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A luminous point is kept inside tank of water at a depth of 10 m below the surface. $\mu$ for water is $\large\frac{4}{3}$ . What must be the minimum diameter of card placed on the surface of water so that it cuts off all the light coming out.

$(a)\;60\;\pi \\ (b)\;25\;\pi \\ (c)\;180\;\pi \\ (d)\;90\;\pi $

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If the angle at which the ray of light the point luminous object is incident at an angle greater than the critical angle the light will not enter air but will be totally reflected into water.
$\therefore $ All the light that enters into air must be incident at an angle less than or equal to critical angle.
We know that $\sin \theta_c=\large\frac{1}{\mu}$
$\sin \theta_c=\large\frac{1}{4/3}$
$\qquad=\large\frac{3}{4}$
$\cos \theta_c =\sqrt {1- \large\frac{3^2}{4^2}}$
$\qquad= \sqrt { \large\frac{16-9}{16}}$
$\qquad= \large\frac{\sqrt 5}{4}$
$\tan \theta_c= \large\frac{\Large\frac{3}{4}}{ \Large \frac{\sqrt 5}{4}}$
$\qquad= \large\frac{3}{\sqrt 5}$
$\tan \theta_c= \large\frac{r}{h}$
$\therefore \large\frac{r}{h} =\frac{3}{\sqrt 5}$
$r=\large\frac{3 \times h}{\sqrt 5}$
$\quad= \large\frac{3 \times 10}{\sqrt 5}$
$\quad= 6 \sqrt 5\;m$
$\therefore$ The area $=\pi \times (6 \sqrt 5)^2$
$\qquad= 180 \pi$
Hence c is the correct answer
answered Jan 20, 2014 by meena.p
 

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