Newton's Rings question (Pearson 35.36)

[rogue_system]

Homework Statement

Newton's rings are visible when a planoconvex lens is placed on a flat glass surface. For a particular lens with an index of refraction of n= 1.50 and a glass plate with an index of n= 1.80, the diameter of the third bright ring is 0.730 mm .

If water (n= 1.33) now fills the space between the lens and the glass plate, what is the new diameter of this ring? Assume the radius of curvature of the lens is much greater than the wavelength of the light.

Relevant Equations

$r_2 = \sqrt {mλR}$
$2t = mλ$ (constructive interference for a thin film)
$2t - (m + \frac 1 2)λ$ (destructive interference for a thin film)

I've gone over the textbook, and ultimately had to look elsewhere for the equation relating the radius of the ring to the radius of curvature of the planoconvex lens.

I'm also assuming that $n_air = 1$.

I've searched high, low, and even my professor's lecture notes and slides, and it's never explained how to attack this problem in her lecture materials, or the textbook, and I'm not even sure if the thin film equations matter here?

What I've gotten thus far

$r_3 = \frac {0.730} 2 = 0.365$

$0.365^2 = 3λR$

$λR = \frac {0.365^2} 3$

And I'm honestly not sure where to go from here? I've seen a few partial online solutions that suggest that snell's law comes into play at some point, but we're not given any angles so I don't see how?

 

[kuruman]

Assume thin film interference at near normal incidence where Snell's law effects are negligible.
What happens to the wavelength of the light when it travels through water instead of air?

This Wikipedia article has all you need.

 

[rogue_system]

Assume thin film interference at near normal incidence where Snell's law effects are negligible.
What happens to the wavelength of the light when it travels through water instead of air?

This Wikipedia article has all you need.

$$n_{lens} λ_{lens} = n_{water} λ_{water}$$

Which gives us

$$λ_{water} = \frac {n_{lens} λ_{lens}} {n_{water}}$$

The part I was missing was how the lens behaved though, the approximation of $$t = \frac {r^2} {2R}$$

Also I have the equations for constructive and destructive swapped above, so for constructive interference in this case we have

$$2t = (m + \frac 1 2)λ$$

Plugging in that for the 3rd ring, m = 2, we can get that $t = \frac 5 4 λ$

Reranging the approximation for t in terms of $r^2$ and plugging in the value I just got

$$r^2 = 2R \frac 5 4 λ$$

Putting in terms of the $D$ value we have
$$D^2 = 10Rλ$$

I did also find an equation giving that

$D^2$ is proportional to $\frac {n_1} {n_2}$ but I don't understand why that would be. (The formula did yield a correct answer, but I am trying to understand how it's derived/works) Based on what I've already done, $D^2$ relates to $λ_{water}$ somehow?

 

[kuruman]

I don't think you should worry about $\lambda_{lens}$. The number of wavelengths you can fit in the lens part of the path is the same regardless of whether there is air or water in the gap.

Putting in terms of the D value we have

What does this $D$ value in the formula represent? I think you need to understand that first.