Angular separation between two wavelengths

Click For Summary

Homework Help Overview

The discussion revolves around calculating the angular separation between two wavelengths, specifically 589.0 and 589.6 nanometers, in the context of diffraction and interference patterns. The problem involves differentiation and the application of relevant equations in optics.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the use of differentiation to derive the angular separation and question the correctness of the initial approach. There is an exploration of the relationship between wavelength separation and angular separation, with attempts to clarify the distinction between first and second order separations.

Discussion Status

The discussion is active, with participants providing feedback on each other's interpretations and approaches. Some guidance has been offered regarding the need to consider second-order effects, indicating a productive exploration of the problem.

Contextual Notes

Participants note specific parameters such as the grating spacing (d = 1/600 mm) and the application of small angle approximations, which are relevant to the calculations being discussed.

socoil23
Messages
4
Reaction score
0
1. using differentiation,find the angular separation in arc minutes in second order for two wavelengths 589.0 and 589.6 nanometers



2. d sin([tex]\theta[/tex])=m ([tex]\lambda[/tex])



3. i try to use D=[tex]\delta[/tex][tex]\theta[/tex]/[tex]\delta[/tex][tex]\lambda[/tex]
by differentating d sin([tex]\theta[/tex])=m ([tex]\lambda[/tex]) i got d/m sec(thetha).this equation can be further change to D=(sinA+sinB)/[tex]\lambda[/tex]sinB.
i assumed D is the separation between the wavelengths.is it the correct way to approach the problem??
 
Physics news on Phys.org
No, the problem is asking for the angular separation Δθ.
 
vela said:
No, the problem is asking for the angular separation Δθ.

thanks.now i know how to solve it. is it Δlambda/Δθ= d/m cosθ? and by using small angle approximation,cos theta = 1.so Δθ= Δlambda*m/d? i forgot to put extra info:d=(1/600)mm
 
I think you need to be a little more careful because the problem asks you to find the separation to second order, not just first order.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
7
Views
1K
Replies
8
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 6 ·
Replies
6
Views
3K
Replies
4
Views
3K
  • · Replies 10 ·
Replies
10
Views
2K