TrueTears question thread

[/0]

For the first formula, yes I think it's the general formula, and is for anti-nodal lines.

is the path difference in each case, so making it equal or will give you nodal or antinodal lines respectively.

I don't think the formula works for single slits... gives the nodal lines in single slit.

In the uni physics book there is no mention of a single-slit formula for antinodal points.

[TrueTears]

Yeah, I have asked kamil about it and he said his uni book does the same, but why does Nelson say it works for single slits? Further more, the proof comes from a single slit experiment in their book :|

[/0]

Yeah, I have asked kamil about it and he said his uni book does the same, but why does Nelson say it works for single slits? Further more, the proof comes from a single slit experiment in their book :|

soz lol do u know which page it's on?

[TrueTears]

Which of the following experiments is most likely to show diffraction effects? Support your answer with appropriate calculations.

A.   500 nm photons passing through a 0.05 mm slit
B. electrons passing through a 0.00015 mm slit.

I think the question is not correctly stated because I think both of the options would not show diffraction. My working is as follows:

For A: Using the ratio where w is the width of the single slit yields:



For B before using the ratio we must first find out the wavelength of the electron. Using De Broglie's wavelength yields

Now using the ratio to find extent of diffraction:

Now clearly B has a much smaller ratio than A which means the extent of diffraction is insignificant, but A's ratio is way below 1. The question asks which is most likely to show diffraction effects but wouldn't niether of them show diffraction? Because A's ratio (0.01) is too much smaller than 1 diffraction is basically insignificant?

[kamil9876]

For the first formula, yes I think it's the general formula, and is for anti-nodal lines.

is the path difference in each case, so making it equal or will give you nodal or antinodal lines respectively.

I don't think the formula works for single slits... gives the nodal lines in single slit.

In the uni physics book there is no mention of a single-slit formula for antinodal points.

Yes i agree with all that.

The book does explain why the formula for single slit antinodal lines is not a simple substitution of n for n+0.5 and I plaigarised that idea in this post:
http://vcenotes.com/forum/index.php/topic,9668.msg159100.html#msg159100

I also modified my proof in the post preceding that one just today(all this is found at the bottom of the post), for those interested. Apologies for the slight incompleteness in my first attempt at the proof.

[dcc]

ok.

[/0]

Thanks kamil ^^

Also, for the diffraction ratio , I know if it is >1 then its significant and <1 means insignificant, what about if it is = 1?

Or can't it ever equal to 1?

Thanks!

Hold on... but

Does the formula break down immediately when ?

[TrueTears]

Thanks kamil ^^

Also, for the diffraction ratio , I know if it is >1 then its significant and <1 means insignificant, what about if it is = 1?

Or can't it ever equal to 1?

Thanks!

Hold on... but

Does the formula break down immediately when ?

Well looking at the wavelength can be larger than w, this will ensure a significant amount of diffraction so it can be larger than 1, but yes since ... hmmm.,,

[/0]

Not only that, but you know for constructive interference, eventually as you get far out and becomes large, for any values of or , so will be undefined, even though it should still be !

[TrueTears]

Not only that, but you know for constructive interference, eventually as you get far out and becomes large, for any values of or , so will be undefined, even though it should still be !

True, haha, I have always ignored the and just focused on the ratio but now that you bring it up... how do you explain it then?

[kamil9876]

Not only that, but you know for constructive interference, eventually as you get far out and becomes large, for any values of or , so will be undefined, even though it should still be !

Im pretty sure it's because the path difference cannot be great enough.

[TrueTears]

When 2 rarefractions meet is this also an antinodal point?

Also what does it mean the direction that a wave travels is always perpendicular to the wavefront?

[TrueTears]

The sound intensity level is defined as where is the sound intensity and is the sound intensity at the threshold of hearing.

However the sound intensity at the threshold of hearing is AT 1000 Hz.

Does this mean the (The sound intensity) also has to be the intensity at 1000 Hz when we sub into the formula or can it be at any frequency?

Thanks

[Mao]

The threshold of hearing varies with frequency. To calculate intensity in decibels, I and Io must be at the same frequency. [the question will provide you with enough information to solve it]

[Mao]

When 2 rarefractions meet is this also an antinodal point?

yes

sound waves are longitudinal waves, hence the zero point is where pressure variation is nought. Rarefractions are below 'normal' pressure, hence two rarefractions will superposition to form a trough, antinode.