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. $2h$
• D. $4h$

Answer 1

Answer: (C)

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

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