- #1

- 85

- 1

It seems like a proof problem to me and I am trying to get a head start.

should I use [tex] D * sin (\theta) = m \alpha[/tex] ?

- Thread starter leolaw
- Start date

- #1

- 85

- 1

It seems like a proof problem to me and I am trying to get a head start.

should I use [tex] D * sin (\theta) = m \alpha[/tex] ?

- #2

OlderDan

Science Advisor

Homework Helper

- 3,021

- 2

Yes, that and what you know about the sine function.leolaw said:

It seems like a proof problem to me and I am trying to get a head start.

should I use [tex] D * sin (\theta) = m \alpha[/tex] ?

- #3

- 85

- 1

that sin of zero degrees is 0

- #4

OlderDan

Science Advisor

Homework Helper

- 3,021

- 2

Yes, but at zero degrees you will never have a minimum. From the geometry of the single slit diffraction setup, to not find any minima after the slit, the angle [itex] \theta [/itex] would have to be 90 degrees for the first minimum. So then what doesleolaw said:that sin of zero degrees is 0

[tex] D * sin (\theta) = m \alpha[/tex]

tell you about D?

- #5

- 85

- 1

- Last Post

- Replies
- 2

- Views
- 3K

- Last Post

- Replies
- 1

- Views
- 2K

- Last Post

- Replies
- 2

- Views
- 2K

- Last Post

- Replies
- 2

- Views
- 3K

- Last Post

- Replies
- 3

- Views
- 8K

- Last Post

- Replies
- 1

- Views
- 540

- Last Post

- Replies
- 9

- Views
- 3K

- Last Post

- Replies
- 3

- Views
- 2K

- Last Post

- Replies
- 0

- Views
- 1K

- Replies
- 9

- Views
- 7K