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Newtons Rings practical experiment
Okay so Tommorow I'll be using Newtons Rings experiment to work out the wavelength of sodium light.
I was reading my notes and something is unclear.
It says once you see the rings under the microscope from the central dark circle you count N number of rings from the left hand side and N number of rings on the right hand side. Yeah I understand that bit. But where am i supposed to be measuring from, is it always from the central dark circle to the edge of each adjacent ring.
From example. Central ring to 2nd ring, Central ring to 3rd ring, Central ring to 4th ring.
Or is it Central ring to 2nd ring, 2nd ring to 3rd ring, 3rd ring to 4th ring.
And is it from the outside edge of one ring to the corresponding outside edge of the next ring that I need to be measuring the distance of?
Thanks for help, it's slightly confusing to me.
I was reading my notes and something is unclear.
It says once you see the rings under the microscope from the central dark circle you count N number of rings from the left hand side and N number of rings on the right hand side. Yeah I understand that bit. But where am i supposed to be measuring from, is it always from the central dark circle to the edge of each adjacent ring.
From example. Central ring to 2nd ring, Central ring to 3rd ring, Central ring to 4th ring.
Or is it Central ring to 2nd ring, 2nd ring to 3rd ring, 3rd ring to 4th ring.
And is it from the outside edge of one ring to the corresponding outside edge of the next ring that I need to be measuring the distance of?
Thanks for help, it's slightly confusing to me.
It is normal in such an experiment to measure the diameters of the light rings, rather than the radius. This is because there is quite a lot of uncertainty in the position of the centre of the system of rings.
A graph of d squared against n (the ring number) has a slope equal to 4R lambda, where R is the radius of curvature of the lens. (d^2 = 4Rn.lambda)
Measure to the centre of each bright ring. If the graph does not pass through the origin it may be that you have miscounted n, or that the lens is not making good contact. Even so, the slope is not affected.
You need to know the curvature of the lens. (Or this can be measured using Boys' method.)
A graph of d squared against n (the ring number) has a slope equal to 4R lambda, where R is the radius of curvature of the lens. (d^2 = 4Rn.lambda)
Measure to the centre of each bright ring. If the graph does not pass through the origin it may be that you have miscounted n, or that the lens is not making good contact. Even so, the slope is not affected.
You need to know the curvature of the lens. (Or this can be measured using Boys' method.)
Okay thanks for the reply, when you say measure the diameter of the ring, so it's from the outside edge of the ring on the left side to the corresponding outside edge on the right side. When I mean sides, i mean the side of the central black dot.
Also I set up the experiment today but all i seen was yellow light from the sodium, i had an interference pattern but it was vertical lines and very faint. How do I ensure I get the correct interference pattern?
Also I set up the experiment today but all i seen was yellow light from the sodium, i had an interference pattern but it was vertical lines and very faint. How do I ensure I get the correct interference pattern?
Its from the middle of the ring on the left to the middle of the same ring on the right. This ensures you are measuring from the maximum or minimum of the light intensity.
The lens must have a large radius of curvature.
Without knowing exactly how you have set this up, it's difficult (impossible) to know what might be wrong. Can you say what equipment you have used and how it is set up?
What sort of lens? What is its radius of curvature? etc.
The lens must have a large radius of curvature.
Without knowing exactly how you have set this up, it's difficult (impossible) to know what might be wrong. Can you say what equipment you have used and how it is set up?
What sort of lens? What is its radius of curvature? etc.
Stonebridge
Its from the middle of the ring on the left to the middle of the same ring on the right. This ensures you are measuring from the maximum or minimum of the light intensity.
The lens must have a large radius of curvature.
Without knowing exactly how you have set this up, it's difficult (impossible) to know what might be wrong. Can you say what equipment you have used and how it is set up?
What sort of lens? What is its radius of curvature? etc.
The lens must have a large radius of curvature.
Without knowing exactly how you have set this up, it's difficult (impossible) to know what might be wrong. Can you say what equipment you have used and how it is set up?
What sort of lens? What is its radius of curvature? etc.
Okay I see the rings now, I have the apparatus set up perfectly and I asked my teacher about the measurement and he told me to measure the dark rings as opposed to the bright rings. I've already started to take measurements.
You've mentioned start from the centre of the bright ring on left to the centre of the bright on the right, isn't that quite inaccurate to get the travelling microscope exactly on the centre of the curve?
I've started taking measurements from the outside edge of the dark curve on the left side to the outside edge of that corresponding curve on right side.
Can i still calculte lambda from this method?
Strictly speaking, it is measured from the middle of the dark (or light) line.
However, if it's easier to get the scope on the the edge of the ring, then do that.
But to get a good value for the diameter of the ring, if you measure the outside edge on the left side, you should measure the inside edge on the right side. Otherwise you will measure the value to be a little larger than it actually is. This gives you a better "d".
However, if it's easier to get the scope on the the edge of the ring, then do that.
But to get a good value for the diameter of the ring, if you measure the outside edge on the left side, you should measure the inside edge on the right side. Otherwise you will measure the value to be a little larger than it actually is. This gives you a better "d".
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