A light meter shows the intensity of a light hanging 6 ft over a circular table is 16 lux. If the table has a 5 ft diameter, what would the illumination be at the edge of the table where actors sit to read a script?

The formula that I got to a previous question is:

let L = intensity of light
let k = constant
let d = distance between the light source and the illuminated object

L = (k/d)^2

I don't know how to begin the first question.

Question

A light meter shows the intensity of a light hanging 6 ft over a circular table is 16 lux. If the table has a 5 ft diameter, what would the illumination be at the edge of the table where actors sit to read a script?

The formula that I got to a previous question is:

let L = intensity of light
let k = constant 
let d = distance between the light source and the illuminated object

L = (k/d)^2

I don't know how to begin the first question.

anonymous · Answer

I attempted to solve this and I did: 
16 = (k/6)^2
took the square root on each side and got
4 = k/6
multiplied by 6 on each side
k = 24

Now I don't know how to incorporate the 5 ft diameter from the question… Any kind of help is greatly appreciated!