5 % of the total energy of a 100 W bulb is converted into visible light. Calculate the average intensity at a spherical surface which is at a distance of 1 m from the bulb. Consider the bulb to be a point source and let the medium be isotropic.[Ans : 0.4 Wm⁻²]
Answers
Answered by
0
Answer:
halo gamecocks special it landlord is still IDK
Answered by
0
Answer:
0.4 Wm⁻²
Explanation:
Power of the bulb = P = 100W (Given)
Distance from the bulb = 1m (Given)
Power of visible radiation -
= P′= 5 × 100/100
= 5W
Intensity of radiation -
= I = P′/4πd²
= 5/4π(1)²
= 0.00398
= 0.4 Wm⁻²
Therefore, the average intensity at a spherical surface which is at a distance of 1 m from the bulb is 0.4 Wm⁻².
Similar questions