Power rating of bulb, P = 100 W

It is given that about 5% of its power is converted into visible radiation.

Therefore ,Power of visible radiation,

P’ = (5/100 ) x 100 = 5 W

Hence, the power of visible radiation is 5W.

Ans (a).

Distance of a point from the bulb, d = 1 m Hence, intensity of radiation at that point is given as:

I = P’/(4πd²) =5/4π(1)² =0.398 W/m²

Ans (b).

Distance of a point from the bulb, d1 = 10 m Hence, intensity of radiation at that point is given as:

I = P’/[4π(d1)²]

=5/4π(10)² =0.00398 W/m²