How Do Newton Rings Help Calculate Lens Curvature?

[bondgirl007]

Homework Statement

These are problems from a Newton rings experiment where a lens was placed on a flat surface and the interference patterns created Newton rings. I measured the diameter of the first five rings and then plotted a graph of d^2 against N (number of the individual ring).

These are my x and y values:

N D^2
1 0.011 mm
2 0.019 mm
3 0.029 mm
4 0.038 mm
5 0.048 mm

From this, I found the slope of the graph and the intercept of the line on the y-axis as I got an equation of y=0.0093x+ 0.0011. From this, I was told to graphically calculate r (radius of curvature of unknown lens) and then calculate r for each ring using the formula d^2 = 4 (lamba) r N + constant where lamba = 589.3 nm and the constant is the y-intercept.

This is what the first question is asking...
1. What is the radius of curvature of this lens? Show your results as separate calculations for each value of N. After doing this, I am required to calculate the power of the lens, assuming the refractive index of the lens is 1.523.

Homework Equations

y=0.0093x+0.0011 which translates to...
d^2 = [4*lamba*r]N + b

F = n'-n/r (for power of the lens after finding r)

The Attempt at a Solution

For the first part where it says graphically calculate r from the slope:

y=0.0093x + 0.0011

I am unsure how to get radius from this...

I tried:
9300 nm^2 = 4 (589.3 nm)(r)
r = 3.95 mm but this seems incorrect...

I would really appreciate help on this!

 

[Simon Bridge]

Homework Statement

These are problems from a Newton rings experiment where a lens was placed on a flat surface and the interference patterns created Newton rings. I measured the diameter of the first five rings and then plotted a graph of d^2 against N (number of the individual ring).

These are my x and y values:

N D^2
1 0.011 mm
2 0.019 mm
3 0.029 mm
4 0.038 mm
5 0.048 mm

Please check the units of D^2 in that table.

Are these D^2 or D?
How did you measure to such a high level of accuracy?

From this, I found the slope of the graph and the intercept of the line on the y-axis as I got an equation of y=0.0093x+ 0.0011.

Since you plotted D^2 vs N, you do not have x and y values. You have N and D^2 values. Your equation should be:

D^2=(0.0093)N+(0.0011) mm²

Never use generic variables.
Always include units.

From this, I was told to graphically calculate r (radius of curvature of unknown lens) and then calculate r for each ring using the formula d^2 = 4 (lamba) r N + constant where lamba = 589.3 nm and the constant is the y-intercept.

What color corresponds to that wavelength?
Does that match the color you used?

This is what the first question is asking...
1. What is the radius of curvature of this lens? Show your results as separate calculations for each value of N.

So you compared the theoretical equation with the equation you determined experimentally.

you have
theory: d^2 = 4λrN + c
experiment: d^2 = (0.0093mm)N + (0.0011mm)
therefore: 4λr=?

I tried:
9300 nm^2 = 4 (589.3 nm)(r)
r = 3.95 mm but this seems incorrect...

... that does seem a tad small. What sort of radius would be plausible?
Note: 0.0093mm² ≠ 9300nm²

 

[bondgirl007]

Thanks for your reply and tidying up my equation!

4λr should equal 0.0093 mm. I tried plugging it in as follows and still get the same answer..

0.0093 mm = 4(0.0005893mm)r
r= 3.95 mm

These measurements were obtained using a Vernier scale, thus they are to three decimal places.
The final answer, which is the surface power of the lens and obtained using the equation (n'-n)/R is around +1.5, which would mean the radius of curvature is around 0.35 m. I am not sure if my units are off with the decimals...

 

[Simon Bridge]

I think you have misplaced the decimal point - yes.
If 4λr=0.093mm, then r=39.5cm - which is the right order of magnitude.

A micrometer scale makes more sense for vernier calipers too.

For the future: you should make some effort to get a crude measurement (scientific estimate) for important properties as a kind-of reality check. The curve of the glass could have been traced, or you could work it out by rocking it and sighting the angle. This is stuff you learn from experience... it gets so you don't even notice yourself doing it half the time.