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}>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,
>>
>

 Sean Breheny
 Amateur Radio Callsign: KA3YXM
 Electrical Engineering Student
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