Question 113 of 295

A diver looking up through the water sees the outside world contained in a circular horizon. The refractive index of water is $\dfrac{4}{3}$, and the diver's eyes are $15cm$ below the surface of the water. Then the radius of the circle is :

A$\dfrac{15\times \sqrt{7}}{3}cm$B$\dfrac{15\times 3}{\sqrt{7}}cm$C$15\times 3\sqrt{7}cm$D$15\times 3\times \sqrt{5}cm$

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Answer: (B) $\dfrac{15\times 3}{\sqrt{7}}cm$

$∴\text{sin}{\theta }_{\text{c}}=\dfrac{1}{\mu }$

$=\dfrac{3}{4}$

$∴\dfrac{\text{r}}{\sqrt{{\text{h}}^{2}+{\text{r}}^{2}}}=\dfrac{3}{4}$

$∴\dfrac{{\text{r}}^{2}}{{\text{h}}^{2}+{\text{r}}^{2}}=\dfrac{9}{16}$

$∴16{\text{r}}^{2}=9{\text{h}}^{2}+9{\text{r}}^{2}$

$7{\text{r}}^{2}=9{\text{h}}^{2}$

$\text{r}=\text{h}\sqrt{\dfrac{9}{7}}$

$=\dfrac{3\text{h}}{\sqrt{7}}$

$=\dfrac{3}{\sqrt{7}}\times 15\text{cm}$