Double Slit defraction with Baseballs

[swansont]

In the standard experiment, how much is the thickness of the slit and its relation to the photon's wavelength?

 

There is no "standard" experiment. The fringe spacing will be z*lambda/d, where z is the distance to the screen, lambda is the wavelength of the light and d is the slit spacing. You can pick and choose values and do the experiment, as long as you are able to resolve the interference pattern. So you want z/d to be >>1, since the wavelength will be somewhere around half a micron for visible light, and you can't resolve that spacing with the naked eye — you might be able to resolve 50 micron spacing. (I think 50 is around the theoretical limit, that's around 500 lines/inch, or 20/mm))

 

http://farside.ph.utexas.edu/teaching/302l/lectures/node151.html

 

 

https://en.wikipedia.org/wiki/Double-slit_experiment

"For example, if two slits are separated by 0.5 mm (d), and are illuminated with a 0.6μm wavelength laser (λ), then at a distance of 1m (z), the spacing of the fringes will be 1.2 mm."

 

So here z/d is ~2000

 

But you can buy gratings with 1 micron spacing and the screen won't have to be as far away.

[michel123456]

 

There is no "standard" experiment. The fringe spacing will be z*lambda/d, where z is the distance to the screen, lambda is the wavelength of the light and d is the slit spacing. You can pick and choose values and do the experiment, as long as you are able to resolve the interference pattern. So you want z/d to be >>1, since the wavelength will be somewhere around half a micron for visible light, and you can't resolve that spacing with the naked eye — you might be able to resolve 50 micron spacing. (I think 50 is around the theoretical limit, that's around 500 lines/inch, or 20/mm))

 

http://farside.ph.utexas.edu/teaching/302l/lectures/node151.html

 

 

https://en.wikipedia.org/wiki/Double-slit_experiment

"For example, if two slits are separated by 0.5 mm (d), and are illuminated with a 0.6μm wavelength laser (λ), then at a distance of 1m (z), the spacing of the fringes will be 1.2 mm."

 

So here z/d is ~2000

 

But you can buy gratings with 1 micron spacing and the screen won't have to be as far away.

Not the spacing.

The thickness of the material used to make the slits.

[swansont]

The material thickness is usually on order of the slit width, or less. I just made one with kitchen aluminum foil and an x-acto knife.

[StringJunky]

The material thickness is usually on order of the slit width, or less

Do they ensure the slit material is optically opaque? I presume they do but thought I'd ask.

[michel123456]

The material thickness is usually on order of the slit width, or less

If i undertsand clearly that means the material thickness and the wavelength are of the same order of magnitude. It is like passing a baseball through a brick wall as thick as the diameter of the ball.

[swansont]

If i undertsand clearly that means the material thickness and the wavelength are of the same order of magnitude. It is like passing a baseball through a brick wall as thick as the diameter of the ball.

 

Standard aluminum foil is 16 microns, or 20x or so wavelengths of red light, which is what I used.

Do they ensure the slit material is optically opaque? I presume they do but thought I'd ask.

 

For transmission gratings, yes. You can also make phase gratings, which transmit but (as the name implies) varies the phase with a different index of refraction

[michel123456]

 

Standard aluminum foil is 16 microns, or 20x or so wavelengths of red light, which is what I used.

 

For transmission gratings, yes. You can also make phase gratings, which transmit but (as the name implies) varies the phase with a different index of refraction

 

 

Standard aluminum foil is 16 microns, or 20x or so wavelengths of red light, which is what I used.

 

For transmission gratings, yes. You can also make phase gratings, which transmit but (as the name implies) varies the phase with a different index of refraction

So, if the baseball is 73mm diameter, the wall with the slit within is 1,5 meter thick.

[swansont]

 

So, if the baseball is 73mm diameter, the wall with the slit within is 1,5 meter thick.

 

It doesn't have to be. It's partly a matter of structural integrity. It's hard to make a double slit by hand that's much thinner. Fabricated gratings are going to be thinner than what I made.

[John Cuthber]

It is possible to repeat the experiment using microwaves with wavelengths of the order of a few cm and slits cut from foil. The "wall" can be thinner than the wavelength- the experiment still works.

It's common to use material that's thicker than the wavelength of light- simply because a foil less than a two thousandth of a mm thick is difficult to work with.

[swansont]

It is possible to repeat the experiment using microwaves with wavelengths of the order of a few cm and slits cut from foil. The "wall" can be thinner than the wavelength- the experiment still works.

It's common to use material that's thicker than the wavelength of light- simply because a foil less than a two thousandth of a mm thick is difficult to work with.

 

Or you can go much thicker, as with x-ray crystallography (or with electrons)

[Lazarus]

Swansont said:

Do you have evidence to back up your claim? Do you have a calculation that predicts the size of the interference pattern? If yes, please present them.

 

Strange said:

Or how about: you show us your calculations?

 

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

 

Here is a simulation of Diffraction of Baseballs. The numbers are the number of times a baseball landed in that square.

 

 

SINGLE EDGE

1232211221 233223133 2 11 32313

211222 35332111 1 33444452 1 1

1444233312 13222211221 12321

1353322322 254333133 2 12111 1132313

4223331 1 35333212 2 33444453 2 2

1655344312 24333211221 12321

2464422432 35433423412 1322211222213

1534344211 46443212 2 33443444 3 3

1655455312 36544423322 1331

2464422442 35333322221 2233222312 3

1534344211 46443212 2 3344233413 4

1655455312 34222322311 1331

2464422432 25544312322 22333333 2 2

1534233211 35333212 2 1433312332314

5434442 2 242222122 1 12321

1353322322 14433312322 22334342 1 1

1323122211 35332111 1 13222 2242414

3323331 1 221112122 1 12321

1232211221 14433211221 22334341

1323122211 23222111 1 12111 1142414

* 211222 221112122 1 12321

 

 

 

DOUBLE SLIT

12323344421221 2332464553333 2 11 11 323162313

2112431222 353356442211 1 334477964553 1 1

14443777353312 132235334421221 123222321

13534576542322 2543585663333 2 12112222424262313

42237533431 1 353367453412 2 334477974655 2 2

16554998464312 243356445421221 123222321

24646688742432 354369667543412 13223434433352213

15344976454211 464478563412 2 334467883747 3 3

16555998575312 365479777643322 13311331

24646688842442 353368555532221 22334456342612 3

15344976454211 464478563412 2 33445678363813 4

16555998575312 342257445432311 13311331

24646688742432 255468667532322 223355663535 2 2

15343866343211 353367453412 2 14334556635472314

54349854642 2 2422463442222 1 123222321

13534576542322 144347556532322 223365754443 1 1

13232545232211 353356442211 1 13223344626382414

33236633431 1 2211342332222 1 123222321

12323344421221 144346545421221 223365744341

13232545232211 232244332211 1 12112222525282414

* 2112431222 2211342332222 1 123222321

 

 

 

Y Single Edge Setup

.

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

. .

. .

b .

u Floor .

m-----------------------------------------------------------------X

p .

s .

. .

. .

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

.

.

 

Even with this crude simulation the diffraction patterns are evident. Also, this explains why the double splits have a stronger diffraction pattern than the Single Slit result. The overlapping patterns make the difference between the highs and lows greater.

 

The mechanical set up is a Single Edge or a Double Slit, 39 feet above the floor with counters for each square foot of a 21 foot by 79 foot rectangle. The uneven edge (bumps) consists of 1 foot squares The Double slits are 3 feet wide, including the “bumps”. The space between the slits is 1 foot. The “bumps” are sloped down from horizontal at 3 different angles, 30, 39 and 51 degrees.. The “bumps” are bent down in the forward direction, then the sides are bent down with the horizontal size still 1 foot square.” Since the Double Slits are equivalent to 4 Single Edges all that is needed for the Double Slit version is to overlay one pattern with another shifted by the distance from one slit to the other which is 4 feet. One hundred baseballs were dropped on each “bump”, evenly spaced.

The calculations

The resulting angle of the bounce, A = s*(h^.5*f + 1)*(h*g + 1)

The horizontal direction vector of the bounce = (1,(j-i)*0.5)

The horizontal distance of the bounce = (39*cos(A)) / sin(A)

Where

A is the angle from horizontal of the bounce.

s is the angle from horizontal of the bump before bending. (30, 39 or 51 degrees)

f is the bend factor to bend the x direction of the bump down. (0.17)

g is the bend factor to bend the sides of the bump down. (0.1)

h is the distance along the x axis of the hit of the baseball on the bump.

i is the distance along the y axis of the hit of the baseball on the bump.

j is the location of the left center of the bump on the y axis.

 

A better simulation program could produce a pattern to match light diffraction as closely as desired but the basic concepts are the same.

[John Cuthber]

Swansont said:

Do you have evidence to back up your claim? Do you have a calculation that predicts the size of the interference pattern? If yes, please present them.

 

Strange said:

Or how about: you show us your calculations?

 

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

 

Here is a simulation of Diffraction of Baseballs. The numbers are the number of times a baseball landed in that square.

 

 

SINGLE EDGE

1232211221 233223133 2 11 32313

211222 35332111 1 33444452 1 1

1444233312 13222211221 12321

1353322322 254333133 2 12111 1132313

4223331 1 35333212 2 33444453 2 2

1655344312 24333211221 12321

2464422432 35433423412 1322211222213

1534344211 46443212 2 33443444 3 3

1655455312 36544423322 1331

2464422442 35333322221 2233222312 3

1534344211 46443212 2 3344233413 4

1655455312 34222322311 1331

2464422432 25544312322 22333333 2 2

1534233211 35333212 2 1433312332314

5434442 2 242222122 1 12321

1353322322 14433312322 22334342 1 1

1323122211 35332111 1 13222 2242414

3323331 1 221112122 1 12321

1232211221 14433211221 22334341

1323122211 23222111 1 12111 1142414

* 211222 221112122 1 12321

 

 

 

DOUBLE SLIT

12323344421221 2332464553333 2 11 11 323162313

2112431222 353356442211 1 334477964553 1 1

14443777353312 132235334421221 123222321

13534576542322 2543585663333 2 12112222424262313

42237533431 1 353367453412 2 334477974655 2 2

16554998464312 243356445421221 123222321

24646688742432 354369667543412 13223434433352213

15344976454211 464478563412 2 334467883747 3 3

16555998575312 365479777643322 13311331

24646688842442 353368555532221 22334456342612 3

15344976454211 464478563412 2 33445678363813 4

16555998575312 342257445432311 13311331

24646688742432 255468667532322 223355663535 2 2

15343866343211 353367453412 2 14334556635472314

54349854642 2 2422463442222 1 123222321

13534576542322 144347556532322 223365754443 1 1

13232545232211 353356442211 1 13223344626382414

33236633431 1 2211342332222 1 123222321

12323344421221 144346545421221 223365744341

13232545232211 232244332211 1 12112222525282414

* 2112431222 2211342332222 1 123222321

 

 

 

Y Single Edge Setup

.

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

. .

. .

b .

u Floor .

m-----------------------------------------------------------------X

p .

s .

. .

. .

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

.

.

 

Even with this crude simulation the diffraction patterns are evident. Also, this explains why the double splits have a stronger diffraction pattern than the Single Slit result. The overlapping patterns make the difference between the highs and lows greater.

 

The mechanical set up is a Single Edge or a Double Slit, 39 feet above the floor with counters for each square foot of a 21 foot by 79 foot rectangle. The uneven edge (bumps) consists of 1 foot squares The Double slits are 3 feet wide, including the “bumps”. The space between the slits is 1 foot. The “bumps” are sloped down from horizontal at 3 different angles, 30, 39 and 51 degrees.. The “bumps” are bent down in the forward direction, then the sides are bent down with the horizontal size still 1 foot square.” Since the Double Slits are equivalent to 4 Single Edges all that is needed for the Double Slit version is to overlay one pattern with another shifted by the distance from one slit to the other which is 4 feet. One hundred baseballs were dropped on each “bump”, evenly spaced.

The calculations

The resulting angle of the bounce, A = s*(h^.5*f + 1)*(h*g + 1)

The horizontal direction vector of the bounce = (1,(j-i)*0.5)

The horizontal distance of the bounce = (39*cos(A)) / sin(A)

Where

A is the angle from horizontal of the bounce.

s is the angle from horizontal of the bump before bending. (30, 39 or 51 degrees)

f is the bend factor to bend the x direction of the bump down. (0.17)

g is the bend factor to bend the sides of the bump down. (0.1)

h is the distance along the x axis of the hit of the baseball on the bump.

i is the distance along the y axis of the hit of the baseball on the bump.

j is the location of the left center of the bump on the y axis.

 

A better simulation program could produce a pattern to match light diffraction as closely as desired but the basic concepts are the same.

So, nothing much to do with diffraction.

[swansont]

Swansont said:

Do you have evidence to back up your claim? Do you have a calculation that predicts the size of the interference pattern? If yes, please present them.

 

 

Is there some reason that the formulas in the link I provided are not sufficient?