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Long Answer Questions · Q2

Q.Describe Fizeau's method to determine the speed of light.

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✓ Free question

Step 1. Fizeau's apparatus: light from source SS reflects off a partially silvered plate GG (tilted 45°45°), passes through a gap in a rotating toothed wheel (NN teeth, NN cuts), travels a known distance dd to a fixed distant mirror MM, and returns along the same path to the observer viewing through GG.

Step 2. As the wheel's angular speed ω\omega is increased from rest, the returning beam eventually vanishes for the first time -- blocked by the tooth adjacent to the gap it originally passed through.

Step 3. At that instant, the wheel has rotated through the angle between one tooth and the next slot, θ=2π2N=πN\theta=\dfrac{2\pi}{2N}=\dfrac{\pi}{N} (total circle divided by the 2N2N teeth-plus-cuts), during the light's round-trip travel time t=2dvt=\dfrac{2d}{v}.

Step 4. Using ω=θ/t\omega=\theta/t: ω=π/N2d/v=πv2dN\omega=\dfrac{\pi/N}{2d/v}=\dfrac{\pi v}{2dN}, so t=πNωt=\dfrac{\pi N}{\omega}.

Step 5. Substituting tt into v=2d/tv=2d/t gives v=2dωπNv=\dfrac{2d\omega}{\pi N}... rearranging the Step 4 relation directly for vv instead gives v=2dωNπv=\dfrac{2d\omega N}{\pi}, Fizeau's working formula. As ω\omega is increased further, the light reappears at 2ω2\omega (through the next slot) and vanishes again at every subsequent odd multiple of the first blocking speed.

✓Final answer

Fizeau's formula for the speed of light is v=2dωNπv=\dfrac{2d\omega N}{\pi}, giving a value close to the modern c=2.998×108 m s−1c=2.998\times10^{8}\ \text{m s}^{-1}.

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