Question

A long horizontal mirror is next to a vertical screen (seen figure).

Parallel light rays are falling on the mirror at an angle a from the vertical. If a vertical object of height his kept on the mirror at a distance $(d > h) \tan \alpha$ . The length of the shadow of the object on the screen would be

• A. $\frac{h}{2}$
• B. $h \tan \propto$
• C. $2 h$
• D. $4 h$

Answer 1

Answer: (C)

From geometry of figure, shadow length is $CD (= H).$

From similar triangles $\Delta BGF$ and $\Delta DEF$,
we have $\frac{DE}{BG}=\frac{FE}{GF}$
$\frac{h'}{h}=\left(\frac{d-x}{x}\right) $
$ \frac{\Rightarrow d}{x}=\frac{h'+h}{h} \cdots\left(i\right)$
Now, from similar triangles $\Delta ABG$ and $\Delta ACE$, we have
$\frac{CE}{AE}=\frac{BG}{AG}$
$\therefore \Delta BG \simeq\Delta FBG$
and $AG =GF=x $
$\Rightarrow \frac{H+h'}{d+x}=\frac{h}{x}$
$\Rightarrow \frac{d}{x}=\frac{H+ h'-h}{h} \cdots\left(ii\right) $
Equating Eqs. $\left(i\right)$ and $\left(ii\right),$ we get
$\frac{h'+h}{h}=\frac{H+h'-h}{h}$
$\Rightarrow 2h =H$
Hence, height of shadow on wall is $2h.$