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Interference Jones and the number of bright bands!

  • Thread starter tlarkin
  • Start date
  • #1
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Homework Statement



Two perfectly flat glass plates are separated at one end by a piece of paper 0.065mm thick. A source of 585nm light illuminates the plates from above
How many bright bands appear to the observer?

Figure: (stars for positioning purposes)
*****eye

****| | | | | incidence light

----------------------glass
*************~~~~~~~~~~~paper
----------------------other piece of glass

Homework Equations



I'm not certain about which equation to use which is my problem
It could be dsin(theta) = m(lambda)/N which is what I've been trying
or it could be dsin(theta) = m(lambda)


The Attempt at a Solution



Here's what I tried

lamda = 585nm
d= .0065 mm
dsin(theta) = m(lambda)/N
(.0065mm)(sin 90 degrees) =585nm/ N
((.0065mm)(1))/585nm = 1/N
N= .09 which I'm pretty sure is wrong

Can anyone help me out on this one? :)
 
Last edited:

Answers and Replies

  • #2
alphysicist
Homework Helper
2,238
1
Hi tlarkin,

In this problem I believe you need to treat the air space as a thin film. With glass on both sides of the air space, what is the condition for constructive interference?
 
  • #3
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Constructive interference occurs when the light paths differ by a multiple of the wavelength and they combine and make a wave with a greater amplitude right? So the m[tex]\lambda[/tex] part of the equation I did was right, right? Bright fringes occur when dsin[tex]\theta[/tex] = m[tex]\lambda[/tex] . But how do I determine the number of bright fringes for that? Would the bright fringes just be m?
 
  • #4
alphysicist
Homework Helper
2,238
1
No, that's not quite right. The expression [itex]d\sin\theta=m\lambda[/itex] is specifically derived for several different experiments (for example, when d is the distance between two sources it is the constructive condition for a 2 slit experiment).

The more general expression is, as you were saying,

[tex]
\mbox{path length difference} = m\lambda
[/tex]

will give a zero degree phase shift between two rays. If the only thing causing interference effects is the path length difference, then this condition would be a constructive condition.

The other condition:

[tex]
\mbox{path length difference} = \left(m+\frac{1}{2}\right)\lambda
[/tex]

will give a 180 degree phase shift between the two rays.

But if the light is incident from above and hitting the glass (approximately) perpendicular to the surface, what is the path length difference between the reflected rays? It's related to thickness of the air space, because one ray reflects back up at the top of the air space and the other reflects at the bottom of the air space . What are the final equations?

Once you have the equations, you still have to identify which one is constructive and which one is destructive. That is because you need to take into account the fact that when light reflects off a higher index material, it experience a 180 degree phase shift. So what is the expression that gives constructive interference for the particular case in this problem?
 

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