Need help: Calculating the intensity of a reflected ray

In summary, the conversation is about calculating the intensity field of a UV lamp due to its reflections. There are four UV lamps surrounding the lamp, each with three reflecting plates. If the geometry is not clear, the poster asks for someone's email address so that they can provide a geometry file. The virtual lamp created by the mirrors is twice the distance from the real lamp as measured from the real lamp along the perpendicular line from the lamp to the mirror.
  • #1
madhukarrapaka
10
0
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 without 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'

Thanks in advance
 
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  • #2
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
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

thanks in advance
 
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  • #4
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.
 
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  • #5
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

thanks in advance
 
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  • #6
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
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.

Thanks in advance
 
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  • #8
Can you calculate the intensity in the centre of the flow field due to output of a single real lamp?
 
  • #9
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)
 
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  • #10
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:

[tex]I = \frac{P}{4 \pi r^2}[/tex]

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

[tex]I = \frac{PL}{4 \pi r^2}[/tex]
 
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  • #11
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
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
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?
 
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  • #14
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
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.

Thanks in advance.

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.
 

Attachments

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  • #16
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
Sorry to bother you Andrevdh
 
  • #18
The info you are looking for are most likely covered in laser theory. Another avenue is plasma tube design (e.g. neon tubes).
 
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  • #19
Try to find help on
http://www.osa.org/" - 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.
 
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1. How do I calculate the intensity of a reflected ray?

To calculate the intensity of a reflected ray, you will need to know the incident angle, the angle of reflection, and the intensity of the incident ray. You can then use the law of reflection, which states that the angle of incidence is equal to the angle of reflection, to calculate the intensity of the reflected ray. The formula for calculating intensity is I = I0 * cos2i), where I0 is the intensity of the incident ray and θi is the incident angle.

2. How do I measure the incident angle and angle of reflection?

The incident angle is the angle between the incident ray and the normal line, and the angle of reflection is the angle between the reflected ray and the normal line. To measure these angles, you can use a protractor or an angle-measuring tool. Place the center of the protractor or angle-measuring tool at the point where the incident ray or reflected ray intersects with the normal line, and then read the angle measurement where the ray intersects with the protractor or angle-measuring tool.

3. Can I use the same formula to calculate the intensity of any reflected ray?

Yes, you can use the same formula to calculate the intensity of any reflected ray as long as you have the necessary information, including the incident angle, angle of reflection, and intensity of the incident ray. Keep in mind that the angle of incidence and the angle of reflection must be measured from the same normal line.

4. How does the intensity of the reflected ray compare to the intensity of the incident ray?

The intensity of the reflected ray is usually less than the intensity of the incident ray. This is because some of the incident light is absorbed or scattered by the reflecting surface, resulting in a decrease in intensity of the reflected ray. The amount of decrease in intensity depends on the properties of the reflecting surface.

5. Can I use the same formula to calculate the intensity of a reflected ray in any medium?

Yes, the formula for calculating the intensity of a reflected ray can be used in any medium, as long as the incident and reflected rays are both in the same medium. If the incident ray is in a different medium than the reflected ray, you will need to use a different formula that takes into account the refractive indices of the two media.

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