Question

The illuminance of a surface varies inversely with the square of its distance from the light source. If the illuminance of a surface is 120 lumens per square meter when its distance from a certain light source is 6 meters, by how many meters should the distance of the surface from the source be increased to reduce its illuminance to 30 lumens per square meter?(a) 3(b) 6(c) 12(d) 15(e) 18

Answer 1

Answer: b) 6

Step-by-step explanation:

Given : The illuminance of a surface varies inversely with the square of its distance from the light source.

i.e. for d distance and l luminance , we have

$d^2\times l=k$ , where k is constant.       (1)

If  the illuminance of a surface is 120 lumens per square meter when its distance from a certain light source is 6 meters.

From (1), we have

$\Rightarrow (6)^2\times 120=k\\\\\Rightarrow\ k=4320$   (2)

For the distance (d) corresponds to the illuminance to 30 lumens per square meter , we have

$d^2\times 30=k$

Put value of k , we get

$d^2\times 30=4320\\\\\Rightarrow\ d^2=\dfrac{4320}{30}=144\\\\\Rightarrow\ d^2=144\\\\\Rightarrow\ d= 12$

Then , the number of meters should the distance of the surface from the source be increased= 12 meters- 6 meters = 6 meters.