Question

A pole $4\text{m}$ high driven into bottom of a lake is $1\text{m}$ above the water. If the incident sun rays make an angle of ${53}^{\circ }$ with the normal at water surface, then the length of the shadow of the pile on the bottom of the lake will be
$[$Given refractive index of water $=\frac{4}{3}]$

• A. $5.58\text{m}$
• B. $4.58\text{m}$
• C. $3.58\text{m}$
• D. $2.58\text{m}$

Answer 1

The correct option is C $3.58\text{m}$


The length of pile is $AC=4\text{m}$
Height of pile above water surface is, $AB=1\text{m}$
$\Rightarrow BC=4-1=3\text{m}$
Applying Snell's law at air - water interface,
$\frac{\sin i}{\sin r}=\mu$
$\Rightarrow \frac{4}{5\sin r}=\frac{4}{3}$
$\Rightarrow \sin r=\frac{3}{5}$
$\Rightarrow r={37}^{\circ }$
From $\Delta DEF,$
$\tan r=\frac{EF}{DE}$
$\Rightarrow \tan {37}^{\circ }=\frac{EF}{3}$
$\Rightarrow EF=3\times \frac{3}{4}=\frac{9}{4}=2.25\text{m}$ ........$\left(1\right)$
Similarly, from $\Delta ABD,$
$\tan ({90}^{\circ }-i)=\frac{AB}{BD}$
$\Rightarrow \tan {37}^{\circ }=\frac{1}{BD}$
$\Rightarrow \frac{3}{4}=\frac{1}{BD}$
$\Rightarrow BD=CE=\frac{4}{3}=1.33\text{m}$ ........$\left(2\right)$
Hence, length of shadow of pile on bottom of lake,
$CF=CE+EF$
$\Rightarrow CF=1.33+2.25$
$[$from $\left(1\right)$ and $\left(2\right)]$
$\Rightarrow CF=3.58\text{m}$

Why this question? It tests your understanding of shadow formation. Tips: The length of shadow of a vertical structure at the bottom of lake is the length of line joining its base and point where refracted ray strikes the bottom.