$(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

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