JEE Main 2014 · 9 Apr · Shift 1 · Physics Q23

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A diver looking up through the water sees the outside world contained in a circular horizon. The refractive index of water is $\dfrac43$ , and the diver's eyes are 15 cm below the surface of water. Then the radius of the circle is :

15\times3\times\sqrt5

15\times3\sqrt7

\dfrac{15\times\sqrt7}{3}

\dfrac{15\times3}{\sqrt7}

Answer and worked solution

Official answer: D

Light from above the surface arrives within the critical cone, with $\sin\theta_c=\frac34$ , so $\tan\theta_c=\dfrac{3}{\sqrt7}$ . The radius is
$r=h\tan\theta_c=\frac{15\times3}{\sqrt7}\text{ cm}.$

Topics:Total internal reflection

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JEE Main 2014 · 9 Apr · Shift 1, Physics question 23