The
Single Slit
To understand this, you need to know how to
analyze the grating by means of phasors, because a grating with N slits of spacing d becomes
a single slit as N --> ¥ and d --> 0 in such a way that the product Nd is the slit width, a. Nd = a. And
if infinity and zero bother you, rest assured that if you had 105
slits in a millimeter width, you would not be able to tell the difference
between the pattern obtained by shining light on this grating and the pattern
obtained by shining the light on a 1 millimeter single slit.
We showed that the first zero for a grating
occurs when the phasor diagram becomes a polygon, and we found that dsinq = l /N, or Ndsinq = l . So for a single slit, the phasor diagram is a circle
at the first zero, and it occurs when asinq = l , where a is the slit
width.
Then as you increase q you
encounter a small relative maximum, then eventually another zero when the
phasor diagram winds up on itself to form two circles- the total phase change
from one side of the slit to the other is 4p radians, and asinq = 2l .
So in general asinq = ml for minima
with a single slit. The same is true for grating minima, replacing a with Nd. But usually no one is
concerned with these in a grating; we are concerned with the big maxima. These
don't occur with a single slit (just the central maximum is large), nor would
they occur with the grating of 105 slits in a millimeter. (Why?)
The central maximum phasor diagram is just a
straight line. Then move a little away from center and we have an arc (grating
with 105 tiny straight segments or single slit with a true arc- no
one can tell the difference). The arc length is the sum of the magnitudes of
the E vectors, and this remains unchanged. The dotted line is the total E for
that particular point. To calculate E, note that radius R has units of E, and
if we bisect f , we
hit the midpoint of the E vector. Hence E = 2Rsin(f /2). The
arc length is the central maximum E (call it Em),
and f in radians is arc length/radius. So R = Em/f . Substituting, we have E
= (2Em/f )sin(f /2).
Intensity is proportional to the
Note also that f is the phase angle between the first an last ray,
so f = (2p /l )asinq , or for
the grating replace a with Nd.
Moving still farther along away from the
central maximum, we come to the first zero, then a small maximum at f = 8.99 radians, a second zero (two circles
superimposed) at f
= 2p rad, etc. The first ray
is on the bottom, the last ray has the arrow (>), and the dotted line is the
total E:
(I did these by eye. The m=2 circle should have half the diameter as the m=1,
and I see that I made it too big.)
Now to answer the question why the 105
slits in a millimeter does not get out of the above pattern of decline and
produce a spectacular maximum, just apply the equation dsinq = ml for m = 1
and d = 1/105 mm = 10-8 m, and you will see that for this
to occur, l needs to be less than 10-8 m (x-rays).
Now undulate back to the other waves stuff.
My other main pages:
mechanics
fluids, heat, electricity and magnetism
quantum
index
Comments, questions: fredrick.gram at tri-c.edu (but use @, not at)