If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.

 

Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

 

 

019. `Query 17

 

Question: `qGeneral College Physics and Principles of Physics Problem 24.2: The third-order fringe of 610 nm light created by two narrow slits is observed at 18 deg. How far apart are the slits?

 

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Confidence Rating:

 

Given Solution:

`aThe path difference for a 3d-order fringe is 3 wavelengths, so light from one slit travels 3 * 610 nm = 1830 nm further.

 

The additional distance is equal to slit spacing * sin(18 deg), so using a for slit spacing we have

 

a sin(18 deg) = 1830 nm.

 

The slit spacing is therefore

 

a = 1830 nm / sin(18 deg) = 5920 nm, or 5.92 * 10^-6 meters.

 

 

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Question: `q**** query gen phy problem 24.7 460 nm light gives 2d-order max on screen; what wavelength would give a minimum?

 

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Confidence Rating:

 

Given Solution:

`aSTUDENT SOLUTION FOLLOWED BY INSTRUCTOR COMMENT AND SOLUTION:

 

The problem states that in a double-slit experiment, it is found that bule light of wavelength 460 nm gives a second-order maximun at a certain location on the screen. I have to determine what wavelength of visible light would have a minimum at the same location. To solve this problem I fist have to calculate the constructive interference of the second order for the blue light. I use the equation dsin'thea=m'lambda. m=2

 

(second order)

dsin'thea=(2)(460nm)

 

=920nm

 

Now, I can determine the destructive interference of the other light, using the equation

dsin'thea=(m+1/2)'lambda=(m+1/2)'lambda m+(0,1,2...)

Now that I have calculated dsin'thea=920nm, I used this value and plugged it in for dsin'thea in the destructive interference equation.(I assumed that the two angles are equal) because the problem asks for the wavelength at the same location.

Thus,

 

920nm=(m+1/2)'lambda. m=(0,1,2,...)

I calculated the first few values for 'lambda.

For m=0 920nm=(0+1/2)'lambda

=1.84*10^nm

 

For m=1 920nm=(1+1/2)'lambda =613nm

 

For m=2 920nm=(2+1/2)'lambda=368 nm

 

From these first few values, the only one of thes wavelengths that falls in the visible light range is 613nm. Therefore, this would be the wavelength of visible light that would give a minimum.

 

INSTRUCTOR COMMENT AND SOLUTION: good. More direct reasoning, and the fact that things like sines are never needed:

 

** The key ideas are that the second-order max occurs when the path difference is 2 wavelengths, and a minimum occurs when path difference is a whole number of wavelengths plus a half-wavelength (i.e., for path difference equal to 1/2, 3/2, 5/2, 7/2, ... of a wavelength).

 

We first conclude that the path difference here is 2 * 460 nm = 920 nm.

 

A first-order minimum (m=0) would occur for a path difference of 1/2 wavelength. If we had a first-order minimum then 1/2 of the wavelength would be 920 nm and the wavelength would be 1860 nm. This isn't in the visible range.

 

A minimum would also occur If 3/2 of the wavelength is 920 nm, in which case the wavelength would be 2/3 * 920 nm = 613 nm, approx.. This is in the visible range.

 

A niminum could also occur if 5/2 of the wavelength is 920 nm, but this would give us a wavelength of about 370 nm, which is outside the visible range. The same would be the case for any other possible wavelength. **

 

 

Self-critique (if necessary):

 

 

 

 

 

 

 

 

 

 

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Question: `q**** query 1 phy problem 35.54 11th edition 35.52 (37.46 10th edition) normal 477.0 nm light reflects from glass plate (n=1.52) and interferes constructively; next such wavelength is 540.6 nm.

 

How thick is the plate?

 

 

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Confidence Rating:

 

Given Solution:

`a** The path difference for constructive interference is an integer multiple of the wavelength. The path difference here is twice the thickness.

 

Wavelengths in glass are 477 nm / 1.52 and 540.6 nm / 1.52.

 

So we know that double the thickness is an integer multiple of 477 nm / 1.52, and also an integer multiple of 540.6 nm / 1.52.

 

We need to find the first integer multiple of 477 nm / 1.52 that is also an integer multiple of 540.6 nm / 1.52.

 

We first find an integer multiply of 477 that is also an integer multiply of 540.6.

 

Integer multiples of 540.6 are 540.6, 1081.2, 1621.8, etc. Dividing these numbers by 477 we obtain remainders 63.6, 127.2, etc. When the remainder is a multiple of 477 then we have an integer multiple of 477 which is also an integer multiple of 540.6.

 

SInce 477 / 63.6 = 8.5, we see that 2 * 477 / 63.6 = 17. So 17 wavelengths of 477 nm light is the first multiple that is equivalent to an integer number of wavelengths of 540.6 nm light.

 

17 * 477 = 8109.

 

Since 8109 / 540.6 = 15, we see that 17 wavelengths of 477 nm light span the same distance as 15 wavelengths of 540.6 nm light.

 

It easily follows that that 17 wavelengths of (477 nm / 1.52) light span the same distance as 15 wavelengths of (540.6 nm / 1.52) light.

 

This distance is 17 * 477 nm / 1.52 = 5335 nm.

 

This is double the thickness of the pane. The thickness is therefore

 

pane thickness = 5335 nm / 2 = 2667 nm.

 

 

 

IF INTERFERENCE WAS DESTRUCTIVE: n * 477 nm / 1.52 = (n-1) * 540.6 nm / 1.52, which we solve:

 

Multiplying by 1.52 / nm we get

 

477 n = 540.6 n - 540.6

 

n * (540.6 - 477 ) = 540.6

 

n * 63.6 = 540.6

 

n = 540.6 / 63.6 = 8.5.

 

This is a integer plus a half integer of wavelengths, which would result in destructive interference for both waves.

 

Multiplying 8.5 wavelengths by 477 nm / 1.52 we get round-trip distance 2667 nm, or thickness 1334 nm. **

 

STUDENT QUESTION

 

I understand that we need an integer multiple to find the plate thickness, but I don’t understand how you find that integer multiple.

 

INSTRUCTOR RESPONSE
 

It's obvious, for example, that 20 is an integer multiple of both 4 and 5.

There are easier ways to find this, but we could have reasoned it out as follows:

2 * 4 = 8, and 8 / 5 is not an integer.
3 * 4 = 12, and 12 / 5 is not an integer.
4 * 4 = 16, and 16 / 5 is not an integer.
4 * 5 = 20 and 20 / 5 is an integer.

So 20 is the smallest number which is an integer multiple of both 4 and 5.

We used a similar brute-force process here.


Alternatively we might have found the result as follows:

We find the least common multiple of 4770 and 5406.

The prime factorization of 4770 is 2 * 3^2 * 5 * 53.

The prime factorization of 5406 is 2 * 3^2 * 17 * 53.

The least common multiple is therefore 2 * 3^2 * 5 * 17 * 53 = 81090, which is 17 * 4770.


In actual practice, with experimental uncertainties, we might not get exact equality, which would limit the usefulness of the least-common-multiple procedure.

We would likely instead look for a multiple of one wavelength whose remainder on division by the other was a sufficiently small fraction of that wavelength.

 

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Question: `q**** query univ phy prob 35.58 / 35.52 11th edition 35.50 (10th edition 37.44): 700 nm red light thru 2 slits; monochromatic visible ligth unknown wavelength. Center of m = 3 fringe pure red. Possible wavelengths? Need to know slit spacing to answer?

 

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 Confidence Rating:

 

Given Solution:

`aSTUDENT SOLUTION: The pure red band at m = 3 suggests that there exists interference between the wavelength of the red light and that of the other light. Since only the red light is present at m = 3 it stands to reason that the wavelength of the other light is a half of a wavelength behind the red wavelength so that when the wavelength of the red light is at its peak, the wavelength of the other light is at its valley. In this way the amplitude of the red light is at its maximum and the amplitude of the other light is at it minimum – this explains why only the red light is exhibited in m = 3.

 

INSTRUCTOR COMMENT

 

At this point you've got it.

 

At the position of the m=3 maximum for the red light the red light from the further slit travels 3 wavelengths further than the light from the closer. The light of the unknown color travels 3.5 wavelengths further. So the unknown wavelength is 3/3.5 times that of the red, or 600 nm.

 

You don't need to know slit separation or distance (we're assuming that the distance is very large compared with the wavelength, a reasonable assumption for any distance we can actually see. **

 

 

 

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