# Need help: Calculating the intensity of a reflected ray

1. Jul 26, 2006

hi all,

may be my problem is a simple one.
I want to calculate the intensity field resulted due to a UV lamp in side glss cylinder located at a distance from the lamp. The lamp is sorrunded 3 sides by reflectors such that 90% of the light power is reflected. now i have to calculate the intensity of the light inside the circularglass cylinder due to the UV lamp.

Here i know how to calculate the intensity with out the reflection, please some one tell me how to include the reflection of the light.

(for clear understanding: I have a circular gas cylinder in xyz plane, and a UV lamp is located at a distance xcm from the cylinder. this UV lamp is sorrounded 3 sides by reflecting plates (like
... \
O o| where O is the circular cylinder
... / o is the lamp (the lines are reflectors)
now how to calculate the intensity inside 'O' due to 'o'

2. Jul 27, 2006

### andrevdh

The effect of the mirrors are to effectively reposition the lamp at the same distance that the lamp is from each mirror to the other side of the mirrors. It is therefore like the light of the lamp that are intercepted by the mirrors create three different lamps on their other sides of the mirrors. Are the mirrors symmetrically positioned around the lamp? How many more mirrors would it require to complete the symmetrical mirror structure?

3. Jul 27, 2006

calculating intensity

QUOTE=andrevdh]The effect of the mirrors are to effectively reposition the lamp at the same distance that the lamp is from each mirror to the other side of the mirrors. It is therefore like the light of the lamp that are intercepted by the mirrors create three different lamps on their other sides of the mirrors. Are the mirrors symmetrically positioned around the lamp? How many more mirrors would it require to complete the symmetrical mirror structure?[/QUOTE]

Hi Andrevdh,

Thanks for the response.
Actually the flow field is sorrounded by 4 UV lamps(located at left, right, top and bottom of the flow filed) and each of these UV lamps is sorrounded by 3 reflecting plates. Overall, These plates form a closed loop over the flowfield.
If the geometry is not clear, please give me ur email id so that i can attach the geometry file
View attachment description.doc

Last edited: Jul 27, 2006
4. Jul 27, 2006

### andrevdh

The light from a lamp falling on the mirror is reflected according to the law incident angle is == reflection angle. Extending the reflected rays backward behind the mirror will cause the rays to intersect at a point behind the mirror equal to the distance that the lamp is in front of the mirror. The effect is therefore to create a virtual lamp the same distance behind the mirror. This virtual lamp therefore emits the (reflected) rays that the mirror is intercepting from the real lamp. Each lamp therefore creates three virtual lamps behind the three surrounding mirrors.

A mirror therefore creates a virtual lamp (a virtual image of the lamp) at twice the distance that the real lamp is from the mirror as measured from the real lamp along the perpendicular line from the lamp to the mirror.

Last edited: Jul 27, 2006
5. Jul 28, 2006

hi andrevdh,

thanks for the response.
so if the lamp has a power of 6 Watt. Then what will be the intensity?
Here i am giving my idea, please tell me whether it is correct or not.
i have calculated the intensity due to the lamp with a power of 6 Watt first, then i have calculated the intensity with a power of 11.4 watt (i.e
6+6*0.9, becos of 90% reflection) then i added these two intensities to get the total intensity.
Is this procedure correct?...if not please let me know the correct one. Please let me know the answer as soon as possible, becos it is very important for me to proceed further in my calculations

Last edited: Jul 28, 2006
6. Jul 28, 2006

### andrevdh

The wattage of an electrical device tells you the rate of electrical energy consumption of the device. In this case the lamp converts the electrical energy to UV light. Since the conversion is not 100% efficient it is not a good idea to use the wattage to calculate the intensity of the lamp.

If the wattage is stated in the specifications as the total UV light output of the lamp, then you may use it to calculate the intensity. So you need to make sure whether the wattage refers to the rate of light or electrical energy production.

Secondly, the intensity of the lamp is dependent on the distance that one is from the lamp. It is a well known fact that the intensity of a source is inversely proportional to the distance from it. So if you talk about the intensity you need to specify the distance from the lamp.

7. Jul 31, 2006

Hi Andrevdh,
the wattage is the total UV light output of the lamp(6 watt). and each lamp is located at a distance of 3.7 cm from the centre of the flow field, lamp diameter is 1.5 cm. The flowfied has a diameter of 7cm.
Please can u give me the solution (i,e the intensity in the flow field with a 90% reflection, due the UV-lamp sorrounded by the mirrors)
Please give me a solution in detail. I am confused how to include the reflection effect.

Last edited: Jul 31, 2006
8. Jul 31, 2006

### andrevdh

Can you calculate the intensity in the centre of the flow field due to output of a single real lamp?

9. Jul 31, 2006

yes. The intensity is given by I = P/2*(Pi)*r*L (that is Intesity = power / Area )

where P= power of the lamp
r= distance from the lamp to a point in the flowfield
L= length of the lamp

This is the intensity at a single point....so in the same way i can calculate the ave. intensity due to the 4 lamps in the flow filed.(this is without the reflection effect)

Last edited: Jul 31, 2006
10. Jul 31, 2006

### andrevdh

Well, I assumed that the total output of the lamp is spread over the surface of a sphere with a radius equal to the distance where the intensity is to be evaluated:

$$I = \frac{P}{4 \pi r^2}$$

If the output is per unit length (watt/meter) of the lamp this will change to

$$I = \frac{PL}{4 \pi r^2}$$

Last edited: Jul 31, 2006
11. Jul 31, 2006

Is this formula including reflection???
Please make me clear about the reflection concept(with my formula ie. I=P/2(PI)r*L and numerics ie, P=6w, 90% reflection). sorry to trouble you, but i am still not clear.

Thanks

12. Jul 31, 2006

### andrevdh

No, we are currently just referring to the real, not virtual, lamps. The virtual lamps are situated further away and their power output will be 0.9P since only 90% of the output is reflected.

13. Jul 31, 2006

thanks Andrevdh,
So in my case as each lamp is sorrounded by 3 mirrors, (for each lamp)i should take three virtual lamps behaind the mirrors(at a same distance as the real lamp and mirror) and i have to calculate the intensity due to these virtual lamps with 0.9P; is this correct????

Last edited: Jul 31, 2006
14. Jul 31, 2006

### andrevdh

Correct. The virtual lamps lie along the perpendicular from the real lamp to the mirror, behind the mirror. One should check that the real point where you calculate the intensity is not in the shadow of the real lamp, but I would think that the designer of the mirror-lamp system would choose such an arrangement that this would not be the case.

15. Aug 7, 2006

Hi Andrevdh,
Sorry to bother u again.
I have a small question regarding the intensity calculation.
I have a Cylindrical tube(base dia=160mm; height=575mm) with a UV lamp placed at the middle of the cylinder'(parallel to its height, and its length=450mm).
How to calculate the intensity due to the UV lamp exactly at the middle of the cylinder(that is at 287.5mm from either the ends of the cylinder).
And how to calculate the total intensity in the volume of the cylinder.

If possible give ma a book name or site URL to refer abt these things(intensity, power etc..of a light source, i am searching in the web, and i am getting a lot of information but not exact information)

Once again thanks.

#### Attached Files:

• ###### fig.doc
File size:
24.5 KB
Views:
78
Last edited: Aug 7, 2006
16. Aug 8, 2006

### andrevdh

These are very specialized questions. Very few people on this planet calculate intensities inside of plasmas - nuclear physicists, astro physicists..The maths will be horrific! I will make an attempt to find some info later in the week, but don't get your hopes up. What on earth are you doing?

17. Aug 8, 2006

Sorry to bother you Andrevdh

18. Aug 9, 2006

### andrevdh

The info you are looking for are most likely covered in laser theory. Another avenue is plasma tube design (e.g. neon tubes).

Last edited: Aug 10, 2006
19. Aug 10, 2006

### andrevdh

Try to find help on
http://www.osa.org/" [Broken] - the homepage of the Optical Society of America. This is probably the richest source of optical information on this planet. Let us know about your findings there.

Last edited by a moderator: May 2, 2017