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.

020. `Query 18

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?

Your solution:

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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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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?

Your solution:

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Given Solution:

`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:

A more accurate solution yields distances of 1.56 m and 0.72 m, for a difference of 0.84 m.

STUDENT QUESTION:

1.57-.72 = 0.85m.

Why subtract?

INSTRUCTOR RESPONSE

We're finding the distance from the 400 nm light on the screen (the violet end of the spectrum), which reinforces at about 17 deg, to the 750 nm light (the red end), which reinforces at 34 deg.

At the given distance the grating will produce a ROY G BIV spectrum from violet to red, spread out from a point .72 m from the center of the pattern, to a point 1.57 m from the center. The spectrum will therefore be 1.57 m - .72 m = .85 m wide.

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Question: `q**** query univ phy 36.61 11th edition 36.59 phasor for 8 slits

Your solution:

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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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