question_answer

A point source of light is placed at the centre of a solid cube of side 'a'. What fraction of area of each face must be covered so that the object is not visible through any face? (R.I. of cube =\[\mu \]):

A) \[\frac{1}{{{\mu }^{2}}-1}\]

B) \[\frac{\pi }{2\left( {{\mu }^{2}}-1 \right)}\]

C) \[\frac{\pi }{4\left( {{\mu }^{2}}-1 \right)}\]

D) \[\frac{{{\pi }^{2}}{{\mu }^{2}}}{4\left( {{\mu }^{2}}-1 \right)}\]

Correct Answer: C

Solution :

\[\sin \,C=\frac{1}{\mu }\]

Also, \[\frac{r}{(a/2)}=\tan \,C\]

\[=\frac{1}{\sqrt{{{\mu }^{2}}-1}}\]

\[\therefore \]      \[r=\frac{a}{2\sqrt{{{\mu }^{2}}-1}}\]

Area,     \[\operatorname{A}=\pi {{r}^{2}}=\frac{\pi {{a}^{2}}}{4({{\mu }^{2}}-1)}\]

Fraction \[=\frac{\pi {{r}^{2}}}{{{A}^{2}}}=\frac{\pi }{4({{\mu }^{2}}-1)}\]