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A bulb is placed at a depth of \mathrm{2\sqrt{7}\ m} in water \mathrm{\left ( \mu _{w}=4/3 \right )} and a floating opaque disc is placed over the bulb so that the bulb is not visible from the surface. What is the minimum diameter of the disc?

Option: 1

\mathrm{8\ m}


Option: 2

\mathrm{12\ m}


Option: 3

\mathrm{15\ m}


Option: 4

\mathrm{20\ m}


Answers (1)

best_answer

Here, refractive index of water, \mathrm{\mu _{w}=\frac{4}{3}}
\mathrm{\sin C=\frac{1}{\mu_w}=\frac{3}{4}................(i)}

From figure,
\mathrm{\sin \mathrm{C}=\frac{\mathrm{R}}{\sqrt{\mathrm{R}^2+(2 \sqrt{7})^2}}................(ii)}
Equating \mathrm{(i)}and \mathrm{(ii)}we get
\mathrm{\frac{3}{4}=\frac{R}{\sqrt{R^2+(2 \sqrt{7})^2}}}
Squaring both sides, we get
\begin{aligned} & \mathrm{\frac{9}{16}=\frac{R^2}{R^2+28} \Rightarrow 9 R^2+28 \times 9=16 R^2} \\ &\mathrm{ 7 R^2=28 \times 9} \\ & \mathrm{ R^2=\frac{28 \times 9}{7}=36 \Rightarrow R=6 m} \end{aligned}
Hence, the minimum diameter of the disc is \mathrm{}2 \mathrm{R}=2 \times 6 \mathrm{~m}=12 \mathrm{~m}

Posted by

sudhir.kumar

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