Question

The illuminance of a small area changes from 900 lumen m−2 to 400 lumen m−2 when it is shifted along its normal by 10 cm. Assuming that it is illuminated by a point source placed on the normal, find the distance between the source and the area in the original position.

Answer 1

The distance between the source and the area in the original position is about 20 cm

Explanation:

Let the luminous source be l, and the distance in the initial position between the source and the area be x.

Given,

Initial illuminance (EA) $= 900 lumen/m^2$

Final illuminance (EB) = $400 lumen/m^2$

Illuminance is given on the initial position by,

$E_{A}=l \frac{\cos \theta}{x^{2}} \ldots \ldots e q^{n}(1)$

Illuminance is given in the final position

$E_{B}=\frac{l \cos \theta}{(x+10)^{2}} \ldots \ldots e q^{n}(2)$

Equating (1) and (2) luminous intensity, we get

$l=\frac{E_{A} x^{2}}{\cos \theta}=\frac{E_{B}(x+10)^{2}}{\cos \theta}$

$x^{2}=400(x+10)^{2}$

$\frac{x}{x+10}=\frac{2}{3}$

$3 x=2 x+20$

$x=20 \mathrm{cm}$

At the initial phase the distance between source and region was 20 cm.

Answer 2

The distance between the source and the area in the original position is 20 cm.

Given:

Initial illuminance = Eᵃ = 900 lumen/m²

Final illuminance = Eᵇ = 400 lumen/m²

Explanation:

Let intensity of luminous be I

Let distance between source and area be x

Illuminance on initial position is given by the formula:

Eᵃ = (l cos θ)/(x²)

l = (Eᵃ x²)/(cos θ) ⇒ equation (1)

Illuminance on final position is given by the formula:

Eᵇ = (l cos θ)/(x + 10)²

I = (Eᵇ (x + 10)²)/(cos θ) ⇒ equation (2)

Now, on equating equation (1) and (2), we get,

(Eᵃ x²)/(cos θ) = (Eᵇ (x + 10)²)/(cos θ)

(Eᵃ x²) = Eᵇ (x + 10)²

On substituting the values, we get,

900 x² = 400 (x + 10)²

900 x² = 400 x² + 40000 + 800 x

500 x² - 800 x - 40000 = 0

5 x² - 8 x - 400 = 0

x/(x + 10) = 2/3

3x = 2x + 20

3x - 2x = 20

∴ x = 20