Question

A cubical vessel with opaque walls is so placed that the eye of an observer cannot see its bottom but can see the entire wall CD. A small object is placed at O at a distance $b=10cm$ from corner D. What minimum depth of water $(\mu =\frac{4}{3})$ should be poured into the vessel which will enable the observer to see the object?

• A. $28.6cm$
• B. $26.7cm$
• C. $22.4cm$
• D. $20cm$

Answer 1

Answer: (B)

$26.7cm$

From figure:-

$BDA={45}^o$ since it is a cube $⟹AB=AD=h$

$OA=h-b$ and $r=ABD={45}^o$

$\mu \sin i=\sin r$

$⟹\sin i=\frac{\sin r}{\mu }=\frac{3}{4\sqrt{2}}$

$⟹\tan i=\frac{3}{\sqrt{(4\sqrt{2}{)}^2-3^2}}=\frac{3}{\sqrt{23}}$

And, $\tan i=\frac{OA}{AB}=\frac{h-b}{h}$

$⟹1-\frac{b}{h}=\frac{3}{\sqrt{23}}$

$⟹1-\frac{10}{h}=\frac{3}{\sqrt{23}}$

$⟹h=\frac{10\times \sqrt{23}}{\sqrt{23}-3}$

$⟹h=26.7cm$

Answer-(B).