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'PICBASIC-L [ot] physics help!'
1999\03\01@194320 by ryan pogge

flavicon
face
Anyone here know some physics?
I need a little help.

It has to do with diffracting a laser beam through a double
slit.

you are given the distance between them=1 fringes when the
apparatus is set up
in the air n=1.  you know the distance from the slits to the
viewing screen,
and you know the wavelength of the laser.

need to figure out how much the distance between m=1 fringes
changes
if the entire setup is immersed in a tank of H20(n=1.33).

do the fringes spread beacause the index of refraction is
greater
or is it more complicated than that?  and how would I
determine the
amount that they spread(or shrink)???

I am totally stumped on this and I know its off topic but
its really important.
Thanks,
Ryan

1999\03\01@205000 by Sean Breheny

face picon face
Hi Ryan,

If I'm correct,changing "n" will have the same effect as changing the
wavelength of light. In other words, going from an "n" of 1 to 1.3 will be
the same as staying with an "n=1" but DEcreasing the wavelength by a factor
of 1.3. This means that the angular separation between fringes will
decrease by a factor of 1.3 (there will be 1.3 times as many fringes per
length).

This comes from the fact that the basic factor in any diffraction problem
is the k*d*sin theta factor (this is the difference in phase between any
two coherent light sources separated by distance d,with wave number k,as
measured by an observer at angle theta from the perpendicular to the line
between the sources). If you are in a medium other than a vacuum, the
wavelength gets smaller,so k gets bigger (k=1/wavelength) by a factor of
n,the index of refraction.

I'm no expert,just took a course in this last semester,and I'm going on my
(not always so good) memory.

Good luck,

Sean

At 07:36 PM 3/1/99 -0500, you wrote:
{Quote hidden}

| Sean Breheny
| Amateur Radio Callsign: KA3YXM
| Electrical Engineering Student
\--------------=----------------
Save lives, please look at http://www.all.org
Personal page: http://www.people.cornell.edu/pages/shb7
spam_OUTshb7TakeThisOuTspamcornell.edu ICQ #: 3329174

1999\03\01@210009 by ryan pogge

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thanks bud,
sounds plausible :)



>Hi Ryan,
>
>If I'm correct,changing "n" will have the same effect as
changing the
>wavelength of light. In other words, going from an "n" of 1
to 1.3 will be
>the same as staying with an "n=1" but DEcreasing the
wavelength by a factor
>of 1.3. This means that the angular separation between
fringes will
>decrease by a factor of 1.3 (there will be 1.3 times as
many fringes per
>length).
>
>This comes from the fact that the basic factor in any
diffraction problem
>is the k*d*sin theta factor (this is the difference in
phase between any
>two coherent light sources separated by distance d,with
wave number k,as
>measured by an observer at angle theta from the
perpendicular to the line
>between the sources). If you are in a medium other than a
vacuum, the
>wavelength gets smaller,so k gets bigger (k=1/wavelength)
by a factor of
>n,the index of refraction.
>
>I'm no expert,just took a course in this last semester,and
I'm going on my
{Quote hidden}

double
>>slit.
>>
>>you are given the distance between them=1 fringes when the
>>apparatus is set up
>>in the air n=1.  you know the distance from the slits to
the
>>viewing screen,
>>and you know the wavelength of the laser.
>>
>>need to figure out how much the distance between m=1
fringes
{Quote hidden}

1999\03\01@211657 by ryan pogge

flavicon
face
ok so what we are saying is that we decrease the angle theta
by 1.3 fold
and then find the new seperation from that?

for example if the angle is .12 degrees it would become .092
degrees
and then you take the tangent of theta and multiply by the
distance from
the slits to the viewing screen to get the new seperation?
well actually
thats from the central max to the m= 1, so if you want the
distance between
the two m=1 fringes it would actually be double that...

am I understanding correctly?
regards, Ryan



>Hi Ryan,
>
>If I'm correct,changing "n" will have the same effect as
changing the
>wavelength of light. In other words, going from an "n" of 1
to 1.3 will be
>the same as staying with an "n=1" but DEcreasing the
wavelength by a factor
>of 1.3. This means that the angular separation between
fringes will
>decrease by a factor of 1.3 (there will be 1.3 times as
many fringes per
>length).
>
>This comes from the fact that the basic factor in any
diffraction problem
>is the k*d*sin theta factor (this is the difference in
phase between any
>two coherent light sources separated by distance d,with
wave number k,as
>measured by an observer at angle theta from the
perpendicular to the line
>between the sources). If you are in a medium other than a
vacuum, the
>wavelength gets smaller,so k gets bigger (k=1/wavelength)
by a factor of
>n,the index of refraction.
>
>I'm no expert,just took a course in this last semester,and
I'm going on my
>(not always so good) memory.
>
>Good luck,
>

1999\03\02@005138 by Sean Breheny

face picon face
Hi Ryan,

Actually,no, what I told you was a bit misleading, I think:

I don't remember the exact expression,but the formula for intensity in
terms of angle is something like:

I = f(p) and p = n*k*d*sin theta

So, a fringe corresponds to a peak in I. We can assign each fringe a "p"
value. When n goes from 1 to 1.3,the p value for each fringe stays the
same,and therefore k*d*sin theta must decrease by 1.3. Since k and d are
constant, sin theta must decrease by 1.3.

Lets say that for n=1 we have a fringe at theta = 0 and theta = 1 and theta
= 1.2 (just as an example,in radians). Now,when n=1.3,these will move to
theta = 0,theta = 0.704,and theta = 0.799. Notice that the zero point
doesn't move (since 0/1.3 = 0) but the others move such that their SINES
are in the ratio 1.3. Do you see what I mean? Qualitatively,the pattern
gets compressed,but I don't think it happens exactly linearly. HOWEVER, if
the target is far away from the slits (say, a couple of feet or so),then
sin theta is approximately equal to theta (since at that distance,all the
angles involved will be small anyway),and the compression will be linear,to
very good approximation (so,you will have very close to 1.3 times as many
fringes per inch,for example).

Incidentally, the regular formula for double slit diffraction is usually
the Fraunhoffer version,which makes an approximation anyway,so your target
should be at least a foot or two away from the slits to agree well with
that formula anyway. If you are within a few multiples of d (i.e.,close
compared to the distance between the slits), then you really need to use
the Fresnel version of the formula anyway. It is somewhat more complicated
since it doesn't assume that the observer is "far away" compared to the
distance between the slits.

I have started to ramble,sorry :) The moral of the story is this: if your
slits are only a few millimeters apart, and you are looking at a pattern a
foot away,then the regular (Fraunhoffer) formula is good,and the number of
fringes per inch on the target will increase by a factor of n.

Sean


At 09:11 PM 3/1/99 -0500, you wrote:
{Quote hidden}

| Sean Breheny
| Amateur Radio Callsign: KA3YXM
| Electrical Engineering Student
\--------------=----------------
Save lives, please look at http://www.all.org
Personal page: http://www.people.cornell.edu/pages/shb7
shb7spamKILLspamcornell.edu ICQ #: 3329174

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