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A decorative object rests at the bottom of a pond of depth 8*sqrt(3) m filled with water of refractive index sqrt(3/2). A diver at the water surface looks at the object such that the line of sight makes 30 deg with the water surface. Find the perpendicular distance (in metres) of the apparent image from the water surface.

1. 1
2. 2
3. 3
4. 4

Correct answer: 4

Solution

Angle of refracted ray in air with normal = 90 - 30 = 60 deg (r = 60 deg). By Snell's law: n*sin(i) = sin(r) => sqrt(3/2)*sin(i) = sin(60 deg) = sqrt(3)/2 => sin(i) = 1/sqrt(2) => i = 45 deg. Apparent perpendicular depth = d * cos³(r)/(n * cos³(i)) = 8*sqrt(3) * (0.5)³ / (sqrt(3/2) * (1/sqrt(2))³) = 8*sqrt(3) * (1/8) / (sqrt(3/2) * 1/(2*sqrt(2))) = sqrt(3) / (sqrt(3/2)/(2*sqrt(2))) = sqrt(3)*2*sqrt(2)/sqrt(3/2) = 2*sqrt(6)/sqrt(3/2) = 2*sqrt(6)*sqrt(2/3) = 2*sqrt(4) = 4 m.

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