BKE3 2 Lighting

What is the fundamental physical process by which all electric lights generate light?

Electron transition.
Electrons absorb energy, jump to a higher energy level (excited state), and release energy as an electromagnetic (EM) wave when they jump back to a lower energy level.

Describe the working principle of an incandescent lamp.

1. Electric current heats the tungsten filament to a high temperature (~2500 degrees Celsius).
   2. Electrons in the solid metal are thermally excited.
   3. Electrons jump to numerous closely-spaced lower energy levels, emitting a continuous spectrum of EM waves.

Why is an incandescent filament lamp filled with an inert gas (e.g., argon or nitrogen)?

To slow down the evaporation of the tungsten filament at high temperatures, thereby preventing the filament from breaking easily and extending the lifetime of the lamp.

State the two major disadvantages of an incandescent lamp and explain why they occur.

1. Low energy efficiency: About 90-95% of emission is invisible infra-red (IR) radiation (wasted as heat).
   2. Short lifetime: The heating effect causes the filament to evaporate unevenly; thinner parts get hotter, evaporate faster, and eventually blow/break.

Describe the step-by-step working principle of a gas discharge lamp (e.g., Fluorescent Tube Lamp / CFL).

1. High voltage accelerates electrons which ionize the inert gas.
   2. The resulting current heats and vaporizes the mercury.
   3. Electrons and ions collide with mercury atoms, exciting the electrons in the mercury atoms.
   4. Mercury electrons jump to lower energy levels, emitting ultra-violet (UV) radiation.
   5. The phosphor coating absorbs the UV radiation and emits visible light (fluorescence).

What is the crucial function of the phosphor coating inside a fluorescent tube lamp (FTL) or CFL?

It absorbs the invisible ultra-violet (UV) radiation emitted by the excited mercury atoms and undergoes fluorescence to emit a broad spectrum of visible light for illumination.

State three drawbacks of using fluorescent tube lamps (FTLs) or compact fluorescent lamps (CFLs).

1. They contain highly toxic mercury vapor (severe environmental/health hazard upon disposal or breakage).
   2. They emit a small amount of ultra-violet (UV) radiation.
   3. They require a short time (10-30 seconds) to reach maximum brightness.

Describe how a light-emitting diode (LED) produces light.

1. A forward d.c. voltage is applied across the p-n junction.
   2. Electrons from the n-type semiconductor and holes from the p-type semiconductor move towards the p-n junction and combine.
   3. The combining electrons jump to a lower energy level, releasing the excess energy as an electromagnetic (EM) wave.

Why is the light emitted by a single LED considered monochromatic, and how do LED lamps produce white light?

The light is emitted within a narrow spectrum determined by the specific energy gap of the p-n junction materials. To produce white light, an LED lamp either combines red, green, and blue LEDs, or uses a phosphor coating to convert blue LED light into white light.

To which wavelength and colour of visible light is the human eye most sensitive?

555 nm (green-yellow light). Visible light of other wavelengths must be emitted at a higher power to yield the same perceived brightness.

Define luminous flux and state its SI unit.

The power output of a light source corrected for the dependence on wavelength of the sensitivity of the human eye.
Unit: lumen (lm).
High-Frequency Trap: Do not confuse this with raw electrical power (W); luminous flux only measures visible light effectively perceived by humans.

Define illuminance (E) and state its formula and SI unit.

The luminous flux falling on a surface per unit area.
Unit: lux (lx) or lm/m^2.
Formula: E = Luminous Flux / Area.
[NOT IN DSE FORMULA SHEET - MUST MEMORIZE]

State Lambert's Cosine Law for illuminance and identify the high-frequency trap regarding the angle.

E = E_perpendicular * cos(theta).
High-Frequency Trap: theta MUST be the angle between the incident light ray and the NORMAL of the surface.
Do not mistakenly use the angle between the light ray and the surface itself!
[NOT IN DSE FORMULA SHEET - MUST MEMORIZE]

State the inverse-square law formula for the illuminance (E) from a point light source.

E = (Luminous Flux * cos(theta)) / (4 * pi * r^2).
High-Frequency Trap: 'r' is the direct straight-line distance from the source to the point on the surface, NOT the vertical height!
[NOT IN DSE FORMULA SHEET - MUST MEMORIZE]
[IMAGE NEEDED: Point source illuminating a surface at an angle | Source: f6 dse book E302.pdf, Fig 2.2l]

What are the strict conditions required for the inverse-square law of illuminance to apply?

1. The light source must be a point source (emits light uniformly in all directions).
   2. Reflection of light from surrounding walls or ceilings must be negligible.
   High-Frequency Trap: The inverse-square law does NOT apply to parallel light beams (e.g., lasers or flashlights).

Define the luminous efficacy of an electric light and state its unit.

The ratio of luminous flux to electrical power input.
Efficacy = Luminous Flux / Input Power.
Unit: lm W^(-1).
It measures how effectively electrical energy is converted into visible light.
[NOT IN DSE FORMULA SHEET - MUST MEMORIZE]

What is the theoretical maximum possible efficacy of any lamp, and when would it occur?

683 lm/W.
This occurs only for an ideal (100% efficient) lamp that converts all electrical power exclusively into monochromatic green-yellow light at exactly 555 nm.

If a green light source and a red light source both have a luminous flux of 30 lumens, which one appears brighter to the human eye?

They appear equally bright.
High-Frequency Trap: Luminous flux ALREADY includes the correction for human eye sensitivity.
Therefore, equal luminous flux guarantees equal perceived brightness, regardless of colour.

If a 10 W green light and a 10 W red light are compared, which appears brighter and why?

The 10 W green light appears brighter.
Because the human eye is much more sensitive to green light (~550 nm) than red light, the same raw power output yields a higher luminous flux (and thus higher illuminance/brightness) for the green light.