IGCSE/GCSE/O & A Level/IB/University Student Forum
Qualification => Subject Doubts => GCE AS & A2 Level => Sciences => Topic started by: kratos009 on November 15, 2009, 04:24:41 am
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Hey guys,
I came across this question, Question 4b ii in the 2003 May/June Physics paper 2. It says
Describe the effect, if any, on the separation and on the maximum brightness of the fringes when the following changes are made:
1. the distance D is increased to 2D, keeping a and lambda constant.
2. the wavelength lambda is increased to 1.5 lambda, keeping a and D constant.
Ok i know what happens to the separation for both the questions but for maximum brightness the answers say it will become less bright in both occasions. Could someone explain to me why the brightness decreases cause i thought the brightness of a fringe is only affected by the width of the slits in the Double slit experiment. ???.
Would appreciate the answer as soon as possible, i got the physics paper 2 tomorrow :o
Thanks a lot. ;D
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Hey guys,
I came across this question, Question 4b ii in the 2003 May/June Physics paper 2. It says
Describe the effect, if any, on the separation and on the maximum brightness of the fringes when the following changes are made:
1. the distance D is increased to 2D, keeping a and lambda constant.
2. the wavelength lambda is increased to 1.5 lambda, keeping a and D constant.
Ok i know what happens to the separation for both the questions but for maximum brightness the answers say it will become less bright in both occasions. Could someone explain to me why the brightness decreases cause i thought the brightness of a fringe is only affected by the width of the slits in the Double slit experiment. ???.
Would appreciate the answer as soon as possible, i got the physics paper 2 tomorrow :o
Thanks a lot. ;D
For 1) It is because the distance of the light wave to the screen is larger. Intensity is inversely proportional to the separation^2. So if your separation increases, the light intensity decreases.
For 2) Intensity is also directly proportional to frequency^2. So since v is constant but lambda is increased. f = v/lambda will show that f has decreased. As f decreases, intensity decreases.
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Oh right i get it. Thanks a lot ;D. You are doing Physics paper 2 tomorrow as well right, if you are good luck :).
Also do you know if we need to know about 1st harmonics and 1st overtones for this exam, paper 2?
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okay um ..there's this ques in may June 2007 ...Q3 c ii....
there asking about the mean separation of water vapor over liq...
in the marking scheme it jst cube roots the previous ans which cums out to be 1600...why??
thank you in advance
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Because i'm assuming that the previous answer gave the volume of the water molecules, so therefore cube rooting the volume would give the length of something in a normal circumstance, but in this case the length is the same as the mean separation of the water molecules. Usually when you cube a length to find the volume so in this case you are doing the opposite to find the length between molecules of water.
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OMG it was dat easy...can i be any more blonder at phy then i already am
thanx:)
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Haha, no problem ;D. Good Luck with Physics if you have Paper 2 tomorrow. ;)
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good luck to u too :)...and yes i have ppr 2 tomorrow!!
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Oh right i get it. Thanks a lot ;D. You are doing Physics paper 2 tomorrow as well right, if you are good luck :).
Also do you know if we need to know about 1st harmonics and 1st overtones for this exam, paper 2?
Yeah, I'm taking the paper tomorrow as well :) And I doubt you need to know about SHM. So, don't stress yourself into studying something out of the boundaries :)
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Oh thanks thats a relief :). Hope paper 2 goes well today ;D.