Investigation on a Light Dependant Resistor

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SUMMARY

This discussion focuses on designing a laboratory experiment using a Light Dependent Resistor (LDR) to measure the intensity of light emitted by a vertical filament lamp across different wavelengths. Key components include a power supply, ray box, prism, and ammeter. The experiment involves using a diffraction grating or prism to separate light wavelengths and calibrate the LDR's response to light intensity. Participants emphasize the importance of controlling variables such as distance from the light source and ambient light interference to ensure valid results.

PREREQUISITES
  • Understanding of Light Dependent Resistor (LDR) operation and characteristics
  • Knowledge of diffraction grating and its application in wavelength measurement
  • Familiarity with black body radiation principles and emission spectra
  • Basic circuit design involving power supplies and ammeters
NEXT STEPS
  • Research the calibration methods for LDRs in light intensity measurements
  • Learn about the use of diffraction gratings for wavelength determination
  • Explore the relationship between resistance and light intensity in LDR circuits
  • Investigate safety precautions and best practices for laboratory experiments involving electrical components
USEFUL FOR

Students and educators in physics and engineering, particularly those involved in laboratory experiments related to optics and light measurement techniques.

  • #31
Megegg, I was thinking of doing a very similar setup, only I'm going to use a potential divider, and work out the voltage over the LDR. My idea is that V out should decrease, as the light is shone onto the LDR. Only problem is, I can't see how I can draw a graph as the filters won't give a specific wavelength, only a range, for example violet has a wavelength range of 380–450 nm. The prism setup would give the same problem.
 
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  • #32
hey I am confused on part b...how the wavelength of the light falling on the LDR is determined...as if u r using a diffraction grating or prism how do u know??
sorrel
 
  • #33
mr kipling - it won't give u the wavelengh - but it will give u the angle - meaning u can work oyut the wavelngh from ur angle - don't know how tho -
 
  • #34
sorrel - I am using coloured filters - which give the value of the colour - so u know the wavelngh u r using
 
  • #35
I've found a website that gives actual values for each wavelength of coloured light. I was also wondering if the filter could come supplied with a wavelength. So I think I'll use this to draw a graph of voltage against wavelength, as with my circuit, voltage should go down as the intensity is increased.

http://eosweb.larc.nasa.gov/EDDOCS/Wavelengths_for_Colors.html"
 
Last edited by a moderator:
  • #36
Hi guys.

I had similar ideas to some of you. I decided to measure the wavelength using coloured filters, i.e blue, red green etc.

However I'm not to sure about how i should set the circuit up. Should i have the LDR in series with the lamp or create a potential divider with the output going to the lamp?

Also as I'm not sure about how to measure intensity i was thinking that i could either measure the current around the LDR as this should increase as resistance decreases. And then i could plot a graph of wavelegth against current. At different wavelegths the current will be higher/lower and so i can use this to show that at certain wavelegths the current (or intensity of light) will be higher/lower.

What do you guys think?
 
  • #37
united fan- i think measuring the current in the circuit is right but i don't think the lamp should be part of the circuit. it should be separate.
hope that helps
 
  • #38
oh and btw thanks megegg!
 
  • #39
sorrel said:
united fan- i think measuring the current in the circuit is right but i don't think the lamp should be part of the circuit. it should be separate.
hope that helps

Oh, ok. cheers mate.

So do you think i should have one circuit with the LDR and the ammetre and then another one with the lamp and a resistor so that the current doesn't get too high?

Thanks for the help
 
  • #40
i too have this assignment, and reading through this thread I'm confused by the how you all seem to be overcomplicating the intensity ^^ the formula is simple

also, you'll want to use diffraction grating to calculate the wavelength of light passing through each colour filter, as you'll have to account for the fact that the method is imperfect, so just getting wavelength figures off the net will create potential error in the data.
 
  • #41
sorrel said:
hey I am confused on part b...how the wavelength of the light falling on the LDR is determined...as if u r using a diffraction grating or prism how do u know??
sorrel

there are several formulae for diffraction gratings that can be found on the net.

unfortunately i can't really help you, as i ignored them all and devised my own formula - it works fine but not many people would use it, so it could be seen by the examining body as collaboration.

try wiki?
 
  • #42
Intensity = power/area

:P

simple really
 
  • #43
one query - part d) the range and precision of any instruments that would be used: what is the average range and precision of a standard digital ammeter? or voltmeter?
 
  • #44
animecrazy said:
one query - part d) the range and precision of any instruments that would be used: what is the average range and precision of a standard digital ammeter? or voltmeter?

Sorry no idea mate. So with the intensity thing are you going to calculate the power around the LDR using an ammetre and voltmetre when different wavelegths shine on the LDR?
 
  • #45
yes. the filament lamp would give off radiation in all directions, so the area would be a sphere, using the distance from the lamp to the LDR as the radius to calculate area.

and it's ok, i found some average values
 
  • #46
animecrazy said:
yes. the filament lamp would give off radiation in all directions, so the area would be a sphere, using the distance from the lamp to the LDR as the radius to calculate area.

and it's ok, i found some average values


alright cheers mate, i'll work on that. I think you can get coloured filters which come with wavelength specifications which would be a lot simpler then diffraction grating i think.
 
  • #47
So what have you guys decided to do for intensity?
 
  • #48
animecrazy said:
yes. the filament lamp would give off radiation in all directions, so the area would be a sphere, using the distance from the lamp to the LDR as the radius to calculate area.

and it's ok, i found some average values

I'm not sure what you mean by that. I think I'm just going to relate a change in voltage to the intensity.
 
  • #49
Mr. Kipling said:
I'm not sure what you mean by that. I think I'm just going to relate a change in voltage to the intensity.


Yeah i was thinking of doing something like that. i will probably measure current and do a graph of wavelegth against current( as current should increase as intensity increases) or wavelegth against resistance.
 
  • #50
Hi, I am also doing this scenario and investigation. I was just reading the reast of the threads and noticed that a lot of people have suggested using the current for intensity, however, I do believe that one of the graphs we need to present/example is Intensity against resistance, or in our case resistance against current. Will that turn out as a bodge job or be a legitimate representation? I haven't tried it yet as only received it today ad was doing some research on how to measure intensity, so what does everyone think?
Cheers
 
  • #51
Bob Spoongfield said:
Hi, I am also doing this scenario and investigation. I was just reading the reast of the threads and noticed that a lot of people have suggested using the current for intensity, however, I do believe that one of the graphs we need to present/example is Intensity against resistance, or in our case resistance against current. Will that turn out as a bodge job or be a legitimate representation? I haven't tried it yet as only received it today ad was doing some research on how to measure intensity, so what does everyone think?
Cheers


Well i was working under the idea that if intensity increases the resistance will decrease and so the current will rise or in other words if intensity rises current will rise and so using current as a substitute for intensity seems plausible.i hope it is at the least as that's what I'm planning to do.
 
  • #52
United fan said:
Well i was working under the idea that if intensity increases the resistance will decrease and so the current will rise or in other words if intensity rises current will rise and so using current as a substitute for intensity seems plausible.i hope it is at the least as that's what I'm planning to do.
Yer, I'm hoping to use that as well. I will be checking with my teacher tomorrow! see what he says. I was looking at the formula for intensity as well and it seems much too complicated for this level so I guess there must be some simpler way. will post back tomorrow with more info. lte's hope it is plausible eh !
 
  • #53
yep hope so too mate. if your teacher does say its alright to use current, let us know mate otherwise I've got a bit of a problem as the exams on thursday
 
  • #54
United fan said:
yep hope so too mate. if your teacher does say its alright to use current, let us know mate otherwise I've got a bit of a problem as the exams on thursday

unlucky. Mine's A-level plan and is in for the 15th so I have a bit of time to get it right, but I will definitely post back to you all asap.
 
  • #55
cheers mate.
 
  • #56
I have just spoken to my Teacher. He said that we are on the right lines, except that we needt o measure the light intensity, not the LDR intensity ( :confused: ) He also said about using the resistance ... the Resistance is inversely proportional to the Light Intensity. That's all the info he could give, without giving me the answers :S It's kinda stumped me not sure about what to put now. I still think that it will work due to the fact that the resistance is measure using Current and Voltage.
Anybody else got any ideas?
 
  • #57
Hmmm. So did he say that intensity = 1/resistance? cause i guess that's not to hard to calculate. in my textbook i found that intensity = power/area and froma point sorce it is inversley proportional i.e intensity =1/r squared.

I would probably use the first one and then measure the power across the LDR divide it by the area from the bulb to LDR though it will be circular. Though am not entierly sure.
 
  • #58
Actually i think i just found that that equation is for sound waves. Hmm this is difficult. I really do think it would be simpler to just measure current or resistance and plot it against wavelegth. do any of you guys know what you're going to do for intensity?
 
  • #59
current is directly proportional to intensity..so the graph of current against wavelength would be the same for graph of intensity against wavelength which means u don't actually need to work out the intensity.
 
  • #60
Yeah i was thinking that but its not exactly proportional. to be honest i don't think they exect us to do any complicated calculations and it will be something like wavelength against resistance or current.
 

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