Optics and a thin wedge

[Malby]

I'm having trouble understanding this problem. I think I'm missing something or misunderstanding the question:

A wedge-shaped film of air is made by placing a small slip of paper between the edges of two pieces of glass as shown below. Light of wave-length 600nm is incident normally on the glass, and interference fringes are observed by reection. If the angle Theta made by the plates is 3x10^-4 rad, how many interference fringes per centimetre are observed? (Hint: Use the small-angle approximation Theta = t/x.)

My thinking is that there is a path difference of 3x10^-4 rad and it is similar to a double slit experiment (or single slit experiment) causing interference on a screen some distance L away. However we haven't been given any such distance.

Here is a diagram: http://i.imgur.com/KX1C4.png

 

[rl.bhat]

Ηι Malby, welcome to PF.

The condition for the constructive interference is

2*t = m*λ, where t is the thickness of the air wedge at a distance x from the point of contact of the glass plates.

t/x = θ. So 2*x*θ = m*λ.

Hence the number of fringes per unit length = m = 2*x*θ/λ.

 

[Malby]

In this case isn't m the fringe number and not the fringe number per unit length? I stumbled upon another answer that said m/x was 2*theta/Lambda which gives you a value. I'm not sure about that though because surely you need to know the distance the "viewing screen" (in this case the glass) is away from the source.

 

[rl.bhat]

In this case isn't m the fringe number and not the fringe number per unit length? I stumbled upon another answer that said m/x was 2*theta/Lambda which gives you a value. I'm not sure about that though because surely you need to know the distance the "viewing screen" (in this case the glass) is away from the source.

2*x*θ = m*λ.

If β is the fringe width ( distance between successive bright or dark fringes) then

2*(x+β)*θ = (m+1)λ

So β = λ/(2θ)

And number of fringes per unit length is

1/β = 2θ/λ

 

[rwisz]

I can understand how this problem may seem confusing at first. However, the key to solving this problem is to understand the concept of thin films and how they interact with light. In this case, the thin wedge of air between the two glass plates acts as a thin film, causing interference patterns when light passes through it.

To solve this problem, we can use the small-angle approximation, which states that for small angles, the tangent of the angle is approximately equal to the angle itself. In this case, we can approximate the angle Theta as t/x, where t is the thickness of the wedge and x is the distance between the two glass plates.

Next, we need to understand how interference fringes are formed in a thin film. When light passes through a thin film, some of it is reflected from the top surface while some is transmitted through and reflected from the bottom surface. These two reflected beams then interfere with each other, creating a pattern of bright and dark fringes.

To calculate the number of fringes per centimeter, we need to use the formula m = (2t/lambda) * cos(Theta), where m is the number of fringes, t is the thickness of the film, and lambda is the wavelength of light. In this case, we are given the wavelength (600nm) and the angle (Theta = 3x10^-4 rad), so we can solve for the number of fringes.

In summary, to solve this problem, we need to use the small-angle approximation and the formula for calculating the number of fringes in a thin film. By understanding these concepts and applying them to the given scenario, we can find the number of interference fringes per centimeter and better understand the behavior of light in this setup.