Assignment 22

course Phy 232

7/12 1730This seems to be the exact same assignment as assignment 20.

020. `Query 18

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Question: `qPrinciples of Physics and General Physics Problem 24.14: By what percent does the speed of red light exceed that of violet light in flint glass?

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Your solution:

I am not in that class.

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Given Solution:

`aThe respective indices of refraction for violet and red light in flint glass appear from the given graph to be about 1.665 and 1.620.

The speed of light in a medium is inversely proportional to the index of refraction of that medium, so the ratio of the speed of red to violet light is the inverse 1.665 / 1.62 of the ratio of the indices of refraction (red to violet). This ratio is about 1.028, or 102.8%. So the precent difference is about 2.8%.

It would also be possible to figure out the actual speeds of light, which would be c / n_red and c / n_violet, then divide the two speeds; however since c is the same in both cases the ratio would end up being c / n_red / ( c / n_violet) = c / n_red * n_violet / c = n_violet / n_red, and the result would be the same as that given above.

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Self-critique (if necessary):

Okay.

Self-critique Rating:3

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Question: `q **** query gen phy problem 24.34 width of 1st-order spectrum of white light (400 nm-750nm) at 2.3 m from a 7500 line/cm grating **** gen phy what is the width of the spectrum?

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Your solution:

I am not in that class.

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Given Solution: 3

`aGOOD STUDENT SOLUTION

We are given that the spectrum is from 400-750 nm. We are also given that the screen is 2.3 meters away and that the grating is 7500 lines/cm. To find this I will find where 400 nm wavelength falls on the screen and also where 750 nm wavelength falls onto the screen. Everything in between them will be the spectrum. I will use the formula...

sin of theta = m * wavelength / d

since these are first order angles m will be 1.

since the grating is 7500 lines/cm, d will be 1/7500 cm or 1/750000 m.

Sin of theta(400nm) =

1 * (4.0 * 10^-7)/1/750000

sin of theta (400nm) = 0.300

theta (400nm) = 17.46 degrees

This is the angle that the 1st order 400nm ray will make.

sin of theta (750nm) = 0.563

theta (750nm) = 34.24 degrees

This is the angle that the 1st order 750 nm ray will make.

We were given that the screen is 2.3 meters away. If we draw an imaginary ray from the grating to to the screen and this ray begins at the focal point for the rays of the spectrum and is perpendicular to the screen (I will call this point A), this ray will make two triangles, one with the screen and the 400nm angle ray and one with the screen and the 750 nm angle ray. Using the trigonomic function; tangent, we can solve for the sides of the triangles which the screen makes up.

Tan of theta = opposite / adjacent

tan of 34.24 degrees = opposite / 2.3 meters

0.6806 = opposite / 2.3 meters

opposite = 1.57 meters

tan of 17.46 degrees = opposite / 2.3 meters

opposite = 0.72 meters

So from point A to where the angle(400nm) hits the screen is 0.72 meters.

And from point A to where the angle(750nm) hits the screen is 1.57 meters.

If you subtract the one segment from the other one you will get the length of the spectrum on the screen.

1.57 m - 0.72 m = 0.85 meters is the width of the spectrum on the screen.

CORRECTION ON LAST STEP:

spectrum width = 2.3m * tan (31.33)) - 2.3m * tan (17.45) = 0.68m

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Self-critique (if necessary):

Okay.

Self-critique Rating:3

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Question: `q**** query univ phy 36.59 phasor for 8 slits

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Your solution:

If you take the phase differences and multiply them by 4 or 2 they will all give you an odd multiple of pi. This means that every phase difference gives you destructive interference. So every slit will interfere, depending on the slit, with the 4th or 2nd slit away.

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Given Solution:

`a** If you look at the phasor diagram for phi = 3 pi / 4 you will see that starting at any vector the fourth following vector is in the opposite direction. So every slit will interfere destructively with the fourth following slit. This is because 4 * 3 pi / 4 is an odd multiple of pi.

The same spacing will give the same result for 5 pi / 4 and for 7 pi / 4; note how starting from any vector it takes 4 vectors to get to the antiparallel direction.

For 6 pi / 4, where the phasor diagram is a square, every slit will interfere destructively with the second following slit.

For phi = pi/4 you get an octagon.

For phi = 3 pi / 4 the first vector will be at 135 deg, the second at 270 deg (straight down), the third at 415 deg (same as 45 deg, up and to the right). These vectors will not close to form a triangle. The fourth vector will be at 45 deg + 135 deg = 180 deg; i.e., horizontal to the left. The next two will be at 315 deg (down and toward the right) then 90 deg (straight up). The last two will be at 225 deg (down and to left) and 360 deg (horiz to the right).

The resulting endpoint coordinates of the vectors, in order, will be

-0.7071067811, .7071067811

-0.7071067811, -0.2928932188

0, 0.4142135623

-1, 0.4142135623

-0.2928932188, -0.2928932188

-0.2928932188, 0.7071067811

-1, 0

0, 0

For phi = 5 pi / 4 each vector will 'rotate' relative to the last at angle 5 pi / 4, or 225 deg. To check yourself the first few endpoints will be

-0.7070747217, -0.7071290944;

-0.7070747217, 0.2928709055;

0, -0.4142040038

and the final endpoint will again be (0, 0).

For 6 pi / 4 you will get a square that repeats twice.

For 7 pi / 4 you get an octagon.

NEW PROBLEM: The longest wavelength is 700 nm and slit spacing is about 1250 nm. The path difference can't exceed the slit spacing, which is less than double the 700 nm spacine. So there are at most central max (path difference zero) and the first-order max (path difference one wavelength).

Note that there will be a second-order max for wavelengths less than about 417 nm. **

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Self-critique (if necessary):

Okay.

Self-critique Rating:3

This is a repeat. Editing error on this end.

Assignment 22

course Phy 232

7/12 1800

023.*********************************************

Question: `qIn your own words explain how the introductory experience with scotch tape illustrates the existence of two types of charge.

Some of the tape pieces attracted each other, some repelled. These means that there must be two different types of charges.

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Self-critique (if necessary):

Okay.

Self-critique Rating:3

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Question: `qIn your own words explain how the introductory experience with scotch tape supports the idea that the force between two charged particles acts along a straight line through those particles, either attracting the forces along this line or repelling along this line.

Depending on how the tape was aligned, they always moved straight towards the other tape or straight away from the other tape.

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Self-critique (if necessary):

Okay

Self-critique Rating:3

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Question: `qIn your own words explain why this experience doesn't really prove anything about actual point charges, since neither piece of tape is confined to a point.

The charge is distributed along the entire distance of the tape, so all of the tape was attracted or repelled from the other piece of tape, causing very complex interactions.

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Self-critique (if necessary):

Okay

Self-critique Rating: 3

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Question: `qIf one piece of tape is centered at point A and the other at point B, then let AB_v stand for the vector whose initial point is A and whose terminal point is B, and let AB_u stand for a vector of magnitude 1 whose direction is the same as that of AB_u. Similarly let BA_v and BA_u stand for analogous vectors from B to A. Vectors of length 1 are called unit vectors. If the pieces attract, then in the direction of which of the two unit vectors is the tape at point A pushed or pulled? If the pieces repel, then in the direction of which of the two unit vectors is the tape at B pushed or pulled? Explain.

If the pieces attract, point A is pulled toward B along the vector AB_v and point B is pulled toward A along BA_v. If the pieces repel, point A is pushed away from B along the vector BA_v and point B is pushed away from A along the vector AB_v. This happens because they are pulled or repelled in a straight line between the two charges.

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Self-critique (if necessary):

Okay.

Self-critique Rating:3

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Question: `qUsing the notation of the preceding question, which you should have noted on paper (keep brief running notes as you do qa's and queries so you can answer 'followup questions' like this), how does the magnitude of the vector AB_v depend on the magnitude of BA_v, and how does the magnitude of each vector compare with the distance between A and B?

The vectors when dealing with forces charges should be equal and opposite to eachother. For this reason, the vectors are equal to the distances between the two points and they have the same magnitude, but opposite directions.

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Self-critique (if necessary):

Okay.

Self-critique Rating:3

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Question: `qUsing the notation of the preceding question, how is the force experienced by the two pieces of tape influenced by the magnitude of AB_v or BA_v?`aThe expected answer is that the force exerted by two charges on one another is inversely proportional to the square of the distance between them. So as the magnitudes of the vectors, which are equal to the separation, increases the force decreases with the square of the distance; and/or if the magnitude decreases the force increases in the same proportinality. The two pieces of tape are not point charges, so this is not strictly so in this case, as some parts of the tape are closer than to the other tape than other parts.

STUDENT COMMENT: Still not sure how to answer this question. I dont know what sort of difference there is between the magnitude vector and the unit vector.

INSTRUCTOR RESPONSE: As noted in the preceding solution, AB_v stands for the vector whose initial point is A and whose terminal point is B, and AB_u stands for a vector of magnitude 1 whose direction is the same as that of AB_u.

To get a unit vector in the direction of a given vector, you divide that vector by its magnitude. So AB_u = AB_v / | AB_v |.

The purpose of a unit vector is to represent a direction. If you multiply a unit vector by a number, you get a vector in the same direction as the unit vector, whose magnitude is the number by which you multiplied it.

The problem does not at this point ask you to actually calculate these vectors. However, as an example:

Suppose point A is (4, 5) and point B is (7, 3). Then AB_v is the vector whose initial point is A and whose terminal point is B, so that AB_v = <3, -2>, the vector with x component 3 and y component -2.

The magnitude of this vector is sqrt( 3^2 + (-2)^2 ) = sqrt(11).

Therefore AB_u = <3, -2> / sqrt(11) = (3 / sqrt(11), -2 / sqrt(11) ). That is, AB_u is the vector with x component 3 / sqrt(11) and y component -2 / sqrt(11). (Note that these components should be written with rationalized denominators as 3 sqrt(11) / 11 and -2 sqrt(11) / 11, so that AB_u = <3 sqrt(11) / 11, -2 sqrt(11) / 11 > ). The vector AB_u is a unit vector: it has magnitude 1.

The unit vector has the same direction as AB_v. If we want a vector of magnitude, say, 20 in the direction of this vector, we simply multiply the unit vector AB_u by 20 (we would obtain 20 * <3 sqrt(11) / 11, -2 sqrt(11) / 11 > = <60 sqrt(11) / 11, -40 sqrt(11) / 11 >).

If you do not completely understand this by reading it here, you should of course sketch this situation and identify all these quantities in your sketch.

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Self-critique (if necessary):

…Okay…

Self-critique Rating:3

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Question: `qQuery introductory set #1, 1-5

Explain how we calculate the magnitude and direction of the electrostatic force on a given charge at a given point of the x-y plane point due to a given point charge at the origin.

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Your solution:

F = k * q1 * Q / r^2. Q1 is the charge at the origin, Q is the charge in the plane, and r is the distance between them. If the charges have opposite charges, the force is an attractive force. If the charges have the same sign, the force is a repulsive force.

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Given Solution:

`a** The magnitude of the force on a test charge Q is F = k q1 * Q / r^2, where q1 is the charge at the origin. The force is one of repulsion if q1 and Q are of like sign, attraction if the signs are unlike.

The force is therefore directly away from the origin (if q1 and Q are of like sign) or directly toward the origin (if q1 and Q are of unlike sign).

To find the direction of this displacement vector we find arctan(y / x), adding 180 deg if x is negative. If the q1 and Q are like charges then this is the direction of the field. If q1 and Q are unlike then the direction of the field is opposite this direction. The angle of the field would therefore be 180 degrees greater or less than this angle.**

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Self-critique (if necessary):

Okay.

Self-critique Rating:3

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Question: `qExplain how we calculate the magnitude and direction of the electric field at a given point of the x-y plane point due to a given point charge at the origin.

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Your solution:

The equation for an electric field is F/Q, so F/Q = k * q1 / r^2. To find the direction of the vector we use arctan(y/x) and if it q1 is negative we simply add 180 degrees to this value.

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Given Solution:

`a** The magnitude of the force on a test charge Q is F = k q1 * Q / r^2, where q1 is the charge at the origin.

The electric field is therefore F / Q = k q1 / r^2. The direction is the direction of the force experienced by a positive test charge.

The electric field is therefore directly away from the origin (if q1 is positive) or directly toward the origin (if q1 is negative).

The direction of the electric field is in the direction of the displacement vector from the origin to the point if q1 is positive, and opposite to this direction if q1 is negative.

To find the direction of this displacement vector we find arctan(y / x), adding 180 deg if x is negative. If q1 is positive then this is the direction of the field. If q1 is negative then the direction of the field is opposite this direction, 180 degrees more or less than the calculated angle. **

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Self-critique (if necessary):

I am having a little trouble with the sign convention, but I will think about it and review some more.

I'm not sure which sign convention you mean, but if it doesn't become clear to you you're welcome to submit a specific question about this.

Self-critique Rating:1

&#See my notes and let me know if you have questions. &#