Open Query-Assignment 20

#$&*

course Phy 232

7/17 7

Question: `q**** query univ phy 36.61 11th edition 36.59 phasor for 8 slitsYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

The phasor diagram starts out at 3*pi/4. Thus, if there are 8 slits, I am assuming the result will be 7*pi/4. However, I am unsure if I completely understand the question.

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

`a** If you look at the phasor diagram for phi = 3 pi / 4 you will see that starting at any vector the fourth following vector is in the opposite direction. So every slit will interfere destructively with the fourth following slit. This is because 4 * 3 pi / 4 is an odd multiple of pi.

The same spacing will give the same result for 5 pi / 4 and for 7 pi / 4; note how starting from any vector it takes 4 vectors to get to the antiparallel direction.

For 6 pi / 4, where the phasor diagram is a square, every slit will interfere destructively with the second following slit.

For phi = pi/4 you get an octagon.

For phi = 3 pi / 4 the first vector will be at 135 deg, the second at 270 deg (straight down), the third at 415 deg (same as 45 deg, up and to the right). These vectors will not close to form a triangle. The fourth vector will be at 45 deg + 135 deg = 180 deg; i.e., horizontal to the left. The next two will be at 315 deg (down and toward the right) then 90 deg (straight up). The last two will be at 225 deg (down and to left) and 360 deg (horiz to the right).

The resulting endpoint coordinates of the vectors, in order, will be

-0.7071067811, .7071067811

-0.7071067811, -0.2928932188

0, 0.4142135623

-1, 0.4142135623

-0.2928932188, -0.2928932188

-0.2928932188, 0.7071067811

-1, 0

0, 0

For phi = 5 pi / 4 each vector will 'rotate' relative to the last at angle 5 pi / 4, or 225 deg. To check yourself the first few endpoints will be

-0.7070747217, -0.7071290944;

-0.7070747217, 0.2928709055;

0, -0.4142040038

and the final endpoint will again be (0, 0).

For 6 pi / 4 you will get a square that repeats twice.

For 7 pi / 4 you get an octagon.

NEW PROBLEM: The longest wavelength is 700 nm and slit spacing is about 1250 nm. The path difference can't exceed the slit spacing, which is less than double the 700 nm spacine. So there are at most central max (path difference zero) and the first-order max (path difference one wavelength).

Note that there will be a second-order max for wavelengths less than about 417 nm. **

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

I do not fully understand the concept of getting shapes, like a square for 6pi/4 and an octagon for 7pi/4.

------------------------------------------------

Self-critique Rating: 1

@& If you sketch a series of vectors of equal length, with the direction changing every time by pi/4, you will get an octagon. You should construct this sketch and verify that it is so.

If you change the direction by 6 pi / 4 = 3 pi / 2 every time, you will get a square. Again you should do this.*@

"

Self-critique (if necessary):

------------------------------------------------

Self-critique rating:

"

Self-critique (if necessary):

------------------------------------------------

Self-critique rating:

#*&!

&#This looks good. See my notes. Let me know if you have any questions. &#