {\phi }_{{BS}^{\prime }}=\frac{h-R+R\cos \theta }{2b-R\sin \theta }$$$$slopeofBC:$$$$\tan {\phi }_{BC}=\frac{-R+R\cos \theta }{b-R\sin \theta }$$$${\phi }_{{BS}^{\prime }}-{\phi }_{OB}={\phi }_{OB}-{\phi }_{BC}$$$$\tan ({\phi }_{{BS}^{\prime }}-{\phi }_{OB})=\tan ({\phi }_{OB}-{\phi }_{BC})$$$$\frac{\frac{h-R+R\cos \theta }{2b-R\sin \theta }+\frac{1}{\tan \theta }}{1-\frac{h-R+R\cos \theta }{2b-R\sin \theta }\times \frac{1}{\tan \theta }}=\frac{-\frac{1}{\tan \theta }-\frac{-R+R\cos \theta }{b-R\sin \theta }}{1-\frac{1}{\tan \theta }\times \frac{-R+R\cos \theta }{b-R\sin \theta }}$$$$\frac{(h-R+R\cos \theta )\sin \theta +(2b-R\sin \theta )\cos \theta }{(2b-R\sin \theta )\sin \theta -(h-R+R\cos \theta )\cos \theta }=\frac{R(1-\cos \theta )\sin \theta -(b-R\sin \theta )\cos \theta }{(b-R\sin \theta )\sin \theta +R(1-\cos \theta )\cos \theta }$$$$\frac{(h-R)\sin \theta +2b\cos \theta }{2b\sin \theta -(h-R)\cos \theta -R}=\frac{R\sin \theta -b\cos \theta }{b\sin \theta +R\cos \theta -R}$$$$\Rightarrow \frac{(\lambda -1)\sin \theta +2\delta \cos \theta }{2\delta \sin \theta -(\lambda -1)\cos \theta -1}=\frac{\sin \theta -\delta \cos \theta }{\delta \sin \theta +\cos \theta -1}$$

Commented by ajfour last updated on 04/Feb/19

$$butifwelet$$$${\mathit{\rho }}^2={(\lambda -1)}^2+{\left(2\delta \right)}^2;\varphi ={\tan }^{-1}\left(\frac{\lambda -1}{2\delta }\right)$$$${\mathit{\mu }}^2=1+{\delta }^{2}and\psi ={\tan }^{-1}\left(\frac{1}{\delta }\right)$$$$\frac{\rho \cos (\theta -\varphi )}{\rho \sin (\theta -\varphi )-1}=\frac{-\mu \cos (\theta +\psi )}{\mu \sin (\theta +\psi )-1}$$$$\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}$$$$\Rightarrow \mu \rho \sin (2\theta -\varphi +\psi )$$$$=\rho \cos (\theta -\varphi )+\mu \cos (\theta +\psi )$$$$\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}.$$

Commented by mr W last updated on 04/Feb/19

$$bestformforresult!$$

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