The lightbulbs have cylindrical filaments much greater in length than in diamete

Question

The lightbulbs have cylindrical filaments much greater in length than in diameter. The evacuated bulbs are identical except that one operates at a filament temperature of 2.220 degree C and the other operates at 2.135 degree C. Find the ration of the power emitted by the hotter lightbulb to that emitted by the cooler lightbulb. With the bulbs operating at the same respective temperatures, the cooler lightbulb is to be altered by making its filament thicker so that it emits the same power as the hotter one. By what factor should the radius of this filament be increased?

Explanation / Answer

assuming black body radiation and same area,
from steffan's law
power per unit are=*T^4

=steffan's constant=5.67*10^(-8)

T=temperature in K

so for bulb 1,temperature in K=2100+273=2373

so for bulb 2,temperature in K=2000+273=2273

ration of power=(2373)^4/(2273)^4=1.1879

(b.)let radius be increased by factor x.

then area be increased by x^2.

so let previous area be A.

then *T1^4*A=*T2^4*(x^2*A)

so x^2=1.1879

x=1.0899 (ans)

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