Question

A 60-Watt bulb illuminates a surface a distance d away with

an illuminance of 3600 lux. What would be the illuminance
on a surface a distance 0.5•d away (i.e., one-half the
distance away) when illuminated by a 120-Watt bulb?
​

Answer 1

Step-by-step explanation:

Relation between power and illumination and distance is as follows

`I = (P)/(R^2)`

For first case

`3600 = (60)/(d^2)`

for second case

`I = (120)/((0.5d)^2)`

`(I)/(3600) = (d^2)/(60) * (120)/(.25d^2)`

I = 3600 x 8

= 28800 lux .