Diffraction & Interference - interference fringes per cm

1. Nov 13, 2015

Isaac Pepper

1. The problem statement, all variables and given/known data
A thin piece of metal foil of thickness 0.0700 mm separates one end of two pieces of optically flat glass. The top sheet of glass is 12.0 cm long. Light of wavelength 529 nm is normally incident on the glass. How many interference fringes are observed per cm in the reflected light?

2. Relevant equations
$\lambda = \frac{\lambda_0}{n}$
$2t = m\lambda$ constructive or destructive depending on phase shift (where t is the thickness 0.0700mm)
$2t = (m+\frac{1}{2})\lambda$
maybe some other equations..I'm not sure!
3. The attempt at a solution
I'm not really sure how to come about solving this problem, and any help would be greatly appreciated.
I don't know if I've understood the question correctly, but I'm thinking there's just a piece of metal separating two pieces of glass, and light coming in normal to the surface of the glass?

Since the refractive index of the glass is less than that of the metal foil there will be no phase shift, and then there will be a phase shift between the foil and the next piece of glass.

Am I on the right track or am I looking at it completely the wrong way?
Thanks !

2. Nov 13, 2015

Mister T

There's a wedge formed by the two pieces of glass, they touch on one end, but are separated by the thickness of the foil on the other. Could have been a figure accompanying the text to make this clear. Light beams incident on the surface will travel different path lengths after having been reflected off the lower piece of glass.

3. Nov 13, 2015

Isaac Pepper

That makes much more sense!
Therefore I need to use the formula $$2t = (m+\frac{1}{2})\lambda$$
Is that correct?
So since the distance between the top and the bottom pieces of glass won't be constant (forms a wedge), call the distance between them h, which varies.
At h=0 (where the two pieces of glace touch), there'll be a dark fringe due to phase shift,
So at m(max) I'll have the number of dark interference fringes?

Hence $2t = (mmax+\frac{1}{2})\lambda$
$$mmax = \frac{2t}{\lambda}-\frac{1}{2} = \frac{2*0.07*10^-3}{529*10^-9}-\frac{1}{2} = 264.1$$

So say 264, do I then multiply that by 2 to get the number of light and dark fringes because it asks for the total fringes, and divide by 12, to get the number of fringes per cm?
$$264*\frac{2}{12} = 44$$

Am I correct in thinking this?

EDIT :: I'm pretty sure this is completely wrong and am currently working through a different solution

Last edited: Nov 13, 2015
4. Nov 13, 2015

Isaac Pepper

Perhaps a different way of looking at it :

Let's first work out the angle that the two pieces of glass form
$$sin\theta = \frac{0.07}{120}$$
Therefore $$\theta = 0.033$$
$cos\theta *12 = 11.999... = 12 = L$ (length of the bottom piece)

$2t = m\lambda$
$\frac{h}{x} = \frac{t}{L}$
Combining the equations:
$\frac{2xt}{L} = m\lambda$

Therefore $$x=m\frac{L\lambda}{2t} = m*\frac{0.12*529*10^-9}{2*0.07*10^-3} = m*0.453mm$$

And so over 12cm there are $453*10^-4*12*10^-2 = 264.65$ fringes.
Which corresponds to my previous answer...(although I made a small mistake above)
This means 264/12 = 22 dark fringes per cm and 22 bright fringes so 44 fringes in total?