Asst 20 query

course Phy 202

7/6/09 8:30PM

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:

The refractions I observed from the graph are 1.7 for the violet and 1.6 for the red. We use the idea that the ratio of velocity is inversely proportional to the refractions.

1.7/1.6= 1.06

1/1.06=.94

1-.96= .06 or 6%. There is a 6% difference.

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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):

We had slightly different numbers but pretty much the same technique. I also like the idea of finding the actual speeds of light. Would n_red and n_violet be the indices I found above from the graph?

They are closer together than your figures, but your estimates are in the right range and you used them correctly.

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:

We pretty much just need to find where the extreme points on our spectrum lie and then find the width as being everything it encompasses. We can start with 400nm and use the equation

Sin(theta)*nm=wavelength/d

Sin(theta)* 400=(4E-7)/750000

I think I am doing this problem wrong. Or I have the equation misinterpreted from my notes. I am a little confused about what I am supposed to do and what to solve for and how.

I would assume that I should solve for both thetas, then use these angles to somehow figure out the spectrum that it includes using trig. I really don’t know how to do this though.

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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:

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

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Self-critique (if necessary):

I understand my initial error of taking sin^-1 of the right hand side finding it to be 17.45. I realize now that you don’t divide by 400, it is just representative. I was pretty confused about that. Also I see that this is one of the angles I need. Then we do the same for the other angle. Once we have these, we can use tan(17.45)=(opp/hypot) We have hypot being 2.4 meters and we should just solve for hypot, then take the difference between the other.

Self-critique Rating: 3

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