Light bulb 1 operates with a filament temperature of 2700 k whereas light bulb 2 has a filament temperature of 2100 k. both filaments have the same emissivity, and both bulbs radiate the same power. find the ratio a1/a2 of the filament areas of the bulbs.

0.3659

Explanation:

The power (p) is given as:

P = AeσT⁴

where,

A =Area

e = transmittivity

σ = Stefan-boltzmann constant

T = Temperature

since both the bulbs radiate same power

P₁ = P₂

Where, 1 denotes the bulb 1

2 denotes the bulb 2

thus,

A₁e₁σT₁⁴ = A₂e₂σT₂⁴

Now e₁=e₂

⇒A₁T₁⁴ = A₂T₂⁴

or

substituting the values in the above question we get

or

=0.3659

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