Why Does Doubling Occur in My Illuminance Calculation?

  • Thread starter Thread starter a b
  • Start date Start date
Click For Summary
SUMMARY

The discussion centers on calculating the illuminance of a floor beneath a spherical lamp with a radius of 6.0 cm and a luminance of 2.0 x 104 cd/m2. The correct illuminance is determined to be 25.13 lux, as per the solution provided in the Irodov book. The user initially calculated an illuminance of 50.27 lux, mistakenly assuming the luminance applied to the entire sphere rather than just the hemisphere facing downwards. The key takeaway is that the luminance provided is already halved, leading to the discrepancy in results.

PREREQUISITES
  • Understanding of illuminance and luminance concepts
  • Familiarity with spherical geometry and solid angles
  • Basic knowledge of photometry and light measurement units
  • Ability to manipulate mathematical formulas involving π and exponents
NEXT STEPS
  • Study the principles of photometry, focusing on illuminance and luminance relationships
  • Learn about solid angles and their applications in lighting calculations
  • Explore the Irodov book for additional exercises on light and optics
  • Investigate the effects of different lamp shapes on illuminance distribution
USEFUL FOR

Students of physics, lighting designers, and anyone interested in understanding the principles of light measurement and calculation in practical scenarios.

a b
Messages
14
Reaction score
0
Hello everybody,
this question is about an exercise but I post it here because it is not homework, it is an exercise that I've done to learn by myself. I hope it is ok.
It is an exercise from Irodov book (exercise 5.12)
The exercise says:
A small spherical lamp, uniformly luminous with radius R= 6.0 cm is suspended at an height h of 3 metres above the floor;
The luminance of the lamp is L=2.0*10^4 cd/m2, indipendent of direction.
Find the illuminance of the floor directly below the lamp.

the solution that the book (and my teacher) gives is

I= \pi\frac{R^{2}}{h^{2}} L = 25.13\, \, lux


I tried to solve it in this way:
the symmetry of the problem gives us many advantages; we can obtain the total luminous flux of the lamp by multiplying the luminance by the total surface and by 2 pi steradians (half of the maximum solid angle, because i assume the sphere doesn't radiate inside itself)

F=L (2 \pi) (4 \pi R^{2} )= L (8 \pi^{^2} R^{2} )

then, to have the illuminance of the floor just below the lamp, we can divide the total flux by the area of the sphere having radius h:

I= L (8 \pi^{^2} R^{2} )/ (4 \pi h^{2})= 2 \pi L \frac{R^{2}}{h^2}=50.27\; lux

but as you can see it is exactly twice the solution given by the book.
I suppose i am wrong, but i cannot understand why. can you help please?
Thank you in advance, sorry for my english
 
Physics news on Phys.org
.Your formula is correct, but the reason why you get twice the results of the book is because the luminance you are given is already the one of the hemisphere facing downwards. The total luminance of the sphere should be twice that value.
 

Similar threads

Replies
3
Views
1K
Replies
12
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 17 ·
Replies
17
Views
2K
Replies
3
Views
2K
Replies
10
Views
4K
Replies
1
Views
1K
Replies
6
Views
3K