Question

A glass cube of edge 1 $cm$ and $\mu =1.5$ has a small spot at the centre. Find areacube face that must covered to prevent the spot from being seen.

• A.
  π5cm2$\therefore A = \dfrac{\pi }{5}c{m^2}$
• B. $A=\frac{\pi }{5}c{m}^2$
• C. $A=\frac{2\pi }{5}c{m}^2$
• D. $A=\frac{3\pi }{5}c{m}^2$
  

Answer 1

Answer: (B)

$A=\frac{\pi }{5}c{m}^2$

$\frac{{\mu }_a}{{\mu }_g}=\frac{\sin i}{\sin r}=\frac{\sin \theta }{\sin {90}^{\circ }}$

$\sin \theta =\frac{1}{{\mu }_g}=\frac{1}{1.5}=\frac{2}{3}$

Putting this in a triangle we get $p=2,h=3$

$b=\sqrt{3^2-2^2}=\sqrt{5}$

$\therefore \tan \theta =\frac{2}{\sqrt{5}}$

From $\Delta ABC$, $\tan \theta =\frac{x}{1/2}=\frac{2}{\sqrt{5}}$

$\therefore x=\frac{1}{\sqrt{5}}$

Area covered, $A=\pi x^2=\pi \times {\left(\frac{1}{\sqrt{5}}\right)}^2$

$\therefore A=\frac{\pi }{5}c{m}^2$

Hence the correct answer is option (A).