I am guessing you are referring to the experiment on this page: http://www.sciencebuddies.org/science-f ... p011.shtml
I will try to answer as many of these as I can. I believe most of these are answered in the webpage already though though.
2. What if the reflected ray isn't equal to θi ? Shall the reflected ray be ordered m=1 automatically ??
-- No, please see below:
"Rays farther from the normal than the reflected beam have order 1, +2, +3, etc. Rays closer to the normal have order −1, −2, −3, etc. In certain cases, for example very small d, some or all of the negative m orders may actually be diffracted through such a large angle that they are on the same side of the normal as the incident light. When the diffracted beam is on the same side of the normal as the incident light, the angle for the diffracted beam is negative.
In other words, if the reflected beam is on the right side of the your laser pointer (using the example on the website), the m order is negative where as if the reflected beam is on the left side of normal, that would be positive.
3. θi is placed to the right of the normal and there is a diffracted ray to the right of θi. What should this diffracted ray's order be ?? And the angle... is it positive or negative ??
- Please see answer to 2.
4. What if the computed d is negative ?? Is this reasonable or absurd ?? Or is there something wrong with substitution of values, especially the signs ??
- d is the spacing of the structure (in this case, the data tracks) so negative space is probably not correct.
5. What if the averaged d-values for some order of diffraction column is negative ?? Is this an error ?? What does this mean ?
6. Since it is mentioned that the d computed using the formula d=mλ/(sinθm-sinθi) is in nm when λ is in nm, how is the computed value used in determining data track spacing ?? Is it that when d is large, then the CD has low storage capacity or the other way around ??
How shall the computed d-values determine storage capacity of CD ??
- That I am afraid is not the entire story, you will also to consider the total number of "space" that's on disc. (A disc may also have multiple layers. This may be helpful: http://www.osta.org/technology/cdqa15.htm
7. In my thesis' Review of Related Literature, there's a part that says, "On a CD, the space between tracks is about 1.6 microns versus spacing on a DVD-R which is about 0.74-0.8 microns."
If so, how can nm be converted to microns ? Is it possible ?
- Yes, nm can be converted into microns. A nm is 1e-9 meter, while a micron is 1e-6 meter, i.e. 1 micron = 1000 nm.
8. Can nm be converted to MB to see if the storage capacity label of a CD matches with the computed ones ? If so, how ?
- Please see answer to #6
9. Lastly, how is the d=mλ/(sinθm-sinθi) formula derived ?? What is the relationship of each variable to each other ?? What are the principles supporting this formula ??
- Please reference the introduction of the experiment for a more detail explanation.
Hope this helps!