find area

A solid iron sphere of radius=1m is just submerged in a water tank. A point source of light is kept vertically above its center at a distance 1m from the water surface.

Find the area of the sphere's shadow. Refractive index of water=1.414.

2 Answers

Sourish Ghosh ·

sinisinr = 1.414
i = 90-r2
Use wolframalpha to solve the first equation.*sin%28x%29+%3D+0

Area: π[2tanr + cot(90+r2)]2

Shaswata Roy ·

I don't have that solution anymore.
Neither do I have that question anymore.

But anyways the solution below is somewhat similar.


The triangles FEI and FCI are congruent.
→ <FCI =<FEI = i
→<HFC = <FCI = i(Alternate angles)

Since FI and FG are tangents drawn from the same point (F) to the circle.

<IFC = <CFG = i+r

<IFH = <IFC +<CFH = i+r + i=2i+r=90°....(2)

Tangents drawn from a point to a circle are of equal lengths.

Using this property,

FI = FJ and GD = GJ


FG = FJ+GJ = FI+GD = FI + GH +HD =2FI+GH (Because HD = FI)

→2sec(r) = 2tan(r) + 2tan(i)
→ 1=sin(r)+sin(i)cos(r)cos(i)
→1 = sin(i)μ+2sin2(i)

This is a quadratic in sin(i).
Out of the 2 roots we need to take the positive root.
Than we can find sin(r) from equation(1).

The length GD = GH+HD=2tan(r)+tan(i)

And area = π(GD)2

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