Investigation on a Light Dependant Resistor

Mr. Kipling

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/W...or_Colors.html

United fan

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?

sorrel

united fan- i think measuring the current in the circuit is right but i dont think the lamp should be part of the circuit. it should be seperate.
hope that helps

sorrel

oh and btw thanks megegg!

United fan

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

animecrazy

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.

animecrazy

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?

animecrazy

Intensity = power/area

:P

simple really

animecrazy

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?

United fan

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?

animecrazy

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

United fan

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.

United fan

So what have you guys decided to do for intensity?

Mr. Kipling

I'm not sure what you mean by that. I think I'm just going to relate a change in voltage to the intensity.

United fan

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.

Bob Spoongfield

Hi, I am also doing this scenario and investigation. I was just reading the reast of the threads and noticed that alot of people have suggested using the current for intensity, however, I do belive 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 recieved it today ad was doing some research on how to measure intensity, so what does everyone think?
Cheers

United fan

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 thats what i'm planning to do.