MCQ
A long horizontal mirror is next to a vertical screen (seen figure). Parallel light rays are falling on the mirror at an angle $\alpha$ from the vertical. If a vertical object of height $h$ is 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
    $h \tan \alpha$
  • $2 \,h$
  • D
    $4 \,h$

Answer

Correct option: C.
$2 \,h$
c
(c)

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

From similar triangles $\triangle B G F$ and $\triangle D E F$,

we have $\quad \frac{D E}{B G}=\frac{F E}{G F}$

$\frac{h^{\prime}}{h}=\left(\frac{d-x}{x}\right)$

$\Rightarrow \quad \frac{d}{x}=\frac{h^{\prime}+h}{h} \quad \dots(i)$

Now, from similar triangles $\triangle A B G$ and $\triangle A C E$, we have

$\frac{C E}{A E}=\frac{B G}{A G}$

$\therefore \quad \triangle A B G \cong \triangle F B G$

and $\quad A G=G F=x$

$\Rightarrow \frac{H+h^{\prime}}{d+x}=\frac{h}{x}$

$\Rightarrow \frac{d}{x}=\frac{H+h^{\prime}-h}{h} \quad \dots(ii)$

Equating Eqs. $(i)$ and $(ii)$, we get

$\frac{h^{\prime}+h}{h}=\frac{H+h^{\prime}-h}{h}$

$\Rightarrow \quad 2 h=H$

Hence height of shadow on wall is $2h$

 

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