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Question Number 44676 by ajfour last updated on 03/Oct/18

Commented by ajfour last updated on 03/Oct/18

$$If\mathit{\mu }\left(x\right)=\frac{{\mu }_{0}L}{L+({\mu }_0-1)x}$$$$Findequationofpathoflight$$$$asittravelsinthemediumwith$$$$refractiveindex\mu \left(x\right).$$$$Hencedetermine\mathit{\beta }intermsof$$$$\mathit{\alpha }and{\mathit{\mu }}_0.$$

Commented by MrW3 last updated on 03/Oct/18

$$\mathit{\mu }\left(x\right)=\frac{{\mu }_{0}L}{L+({\mu }_0-1)x}$$$$\frac{d\mu }{dx}=-\frac{({\mu }_0-1){\mu }_{0}L}{{[L+({\mu }_0-1)x]}^2}=-\frac{({\mu }_0-1)\mu }{L+({\mu }_0-1)x}$$$$\mu \sin \theta =constant$$$$\sin \theta \frac{d\mu }{dx}+\mu \cos \theta \frac{d\theta }{dx}=0$$$$-\frac{({\mu }_0-1)\mu }{L+({\mu }_0-1)x}+\mu \frac{\cos \theta }{\sin \theta }\times \frac{d\theta }{dx}=0$$$$-\frac{({\mu }_0-1)}{L+({\mu }_0-1)x}+\frac{d\left(\ln \sin \theta \right)}{dx}=0$$$$let\mathrm{\Phi }=\ln \sin \theta$$$$-\frac{({\mu }_0-1)}{L+({\mu }_0-1)x}+\frac{d\mathrm{\Phi }}{dx}=0$$$$d\mathrm{\Phi }=\frac{({\mu }_0-1)dx}{L+({\mu }_0-1)x}$$$${\int }_{{\mathrm{\Phi }}_0}^{\mathrm{\Phi }}d\mathrm{\Phi }={\int }_0^x\frac{({\mu }_0-1)dx}{L+({\mu }_0-1)x}$$$$\mathrm{\Phi }-{\mathrm{\Phi }}_0={\left[\ln \{L+({\mu }_0-1)x\}\right]}_0^x$$$$\ln \frac{\sin \theta }{\sin {\theta }_0}=\ln \frac{L+({\mu }_0-1)x}{L}$$$$\sin \theta =\sin {\theta }_0\frac{L+({\mu }_0-1)x}{L}$$$$atx=L:$$$$\sin {\theta }_1={\mu }_0\sin {\theta }_0$$$$\Rightarrow {\theta }_1={\sin }^{-1}\left({\mu }_0\sin {\theta }_0\right)$$$$\sin \theta =\sin {\theta }_0\frac{L+({\mu }_0-1)x}{L}=\sin {\theta }_0[1+({\mu }_0-1)\frac{x}{L}]=\lambda \left(x\right)$$$$\lambda \left(x\right)=\sin {\theta }_0[1+({\mu }_0-1)\frac{x}{L}]$$$$d\lambda =\sin {\theta }_0\frac{{\mu }_0-1}{L}dx$$$$\tan \theta =\frac{\sin \theta }{\sqrt{1-{\sin }^2\theta }}=\frac{\lambda }{\sqrt{1-{\lambda }^2}}$$$$\frac{dy}{dx}=\frac{\lambda }{\sqrt{1-{\lambda }^2}}$$$$dy=\frac{\lambda }{\sqrt{1-{\lambda }^2}}dx=\frac{\lambda }{\sqrt{1-{\lambda }^2}}\times \frac{L}{({\mu }_0-1)\sin {\theta }_0}\times d\lambda$$$$\frac{({\mu }_0-1)\sin {\theta }_0}{L}dy=\frac{\lambda }{\sqrt{1-{\lambda }^2}}d\lambda$$$$\frac{({\mu }_0-1)\sin {\theta }_0}{L}{\int }_0^{y}dy={\int }_{{\lambda }_0}^{\lambda }\frac{\lambda }{\sqrt{1-{\lambda }^2}}d\lambda$$$$\frac{({\mu }_0-1)\sin {\theta }_0}{L}y=-\frac{1}{2}{\int }_{{\lambda }_0}^{\lambda }\frac{1}{\sqrt{1-{\lambda }^2}}d(1-{\lambda }^2)$$$$\frac{({\mu }_0-1)\sin {\theta }_0}{L}y=-{\left[\sqrt{1-{\lambda }^2}\right]}_{{\lambda }_0}^{\lambda }=\sqrt{1-{\lambda }_0^2}-\sqrt{1-{\lambda }^2}$$$${\lambda }_0=\sin {\theta }_0$$$$\frac{({\mu }_0-1)\sin {\theta }_0}{L}y=\cos {\theta }_0-\cos \theta =\cos {\theta }_0-\sqrt{1-{\sin }^2{\theta }_0{[1+({\mu }_0-1)\frac{x}{L}]}^2}$$$$\Rightarrow y=\frac{L}{({\mu }_0-1)\sin {\theta }_0}\{\cos {\theta }_0-\sqrt{1-{\sin }^2{\theta }_0{[1+({\mu }_0-1)\frac{x}{L}]}^2}\}$$$$$$$$atx=0:\mu \left(0\right)={\mu }_0$$$$fromairintomedium$$$$\sin \alpha ={\mu }_0\sin {\theta }_0$$$$\Rightarrow \sin {\theta }_0=\frac{\sin \alpha }{{\mu }_0}$$$$atx=L:\mu \left(1\right)=1$$$$frommediumintoair$$$$\mu \left(1\right)\times \sin {\theta }_1=\sin \beta$$$$\Rightarrow \beta ={\theta }_1={\sin }^{-1}\left({\mu }_0\sin {\theta }_0\right)={\sin }^{-1}\left(\sin \alpha \right)=\alpha$$$$\Rightarrow y=\frac{{\mu }_{0}L}{({\mu }_0-1)\sin \alpha }\{\sqrt{1-{\left(\frac{\sin \alpha }{{\mu }_0}\right)}^2}-\sqrt{1-{\left(\frac{\sin \alpha }{{\mu }_0}\right)}^2{[1+({\mu }_0-1)\frac{x}{L}]}^2}\}$$

Commented by ajfour last updated on 03/Oct/18

$$Awesome,Sir.$$

Commented by MrW3 last updated on 03/Oct/18

$$thankyouforcheckingsir!$$

Commented by ajfour last updated on 03/Oct/18

$$Sir,ouranswersdifferonlyby$$$$afactorofc=\sin \alpha .$$

Commented by MrW3 last updated on 03/Oct/18

$$youarerightsir.Ioverlookedsomething.$$

Answered by ajfour last updated on 03/Oct/18

$$\sin \alpha ={\mu }_0\sin {\theta }_0=\sin \beta =\mu \sin \theta =c$$$${\left(\frac{dy}{dx}\right)}^2=\tan {}^2\theta =\frac{\sin {}^2\theta }{1-\sin {}^2\theta }$$$$\frac{dy}{dx}=\frac{c/\mu }{\sqrt{1-\frac{c^2}{{\mu }^2}}}=\frac{c}{\sqrt{{\mu }^2-c^2}}$$$$y={\int }_{{\mu }_0}^{\mu }\frac{c}{\sqrt{{\mu }^2-c^2}}\times \left(\frac{dx}{d\mu }\right)d\mu$$$$\mathit{\mu }=\frac{{\mathit{\mu }}_{0}L}{L+({\mathit{\mu }}_0-1)x}$$$$({\mu }_0-1)x=\frac{{\mu }_{0}L}{\mu }-L$$$$\Rightarrow \frac{dx}{d\mu }=-\frac{{\mu }_{0}L}{{\mu }^2({\mu }_0-1)}$$$$y=-{\int }_{{\mu }_0}^{\mu }\frac{c}{\sqrt{{\mu }^2-c^2}}\left(\frac{{\mu }_{0}L}{{\mu }^2({\mu }_0-1)}\right)d\mu$$$$=\frac{{\mu }_{0}cL}{{\mu }_0-1}{\int }_{\mu }^{{\mu }_0}\frac{d\mu }{{\mu }^2\sqrt{{\mu }^2-c^2}}$$$$let\mu =c\sec \varphi$$$$\Rightarrow d\mu =c\sec \varphi \tan \varphi d\varphi$$$$y=\frac{{\mu }_{0}cL}{{\mu }_0-1}{\int }_{\varphi }^{{\varphi }_0}\frac{c\sec \varphi \tan \varphi d\varphi }{c^2\sec {}^2\varphi \left(c\tan \varphi \right)}$$$$=\frac{{\mu }_{0}L}{c({\mu }_0-1)}[\sin {\varphi }_0-\sin \varphi ]$$$$\mathit{y}=\frac{{\mathit{\mu }}_0\mathit{L}}{\mathit{c}({\mathit{\mu }}_0-1)}[\frac{\sqrt{{\mathit{\mu }}_0^2-{\mathit{c}}^2}}{{\mathit{\mu }}_0}-\frac{\sqrt{{\mathit{\mu }}^2-{\mathit{c}}^2}}{\mathit{\mu }}]$$$$with\mathit{\mu }=\frac{{\mathit{\mu }}_0\mathit{L}}{\mathit{L}+({\mathit{\mu }}_0-1)\mathit{x}}\mathrm{\&}\mathit{c}=\sin \mathit{\alpha }.$$$$$$

Commented by MrW3 last updated on 03/Oct/18

$$niceandstraightforward!$$

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