Question

The illustration shows a lamp and a candle illuminating a screen positioned between them at 0.34 m. If the lamp is moved to a distance r = 0.588 m, how much brighter must the lightbulb be than the candle to have the same illuminance on the screen?

Answer 1

For the initial and final distance of the bulb, according to the inverse square rule, we have:

$$\dfrac{I_2}{I_1}=\dfrac{r_2^2}{r_1^2}=\dfrac{0.588^2}{0.34^2}=3.$$