Find the phase difference (in radians)

[CalinDeZwart]

Homework Statement

In a Newton's rings apparatus, find the phase difference (in radians) when an air wedge of 500nm thickness is illuminated with red light (lambda = 640nm).

t = 500nm
lambda = 640nm
radians = degrees × pi / 180°

Answer) 13

Homework Equations

2pi x path difference / lambda

The Attempt at a Solution

[/B]
I tried (2pi x 500nm / 640) = (4.9), and; (2pi x (2x500nm) / 640) = (9.8)

Neither value is appropriate when converted, what am I missing?

Thanks.

 

[ehild]

Homework Statement

In a Newton's rings apparatus, find the phase difference (in radians) when an air wedge of 500nm thickness is illuminated with red light (lambda = 640nm).

t = 500nm
lambda = 640nm
radians = degrees × pi / 180°

Answer) 13

Homework Equations

2pi x path difference / lambda

The Attempt at a Solution

[/B]
I tried (2pi x 500nm / 640) = (4.9), and; (2pi x (2x500nm) / 640) = (9.8)

Neither value is appropriate when converted, what am I missing?

Thanks.

You forgot the phase change at the interfaces.

 

[CalinDeZwart]

You forgot the phase change at the interfaces.

Thanks for your reply,

Can you please clarify what you mean?

Is it something to do with the light reflecting off the glass and the lens before being reaching the scope?

 

[ehild]

Thanks for your reply,

Can you please clarify what you mean?

Is it something to do with the light reflecting off the glass and the lens before being reaching the scope?

Yes, part of the incident light reflects from the air and the other part goes into the air wedge then reflects from glass, then goes through the air edge again, enters the glass lens and interferes with the directly reflected ray. What are the phase changes at reflection at the interfaces?

 

[haruspex]

Is it something to do with the light reflecting off the glass and the lens before being reaching the scope?

Further to ehild's reply, it might help to understand why the phase change applies to the reflection at the air-to-glass transition but not at the other.
You can think of the medium with the higher refractive index as resisting the oscillation. It does so by adding a wave (which propagates in both directions) which, at the interface, opposes the original oscillation.
Conversely, when entering a lower refractive index medium, the light finds it easier to oscillate, so the added wave is in phase.
The same principle applies in sound (open and closed ends of tubes), vibrating strings, electricity...

See e.g. https://en.m.wikipedia.org/wiki/Reflection_phase_change

 

[CalinDeZwart]

Yes, part of the incident light reflects from the air and the other part goes into the air wedge then reflects from glass, then goes through the air edge again, enters the glass lens and interferes with the directly reflected ray. What are the phase changes at reflection at the interfaces?
View attachment 225566

Okay cool.
So my understanding so far is that light goes from air to glass (n 1.0 to n 1.5 - resulting in a phase change), then back from glass to air (n 1.5 to n 1.0 - no phase change because index is lower?), then same again for the lens?

If I need to take into account the index, how do I apply that to the formula in the OP?

 

[ehild]

Okay cool.
So my understanding so far is that light goes from air to glass (n 1.0 to n 1.5 - resulting in a phase change), then back from glass to air (n 1.5 to n 1.0 - no phase change because index is lower?), then same again for the lens?

If I need to take into account the index, how do I apply that to the formula in the OP?

When the light reflected from the bottom glass plate and going through the air wedge (the green ray) reaches the lens, greater part of it enters the lens and that part interferes with the other (blue) ray, directly reflected from the interface between the lens-air interface.
The other part of the green ray reflects again and turns backward, but it does not meat the green ray.
The reflection coefficient is defined as the ratio of the reflected electric field/ incident electric field. When light enters from material 1, and reflects from material 2, the reflection coefficient at the interface is (n₁-n₂)/(n₁+n₂) . It is negative if the light enters from a lower index material to a higher index one and that corresponds to pi phase change if the reflected electric wave.
If the light enters from higher index material and reflects from a lower-index one, the reflection coefficient is positive, no phase change.
You do not need to work with the reflection coefficient, use only the sign of it.

 

[CalinDeZwart]

What is the sign?
Is it something I have to add to *2pi x path difference / lambda* (am I on the right track with my formula?)
I understand the concepts of reflection and refraction, but I am bad at equations and reasoning.

The question itself is also really confusing, am I just measuring the air between the two glass indexes?

 

[ehild]

What is the sign?
Is it something I have to add to *2pi x path difference / lambda* (am I on the right track with my formula?)
I understand the concepts of reflection and refraction, but I am bad at equations and reasoning.

The question itself is also really confusing, am I just measuring the air between the two glass indexes?

You need to consider the air gap between the glass plate and glass lens. The green wave goes through the air gap, the blue one does not. As you said the phase of the green wave changes by 2pi x path difference / lambda, + the phase change at the air - glass interface. The phase of the blue wave does not change as it has zero path, and when reflected from the air, the reflection coefficient is positive.
The electric field of a wave can be described as Eo sin(ωt-2πx/λ). The sign changes to the opposite at reflection when the light enters from low refractive index material onto a high refractive index one. So the reflected wave becomes -Eo_r sin(ωt-2πx/λ). But it is equivalent to Eo_r sin(ωt-2πx/λ+ψ). Basically, you have the relation -sin(α)=sin(α+ψ). You certainly are familiar with the trigonometric functions. From the Addition Law sin(α+ψ)=sin(α)cos(ψ)+cos(α)sin(ψ), now it is equal -sin(α). So sin(ψ)=0 and cos (ψ)=-1. What is ψ then?