---
title: "Walkway illuminance from 4750 cd lamps"
subject: "Industrial Power Distribution and Illumination"
chapter: 8
key: "semester_6/industrial_power_distribution_and_illumination:walkway-lamp-illuminance"
source: https://www.bctengineer.com/q/semester_6/industrial_power_distribution_and_illumination/walkway-lamp-illuminance
site: BCT Engineering
---

# Walkway illuminance from 4750 cd lamps

Industrial Power Distribution and Illumination · Chapter 8: Outdoor Lighting Design

- Asked in: 2077 Chaitra, 8 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/walkway-lamp-illuminance
- Whole chapter: https://www.bctengineer.com/md/semester_6/industrial_power_distribution_and_illumination/chapter-8

## Question

A walkway is illuminated by 250W lamps each having a luminous intensity of 4750 candela in all directions below the horizontal. Each lamp is installed at a height of 6m and the distance between them is 16 meters. Calculate the illuminance contributed by each lamp:
a) (i) directly underneath, (ii) 16 meters from the base,
b) The total illuminance at Midway between the base of each lamp post.

## Short answer

Illuminance E = I h / d³. (a)(i) Directly under lamp: E = 4750/6² = 131.94 lux. (a)(ii) 16 m away: E ≈ 5.71 lux. (b) Midway (8 m from each of two lamps): each contributes 28.5 lux, total 57.0 lux.

## Answer

Each lamp has I = 4750 cd in all directions below the horizontal, mounting height h = 6 m, spacing 16 m.

For a point at horizontal distance x from the base:
```text
d = √(h² + x²)
cos θ = h / d
E = I cos θ / d² = I h / d³
```

#### (a)(i) Directly underneath (x = 0)
```text
E = I / h² = 4750 / 6² = 4750 / 36
  = 131.94 lux
```

#### (a)(ii) 16 m from the base (x = 16 m)
```text
d² = 6² + 16² = 36 + 256 = 292
d  = 17.09 m
cos θ = 6 / 17.09 = 0.351
E = 4750 × 6 / 292^1.5
  = 28 500 / 4989.8
  = 5.71 lux
```

#### (b) Total illuminance midway between two lamp posts (x = 8 m)
Each of the two adjacent lamps is 8 m away horizontally:
```text
d² = 6² + 8² = 100,  d = 10 m
cos θ = 6 / 10 = 0.6
E (one lamp) = 4750 × 0.6 / 100 = 28.5 lux
E (total)    = 2 × 28.5         = 57.0 lux
```
(The next lamps on either side, 24 m away, add only about 1.88 lux each, so they are normally neglected.)

```text
   L1                    L2
   ●──────── 16 m ───────●
   │6 m                  │
 ──┴──────────┬──────────┴──
   P1       Pmid(8 m)    P2
```

**Answer:** (a)(i) 131.94 lux, (a)(ii) 5.71 lux per lamp; (b) total at midway = **57.0 lux** (28.5 lux from each lamp).

## Check yourself

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

1. At horizontal distance x from the base of a lamp at height h, what is E?
   A. I h / d^3, where d = sqrt(h^2 + x^2)
   B. I / x^2
   C. I x / d^3
   D. I d / h^2
   Answer: A. E = I cos(theta)/d^2 with cos(theta) = h/d gives I h/d^3. (Key point: Formula for illuminance)

2. What is the illuminance directly beneath a 4750 cd lamp at 6 m?
   A. 131.94 lux
   B. 791.7 lux
   C. 57 lux
   D. 5.71 lux
   Answer: A. E = I/h^2 = 4750/36 = 131.94 lux. (Key point: Directly underneath)

3. What is the illuminance at x = 16 m from the base?
   A. 28.5 lux
   B. 5.71 lux
   C. 131.9 lux
   D. 0.35 lux
   Answer: B. d^2 = 292, so E = 4750 x 6/292^1.5 = 5.71 lux. (Key point: At 16 m from base)

4. Why is the total midway between posts 57.0 lux?
   A. Two adjacent lamps each 8 m away give 28.5 lux each; farther lamps are neglected
   B. A single lamp gives 57 lux
   C. Four lamps give 14.25 lux each
   D. The 8 m is the height
   Answer: A. At x = 8 m, d = 10 m and cos = 0.6, so each lamp gives 4750 x 0.6/100 = 28.5 lux; two give 57 lux. (Key point: Midway total)

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