Which pattern is correct?
With green light, does the ring size ... ?

#### Newton's Rings

Newton's rings is the name given to the bulls-eye type interference pattern obtained when a spherical piece of glass (a watch glass, for instance) sits on a flat piece of glass. This creates a thin film between the two pieces of glass with a thickness depending on the distance from contact point. If you shine light down from above you see a pattern of bright and dark rings from constructive and destructive interference.

Newton, one of the main supporters of the particle theory of light, did not seem to realize that Newton's rings provided evidence to support the wave theory.

Our thin film is a film of air between two pieces of glass. Is the center of the pattern, where the pieces of glass touch, bright due to constructive interference or dark due to destructive interference?

1. bright
2. dark

The center of the pattern is dark. Again, look at the first few steps in the seven-step method.

Step 1. Because the air film has a lower index of refraction than glass, the wave reflecting off the top of the film does not experience a phase shift.

Δt = 0

Step 2. Because the glass has a higher index of refraction than the air, the wave reflecting off the bottom surface of the film has a half-wavelength shift. It also travels an extra distance of 2t.

Δb = 2t + λn/2

Step 3. The relative shift is thus:

Δ = Δb - Δt = 2t + λn/2

As t goes to zero the relative shift approaches half a wavelength, so at the center where the film is very thin we get destructive interference so the pattern is dark.

Let's say you create a Newton's rings pattern with red light. When you switch to green light will the rings in the pattern be larger, smaller, or remain the same size?

1. larger
2. smaller
3. the same size

Green light has a smaller wavelength than red light. With green light you don't have to go as far from the center to find the film thickness that gives constructive (or destructive) interference, so the rings are smaller with green light than with red.