Newton's Rings and Gap Size - See Attachment

[PeachBanana]

Homework Statement

A shallowly curved piece of glass
is placed on a flat one. When
viewed from above, concentric
circles appear that are called
Newton’s rings. In order to see
bright rings, the gap can be:

1. 0
2. 1/4λ
3. 1/2λ
4. λ

Homework Equations

1/2 λ occurs when n2>n1

The Attempt at a Solution

Why isn't the answer 1/2λ? For ray 1, when air travels into glass there is a 180° phase change. When glass travels back into air, there is no phase change. For Ray 2, it looks like air travels into air (although through glass) back into air so I didn't think there would be a phase change for that.

 

[TSny]

For ray 1, when air travels into glass there is a 180° phase change. When glass travels back into air, there is no phase change. For Ray 2, it looks like air travels into air (although through glass) back into air so I didn't think there would be a phase change for that.

Your wording has me a bit confused. The glass and the air are not traveling, just the light is traveling. Also, I'm not sure which ray you are calling Ray 1. Anyway, what's important are the two reflections that are labeled by B and C.

Is there a phase change at B? At C?

 

[PeachBanana]

Sorry about that. I meant to say light is traveling.

Phase change at B - Yes.
Phase chance at C - No

 

[TSny]

Phase change at B - Yes.
Phase chance at C - No

That's not quite right. Can you state the general rule for deciding whether or not there is a phase change?

 

[TSny]

1/2 λ occurs when n2>n1

Sorry, I see you already did state the rule! So, consider reflection B. Which medium is n1 and which is n2?

 

[PeachBanana]

I think my main problem is deciding the values of n2 and n1.

n1 = air = 1.00
n2 = glass ≈ 1.5?

point c

n1 = air = 1
n2 = air = 1

I'm unsure about part "C" because I know light travels through the glass but ends up in air.

 

[TSny]

At a point of reflection (say B), n1 is the medium in which the light is traveling just before it strikes the reflecting surface and n2 is the medium that the light would have traveled into if it had not reflected.

 

[PeachBanana]

Ohhhh, ok. That would mean n1 = glass ≈ 1.5. n2 = air = 1. n2 < n1 no phase change

 

[TSny]

Right.

 

[PeachBanana]

Since that is the case, at point "C" there is a phase change. I'm still trying to understand why the answer is 1/4λ.

 

[TSny]

OK, good. So, the two reflections together amount to a half-wavelength phase difference. In order to get them back in phase, what extra distance does one wave have to travel compared to the other wave (in terms of the wavelength)?

 

[PeachBanana]

Does the other wave have to go an extra "1/2" because that way both waves will be in phase and lead to constructive interference?

 

[TSny]

Exactly! So, how would you express the extra distance traveled by one of the waves in terms of the gap distance?

 

[PeachBanana]

I'm going to pretend I don't know what the answer is, haha.

Instinctively I would add them to get λ but I'm trying to look for an equation so I can algebraically see why the answer is 1/4λ.

 

[TSny]

Can you visualize the "extra distance" traveled by the wave that reflected at C?

 

[PeachBanana]

Yes. Would it make sense if I thought about it like "light has to travel the wedge distance twice?"

 

[TSny]

Yes.

 

[PeachBanana]

Yay!

 

[TSny]

Yahoo!