---
title: "Lamp suspension height and illumination at offset"
subject: "Industrial Power Distribution and Illumination"
chapter: 7
key: "semester_6/industrial_power_distribution_and_illumination:lamp-height-250cp"
source: https://www.bctengineer.com/q/semester_6/industrial_power_distribution_and_illumination/lamp-height-250cp
site: BCT Engineering
---

# Lamp suspension height and illumination at offset

Industrial Power Distribution and Illumination · Chapter 7: Illumination Design Principles

- Asked in: 2074 Bhadra, 4 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/industrial_power_distribution_and_illumination/lamp-height-250cp
- Whole chapter: https://www.bctengineer.com/md/semester_6/industrial_power_distribution_and_illumination/chapter-7

## Question

The illumination at a point on a working plane directly below the lamp is to be 100 lumen/m². The lamp gives 250 c.p. uniformly below the lamp horizontal plane. Determine the height at which the lamp is suspended. Also find illumination at a point on the working table 1.2m away from the vertical axis of the lamp.

## Short answer

Lamp height = 1.58 m for 100 lux directly below (h = √(I/E)). Illuminance at 1.2 m from vertical axis = 50.54 lux using cosine law (E = I·h/d³).

## Answer

**Given:** luminous intensity I = 250 cd (c.p.) in all directions below the horizontal; illuminance directly below the lamp E₀ = 100 lux (lm/m²).

#### Height of the lamp

For a point directly below the lamp, θ = 0 and distance = h:

```text
E₀ = I / h²
h  = √(I / E₀) = √(250 / 100) = √2.5 = 1.581 m
```

#### Illuminance at 1.2 m from the vertical axis

By the cosine law, E = (I / d²) × cos θ, where d is the distance from lamp to point and cos θ = h/d.

```text
d     = √(h² + x²) = √(2.5 + 1.2²) = √3.94 = 1.985 m
cos θ = h/d = 1.581/1.985 = 0.7966   (θ = 37.2°)
E     = I × h / d³
      = 250 × 1.581 / 1.985³
      = 395.3 / 7.821
      = 50.54 lux
```

**Answer: lamp height = 1.58 m above the working table; illuminance at 1.2 m from the axis = 50.54 lux (lm/m²).**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Directly below the lamp, E = 100 lux and I = 250 cd. Which gives the height?
   A. h = I x E = 25000 m
   B. h = sqrt(I / E) = 1.58 m
   C. h = I / E = 2.5 m
   D. h = E / I = 0.4 m
   Answer: B. E = I / h^2 for a point directly below, so h = sqrt(250/100) = 1.58 m. (Key point: Height from illuminance below lamp)

2. For a point off the axis, which relation is used?
   A. E = I / d^2 only
   B. E = (I / d^2) x cos(theta), with cos(theta) = h/d
   C. E = I x cos(theta)
   D. E = I x d^2 x sin(theta)
   Answer: B. Illuminance on a horizontal plane falls with inverse square of distance and with the cosine of the angle of incidence. (Key point: Cosine law)

3. At 1.2 m horizontal distance (d = 1.985 m, cos(theta) = 0.797), what is the illuminance?
   A. 100 lux, unchanged
   B. About 50.5 lux
   C. About 126 lux
   D. About 80 lux
   Answer: B. E = 250 x 1.581 / 1.985^3 = 50.54 lux, about half the value below the lamp. (Key point: Result at 1.2 m)

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