A lamp of 80 watt with efficiency of 80% of watt/CP is suspend... | Technical Ability MCQ | Examoogle

Technical AbilityIllumination Engineering

A lamp of 80 watt with efficiency of 80% of watt/CP is suspended, The illumination at a point on a working plane directly below the lamp is 25 lumens/m 2 . Determine the height at which the lamp is suspended?

4 meters

2 meters

3.2 meters

1.6 meters

Correct Answer

⚙️ TE • Technical Concept & PrincipleTechnical AbilityIllumination Engineering

Option B

2 meters

Quick Summary:

The illumination at a point directly below a light source is given by the Inverse Square Law. Given the lamp power $P = 80$ W and an efficiency of $0.8$ W/CP, the Candle Power ( $CP$ or $I$ ) is calculated as $\frac{80}{0.8} = 100$ CP. Using the illumination formula $E = \frac{I}{h^2}$ , we solve for the height $h$ .

⚙️TETechnical SolutionConcept & Principle

💡 Explanation

🔢 Key Formulas

$I = \frac{\text{Power}}{\text{Efficiency}} = \frac{P}{\eta}$ — Calculation of Candle Power (I)

$E = \frac{I}{h^2}$ — Inverse Square Law for illumination directly below the lamp

⚙️ Working Principle

The illumination $E$ (measured in lumens/m $^2$ or lux) on a surface perpendicular to the light rays is inversely proportional to the square of the distance $h$ from the source. Since the point is directly below the lamp, the angle of incidence is $0°$ , and the cosine rule factor $\cos(\theta)$ equals 1, simplifying the relation to $E = \frac{I}{h^2}$ .

📌 Key Points

▸
Candle Power ( $I$ ) represents the luminous intensity of the source in a specific direction.
▸
The Inverse Square Law holds strictly for point sources where the surface is normal to the light rays.
▸
Unit of illumination $E$ is lux, which is equivalent to 1 lumen/m $^2$ .

✅ Advantages

▸
Simple mathematical model for point sources
▸
Provides a quick estimation for architectural lighting design

❌ Disadvantages / Limitations

▸
Assumes the lamp acts as a point source
▸
Does not account for inter-reflections from walls or ceilings

🛠️ Applications / Uses

▸
Direct lighting calculation for simple workspace layouts
▸
Fundamental physics problems in electrical engineering

📄 Additional Information

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Calculation: $I = \frac{80}{0.8} = 100$ CP. $25 = \frac{100}{h^2} \implies h^2 = \frac{100}{25} = 4 \implies h = 2$ meters.
▸
Option A (4m) results in $E = 6.25$ lux. Option C (3.2m) results in $E \approx 9.76$ lux. Option D (1.6m) results in $E = 39.06$ lux.

📊 Diagram / Illustration

✅

B is correct — The height is calculated as 2 meters using the Inverse Square Law of illumination.

Core Concepts Used

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Inverse Square Law Luminous Intensity Illumination Calculation

💡 EXAM TIP

In illumination problems, always check if the point is directly below the source ( $\theta = 0$ ) or at an angle ( $\theta > 0$ ); if angled, you must use $E = \frac{I}{d^2} \cos(\theta)$ where $d$ is the slant distance.

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