A long horizontal mirror is next to a vertica | vedclass

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(C) From the geometry of the figure,the shadow length on the screen is $CD = H$.From similar triangles $	riangle BGF$ and $	riangle DEF$,we have:$\f

Vedclass · Accepted Answer

$2 \,h$

Vedclass · Answer

(C) From the geometry of the figure,the shadow length on the screen is $CD = H$.From similar triangles $	riangle BGF$ and $	riangle DEF$,we have:$\frac{DE}{BG} = \frac{FE}{GF} \Rightarrow \frac{h'}{h} = \frac{d-x}{x} \Rightarrow \frac{d}{x} = \frac{h' + h}{h} \quad \dots(i)$Now,from similar triangles $	riangle ABG$ and $	riangle ACE$,we have:$\frac{CE}{AE} = \frac{BG}{AG} \Rightarrow \frac{H + h'}{d + x} = \frac{h}{x} \Rightarrow \frac{d+x}{x} = \frac{H + h'}{h} \Rightarrow \frac{d}{x} + 1 = \frac{H + h'}{h} \Rightarrow \frac{d}{x} = \frac{H + h' - h}{h} \quad \dots(ii)$Equating equations $(i)$ and $(ii)$:$\frac{h' + h}{h} = \frac{H + h' - h}{h}$$h' + h = H + h' - h$$H = 2h$Thus,the height of the shadow on the wall is $2h$.

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