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← JEE Main 2022 · 26 Jun · Shift 2
PhysicsInteger+4 / -1Question 58 of 90

JEE Main 2022 · 26 Jun · Shift 2, Physics question 58

A small bulb is placed at the bottom of a tank containing water to a depth of 7\sqrt7 m. The refractive index of water is 43\dfrac43. The area of the surface of water through which light from the bulb can emerge out is xπx\pi m2^2. The value of xx is ______.

Answer and worked solution

Official answer: 9

Critical angle. sin⁡C=34\sin C=\frac34, tan⁡C=37\tan C=\frac3{\sqrt7}.

Radius. r=htan⁡C=7×37=3r=h\tan C=\sqrt7\times\frac3{\sqrt7}=3 m; area 9π9\pi m2^2.

Answer: 9