Investigation on a Light Dependant Resistor

[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 don't think the lamp should be part of the circuit. it should be seperate.
hope that helps

 

[sorrel]

oh and btw thanks megegg!

 

[United fan]

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 seperate.
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

 

[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]

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?

 

[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]

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?

 

[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]

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.

 

[United fan]

So what have you guys decided to do for intensity?

 

[Mr. Kipling]

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.

 

[United fan]

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.

 

[Bob Spoongfield]

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

 

[United fan]

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.

 

[Bob Spoongfield]

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 !

 

[United fan]

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

 

[Bob Spoongfield]

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.

 

[United fan]

cheers mate.

 

[Bob Spoongfield]

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 ( ) 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?

 

[United fan]

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.

 

[United fan]

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?

 

[sorrel]

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.

 

[United fan]

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.

 

[Bob Spoongfield]

Yer I think that they just want us to think complicated, but actually realize that it is something a lot simpler. I am just going to use current against wavelength and resistance against wavelength and see which one looks more like the one it should

 

[andy262]

i am doing the same palnning exercise and my teacher suggested using a spectrometer but I am not sure how it works can anyone help?

 

[United fan]

does anyone know the accuracy and range of the standard voltmetres and ammetres used in schools? as we need to write this too i think

 

[Justicazza]

i am confused over whether a diffraction grating would work when determining a wavelngth of a specific filter - http://www.physics.smu.edu/~scalise/emmanual/diffraction/lab.html refers to a assumed white light source so if i use the diffraction grating whilst the filter is in place over the source would i still get the spectrum of colors?

its a toss up between going for the inaccurate method of predetermined wavelength filters and diffraction gratings - anyone know of another known method of measuring the wavelength of the light through a filter which can be used as an AS level explanation?

 

[Bob Spoongfield]

i am doing the same palnning exercise and my teacher suggested using a spectrometer but I am not sure how it works can anyone help?

Try the Phillip Harris website and Rapid Electronics... they are both suppliers for schools

 

[RBanana]

All you have to do is separate the light using filters, then place the LDR in each colour of light (make sure it's the same distance from light source etc to keep it a fair test) and then you can find the resistance using a multimeter (but it on the ohm setting).
Find a transfer graph from a LDR manufacturer (resistance against intensity)- this will count for your biblography. Then using this graph, you can use the resistance you found for each colour, and find out intensity.
To find wavelength just use Young's Double Slit interference for each colour.
Lambda= ax/d where a is distance between the slits, x is fringe separation and d is distance from the slits to the screen.
That's basically what I've done, add more detail etc. My teacher was saying how realistically it would be hard to do, but it doesn't matter because we're only doing a plan.
After all, it is only worth like 2% of the whole A level...

 

[fblade]

for (b) my teach said to read it off the filters. And (c) use the Current as the Intensity. I have also found another equation: R = 500/Intensity. Could I measure the Resistance and find the intensity via the equation?

Im not sure on how to answer (d) the range and precision of any instruments that would be used?

Thanks.

 

[MattyB]

If you use a single slit first to get a diffraction patern then the Youngs Double slit interference will work for the lamp as it creates light in single waves from one point

hope this helps!

 

[DKOli]

lol intensity=power/area - hope no1 used that 2day. How about you guys learn physics first before posting.

r=500/i lux kOhm.

keep it simple. The area for that equation is not going to be using a raduis of distance between source to ldr. That would find intensity within sphere NOT IN LDR - to do that u'd need inverse square rules etc etc. Sorry the post is late but i found this forum right before beginning my planning exam.