(1) Figure C is a straight line, so the light intensity is proportional to the square of the distance;

The expression is: I = kd2.

The slope is: k ≈ 0.035 W.

So the answer is: kd2, 0.035? (0.032-0.038);

(2) The light intensity is equal to the light energy flowing through the unit area in unit time, so the light power is:

P0=SI=4πd2I

So the answer is: P0 = 4 π d2i.

(3) The luminous power is P0 = Si = 4 π d2i = 4× 3.14× (13.5× 0.01) 2×1.97 ≈ 0.45 W..

Luminous efficiency: η=≈0.45W8W× 100%≈6%.

So the answer is: 6.