Wavelength of light and interference pattern

[abeltyukov]

Homework Statement

The interference pattern on a screen 1.2 m behind a 710 line/mm diffraction grating is shown in Figure P22.45 (http://i137.photobucket.com/albums/q208/infinitbelt/p22-45alt.gif ), in which y1 = 52.3 cm and y2 = 107.6 cm. What is the wavelength of the light?


2. The attempt at a solution

I believe that is a single slit diffraction grating. Thus the equation is a*sin(theta) = m(lambda), right? I then have sin(theta) = y/L Where do I go from there?

I tried having two equations (one for y1 and one for y2) and setting them equal to one another by setting them individually equal to a. But that did not get me the right answer.


Thanks!

 

[hage567]

Can you show more of your calculation?

 

[abeltyukov]

Can you show more of your calculation?

I had:

a*(y1 / L) = m1*lambda
a*(y2 / L) = m2*lambda

(lambda*L) / y1 = (2*lambda*L) / y2

lambda = [(y2)(L)] / [(y1)(L)]


Is that right?


Thanks!

 

[hage567]

y2 is not the total distance to the second maximum from the centre, it is the distance between the first order and second order maxima. Also, be careful of the approximation of sin(theta) = y/L. That doesn't work when the angle gets too large. So that is something to consider.

 

[abeltyukov]

y2 is not the total distance to the second maximum from the centre, it is the distance between the first order and second order maxima. Also, be careful of the approximation of sin(theta) = y/L. That doesn't work when the angle gets too large. So that is something to consider.

So should y2 be y2 + y1? I am still a bit confused.

Thanks!

 

[mezarashi]

This experiment uses a diffraction grating, and you need to use the equations for that. You should focus on the first (closest to center) maxima. As hage567 has mentioned, the equations are only approximations based on the fact that (sin x = x) for small values of x. If you are using large angles, then your equations will not be accurate.

This page has a good geometric explanation for the equation you mentioned.
http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/gratcal.html

 

[abeltyukov]

This experiment uses a diffraction grating, and you need to use the equations for that. You should focus on the first (closest to center) maxima. As hage567 has mentioned, the equations are only approximations based on the fact that (sin x = x) for small values of x. If you are using large angles, then your equations will not be accurate.

This page has a good geometric explanation for the equation you mentioned.
http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/gratcal.html

Thanks! I got the right answer but am still not sure why. Here's what I did:

y = ((lambda)*(L)) / d

((0.523 * 1.408E-6) / 1.2) - (1.076E-7 - 0.523E-7) = 558.5 nm.


Thanks!