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2007 AS Physics Planning Exercise Help!
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wat proceedors u follow?
and intensity resistance graph is inversly proportional so does that mean when resistance is 6 ohmes intensity is at 4
could i just find the resistance and use that ratio of 6-4 to get intensity?
man this is hard lol the chemistry planning is 40 times easier than this lol
help me!!!
and intensity resistance graph is inversly proportional so does that mean when resistance is 6 ohmes intensity is at 4
could i just find the resistance and use that ratio of 6-4 to get intensity?
man this is hard lol the chemistry planning is 40 times easier than this lol
help me!!!
londonkid89
The reason the filters are being used because it allows a certain wavelength of light to pass through (e.g. a red filter will only allow red light to pass through) and we're finding out what the effect the wavelength of light has on the intensity.
I know the filters are used to seperate the different wavelengths but my point is that the filters may lower the intensity of the light aswell as seperating the wavelengths and we do not want this.
@londonkid: ohhh k thank you lol...i think i might try that out as a prelim exp later... i have a luxmeter in skl as well, but i cant figure a way of using that, as the intensity would be there w/out ldr usage
@jimmy: i think thats the whole point of the exp. diff wavelengths prob have diff intensity, and thats what they want us to find out? the filter will only let the specific w/length through, and when u plot it, what diff will it make?
@jimmy: i think thats the whole point of the exp. diff wavelengths prob have diff intensity, and thats what they want us to find out? the filter will only let the specific w/length through, and when u plot it, what diff will it make?
lucic17
@londonkid: ohhh k thank you lol...i think i might try that out as a prelim exp later... i have a luxmeter in skl as well, but i cant figure a way of using that, as the intensity would be there w/out ldr usage
Could someone explain what a lux meter is?
@jimmy: i think thats the whole point of the exp. diff wavelengths prob have diff intensity, and thats what they want us to find out? the filter will only let the specific w/length through, and when u plot it, what diff will it make?
If your doing a prelim exp. try this and see what you get:
Blue filter with Dim light and Bright light
Red filter and Bright light
You're all making some of this too complicated.
Diffraction grating lets through the light you want in that instance. Remember this is an experiment, so you will take a set of values, and plot them on a graph against intensity.
Diffraction grating only lets through certain wavelengths. Now when comparing it to natural light, you want a comparable value of sunlight to put your resulsts against yet? Go outisde and do it then. Get some readings for hte wavelength of natural light with intensity. And you STILl use the diffraction grating here, so that you're isolating the same wavelengths. This will tell you if the intensity is the same and thus whether the lights in the greenhosue aare comparable to the sun.
Potential divider. LDR as R2. I dunno what to do with Vout, to change it into intensity; my only problem.
Diffraction grating lets through the light you want in that instance. Remember this is an experiment, so you will take a set of values, and plot them on a graph against intensity.
Diffraction grating only lets through certain wavelengths. Now when comparing it to natural light, you want a comparable value of sunlight to put your resulsts against yet? Go outisde and do it then. Get some readings for hte wavelength of natural light with intensity. And you STILl use the diffraction grating here, so that you're isolating the same wavelengths. This will tell you if the intensity is the same and thus whether the lights in the greenhosue aare comparable to the sun.
Potential divider. LDR as R2. I dunno what to do with Vout, to change it into intensity; my only problem.
I know I said this earlier...
But, after speaking to my teacher, he said that I didn't need to do measure the power, however I can find the resistance of the LDR and then calculate intensity, as resistance is inversely proportional to intensity (I think baller2k3 was asking about this before), so apologies for that power mistake.
londonkid89
I'm not sure about a graph bit, but I do know that intensity is the rate of which a wave transfers energy (power) to a square metre of a surface. (Units are usually kW/m^2). By taking values of the current, I, and the potential difference of that LDR, V, from the potential divider circuit, use the formula P = V I to calculate this power.
I'm not sure about a graph bit, but I do know that intensity is the rate of which a wave transfers energy (power) to a square metre of a surface. (Units are usually kW/m^2). By taking values of the current, I, and the potential difference of that LDR, V, from the potential divider circuit, use the formula P = V I to calculate this power.
But, after speaking to my teacher, he said that I didn't need to do measure the power, however I can find the resistance of the LDR and then calculate intensity, as resistance is inversely proportional to intensity (I think baller2k3 was asking about this before), so apologies for that power mistake.
Nope, its just intensity inversely proportional to resistance. Found a source here. 2/3 down the page. (Ok it says resistance inversely proportional to intensity but its the same the other way round)
yeh guys i understand everything now i done hours of studying and now i am currently doing the planning
thanks for the help to some here
yes you guys have same understanding as me intensity is inversly proportional to resistance and yes to adam811 4ohmes =1/4 intensity thats all u need to do
working out intensity is the easy bit its just wavelength which is annoying becuase there is soo much to write about it (diffraction grating and spectrometer) mannnn 500 words is little
thanks for the help to some here
yes you guys have same understanding as me intensity is inversly proportional to resistance and yes to adam811 4ohmes =1/4 intensity thats all u need to do
working out intensity is the easy bit its just wavelength which is annoying becuase there is soo much to write about it (diffraction grating and spectrometer) mannnn 500 words is little
mustafa
hi i have written my plan up but its like 900 words 400 worrds more than required is there consequences for writing above the limit???
yeh man you will lose alot of marks
im trying at the moment to fit everything in and its hard. take some bits out of your work and compress it dont risk losing marks remember its only 16 mark question
try getting full marks
Adam811
so basically: say your resistance was 4 ohms, your intensity would be 1/4, i.e. inversely proportional
Shoud there not be a constant invovled?
is that right? if it is does that mean you use the readings of intensity and the wavlengths of the colour filters to plot the graph?
Thats what I thought too.
fblade
Shoud there not be a constant invovled?
Thats what I thought too.
Thats what I thought too.
no a constant doesnt need to be involved because if you read the question it says how does it VARY with wavelength
this means that we dont need to mention the constant bit but u can if u want to if u have enough space to write more words but i dont realy think u need it.
Adam811
ive got most of it sorted but i dont understand how you find the resistance of the LDR using a potential divider. i know that theres an equation to find voltage out but not sure how that relates to the resistance of the actual LDR?
man you guys are doing this tooo complicated remember we are not in uni
u can use potential divider circuit but why not just use ordinary series ciruit where ldr is across multimeter measuring resistance.-
is anyone talking about spectrometer in this planning???????
now can any help me find a graph of intensity againt wavelength please!!!!!!
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