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AS Physics P22 Related.

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Lets say if someone wrote learns most of the definitions and formulas..has an idea of all as levels chapters and it's parts...but has only limited past paper practice... about 9 or 10 past papers practised. How good is his preparation for the exam?
 
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Its good enough.. past paper practice is needed only to know HOW to write and answer not WHAT the answer is. As long as that person has adequate knowledge about any question he sees its good enough. :)
 
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wavelength=ax/D can anyone explain all the conditions relating to this formula? like increasing or decreasing the separation of slit and width of each slit and or any other condition if so!! plz be quick ITS MY EXAM TOMORROW !!!!!!!!!!!!!!!!!!
 
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wavelength = ( distance b/w two dark fringes * Distance B/w two slits ) / (Distance B/w slits and screen ) :)
 
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wavelength=ax/D can anyone explain all the conditions relating to this formula? like increasing or decreasing the separation of slit and width of each slit and or any other condition if so!! plz be quick ITS MY EXAM TOMORROW !!!!!!!!!!!!!!!!!!

Wavelength is increased: Fringe separation increases. Contrast is unchanged.
Wavelength is reduced: Fringe separation reduces. Contrast is unchanged.

D is increased: Fringe separation increases. Contrast decreases.
D is decreased: Fringe separation decreases. Contrast increases.

a is increased: Fringe separation decreases and vice versa.

If size of one slit only is increased, with 'a' unchanged. Interference pattern observed over a limited area. Bright fringes become more bright.
If one slit is completely closed, no interference pattern.
 
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Wavelength is increased: Fringe separation increases. Contrast is unchanged.
Wavelength is reduced: Fringe separation reduces. Contrast is unchanged.

D is increased: Fringe separation increases. Contrast decreases.
D is decreased: Fringe separation decreases. Contrast increases.

a is increased: Fringe separation decreases and vice versa.

If size of one slit only is increased, with 'a' unchanged. Interference pattern observed over a limited area. Bright fringes become more bright.
If one slit is completely closed, no interference pattern.
i would like to correct one of your statements regarding change in wavelength.
Remember that for a sufficient change in wavelength, frequency is affected and so is the intensity of the light waves
Wavelength is inversely proportional to frequency.
Intensity is directly proportional to (frequency)^2
So if wavelength is decreased, frequency increases, and so do intensity so the contrast will be more.
If wavelength is increased, frequency decreases, so do intensity and thus contrast decreases.
 
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i would like to correct one of your statements regarding change in wavelength.
Remember that for a sufficient change in wavelength, frequency is affected and so is the intensity of the light waves
Wavelength is inversely proportional to frequency.
Intensity is directly proportional to (frequency)^2
So if wavelength is decreased, frequency increases, and so do intensity so the contrast will be more.
If wavelength is increased, frequency decreases, so do intensity and thus contrast decreases.
Frequency? don't you mean amplitude?
 
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i would like to correct one of your statements regarding change in wavelength.
Remember that for a sufficient change in wavelength, frequency is affected and so is the intensity of the light waves
Wavelength is inversely proportional to frequency.
Intensity is directly proportional to (frequency)^2
So if wavelength is decreased, frequency increases, and so do intensity so the contrast will be more.
If wavelength is increased, frequency decreases, so do intensity and thus contrast decreases.
Intensity is directly proportional to AMPLITUDE squared........not frequency........
 
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well what i learnt is that intensity is proportional to the square of amplitude only :p :( ......... its scary when you get to know this the night before the exam ? :p ;)
 
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yeah its amplitude squared aswell as frequency squared and hence intensity is inversely proportional to wavelength square :p
 
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