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?

Distance between the surface and source of light will be increased by 6 meters.

Step-by-step explanation:

The illuminance of a surface varies inversely with the square of ts distance from the light source.

Let illuminance of the surface = x lumens per square meter

and distance from a light source = y meter.

Now x ∝

Or [k = proportionality constant]

Now we will find the value of k.

k = xy²

k = 120×(6)²

k = 4320

We have to calculate the distance of the source if illuminance of the surface is 30 lumens per square meter.

y² =

y² = 144

y = √144 = 12 meters

So the source of the light will be shifted away from the surface = 12 - 6 = 6 meters.

false

step-by-step explanation:

whenever we assume a point on a plane then according to geometry it will definitely have 2 dimensions and if it assumed in space then it will have 3 dimensions.

< 6 cm to > 26 cm i believe

step-by-step explanation: