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

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Solution

$l = 10$ $c m$
$u = 4 / 3$
In $△ A O C$
$tan 45^{o} = \frac{A O}{O C}$
$l = \frac{A O}{O C}$
$A O = O C = 10$ $c m$
Shift $= 10$ $c m$
Now,
$u = \frac{R e a l d e p t h}{A p p a r e n t d e p t h}$
Let Real depth of object be $x$
So,
$\frac{4}{3} = \frac{x}{x - 10}$
$\Rightarrow 4 x - 40 = 3 x$
$\Rightarrow x = 40$ $c m$
So, water should be poured upto $40$ $c m$ into the vessel.

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l=10 cmu=4/3 In △AOCtan45o=AOOCl=AOOCAO=OC=10 cmShift =10 cmNow,u=RealdepthApparentdepthLet Real depth of object be xSo,43=xx−10⇒4x−40=3x⇒x=40 cmSo, water should be poured upto 40 cm into the vessel.