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Question Number 47497 by MrW3 last updated on 10/Nov/18

Commented by MrW3 last updated on 10/Nov/18

$$Findtheareaoftheshadowofthe$$$$cylinderproducedbyapointsourceof$$$$lightS.$$$$$$$$seealsoQ47449.$$

Answered by mr W last updated on 12/Nov/18

Commented by mr W last updated on 12/Nov/18

$$Ifthesidewallofthecylinderwere$$$$transparent,wewouldonlyseethe$$$$shadowof{cylinder}^{\prime }stop.Sincethe$$$$topisacircle,itcanbeeasilyseen,$$$$thatitsshadowhasalsotheshapeof$$$$acircle.$$$$\frac{L}{d+r}=\frac{H}{H-h}$$$$\Rightarrow L=\frac{H(d+r)}{H-h}$$$$\frac{2e}{2r}=\frac{H}{H-h}$$$$\Rightarrow e=\frac{Hr}{H-h}=b$$$$\cos \theta =\frac{r}{d}$$$$\Rightarrow \theta ={\cos }^{-1}\frac{r}{d}$$$$A_{shadow}=e(L-e)\sin \theta +\pi e^2-\theta e^2-\pi r^2$$$$A_{shadow}=\frac{Hr}{H-h}[\frac{H(d+r)}{H-h}-\frac{Hr}{H-h}]\sin \theta +(\pi -{\cos }^{-1}\frac{r}{d})\frac{H^2{r}^2}{{(H-h)}^2}-\pi r^2$$$$\Rightarrow A_{shadow}=\frac{H^{2}r}{{(H-h)}^2}[\sqrt{d^2-r^2}+r(\pi -{\cos }^{-1}\frac{r}{d})]-\pi r^2$$

Commented by ajfour last updated on 14/Nov/18

$$ThankyouSir.Iwasmislead!$$

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