Environment, Ventilation & Hazards

Mine Lighting (Illuminance)

Illuminance from luminous flux over an area, the inverse-square law for point sources, and the flux needed for a target lux level.

PART 1

Topic Breakdown & Traps

The Engineering Principle

Adequate illuminance (lux = lumens per m²) is essential for safe working. Spread over a surface,

E = Φ/ A

. From a point source the inverse-square law gives

E = I / d^{2}

(illuminance falls with the square of distance). Designing a layout, the lumen method inflates the required flux by dividing by the utilisation and maintenance factors.

The Core Formula Matrix

Illuminance (area):

E = \frac{Φ}{A} [lux]

.

Inverse-square law:

E = \frac{I}{d ^{2}}

(

I

in candela,

d

in m).

Lumen method (required flux):

Φ = \frac{E \times A}{U F \times M F}

The ‘IIT Traps’

⚠
**Illuminance falls as $1/ d^{2}$ ** — doubling the distance quarters the light.
⚠
Lux = lumens / m²; don't confuse luminous flux (lm) with illuminance (lux).
⚠
Divide by UF and MF (both < 1) so the installed flux is larger than the bare requirement.

PART 2

Progressive 3-Tier Question Suite

Q1BASIC1 Mark · MCQ

A luminaire delivering

2000 lm

uniformly over

20 m^{2}

gives an illuminance of:

Q2MEDIUM2 Marks · NAT

A point source of

200 cd

illuminates a surface

4 m

away (normal incidence). The illuminance is ______ lux. (Round off to two decimal places.)

lux

Q3HARD2 Marks · NAT

To achieve

150 lux

over

50 m^{2}

with utilisation factor

0.6

and maintenance factor

0.8

, the required luminous flux is ______ lm. (Round off to the nearest whole number.)