Optics and Magnetics Questions

[Jay2910]

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Hey guys,
Here are some random physics problems that I didn't quite understand. Help would be appreciated greatly!


http://i270.photobucket.com/albums/jj118/sunshine9255/IndexofRefraction.jpg

For the 1st problem:
I thought, the steeper the line, the more the index of refraction . .. so in my opinion, A was steeper than C which was steeper than B. And I know, the higher the index of refraction, the steeper the line gets, so I chose D but the answer is C. Why is that?

For 65 . . ..
I got the equation qED=.5mv^2
When I plug in the changes though, I keep getting, that it will turn out to be the same distance, but the answer is B. Doubling the velocity quadruples the distance while halving the mass . .. cuts the distance by 2 . . .but you also have to take the existing 1/2( from the right side into account).
http://i270.photobucket.com/albums/jj118/sunshine9255/Magnetism.jpg

For 53:
What does this mean?
"At the bottom of the thin film part of the wave is transmitted into air and part is reflected back up"
And what does inverted mean? I know it means to turn upside down but how do you turn a ray upside down?

 

[Deadlifts]

1st problem, You are thinking about the wrong side regarding "steepness." Draw a normal line perpendicular to the surface, if the refracted ray across the surface goes TOWARDS that normal, it's a higher index. If it refracts AWAY from the normal, it is a lower index. Mathematically, this is Snell's law, n1sin(angle of incident) = n2sin(angle of refracted). So you can tell that nB > nA and nB > nC. You can then determine which has a greater angle and see that nC>nA. Therefore, increasing indices are listed as b) ACB
Recap: Low to high n, TOWARD normal. High to low n, AWAY from normal. If not given numbers, simply use proportional reasoning with Snell's law.

Additionally, if a higher index had a steeper angle, Snell's law would make no sense. If n1 < n2, theta1 > theta2.

For question 65, you had the answer right until the considered the 0.5 term of 0.5mv^2. Your constants in this equation are q,E and that 0.5 multiplier. It doesn't change based on your inputs, therefore don't use it in the calculation. This is just proportional reasoning (like a ton of the physics questions on MCAT).

Equation for purposes of 65 simplifies to: D= mv^2. D= (1/2)(2^2) = 2

#53, light behaves as both a ray and a WAVE (wave in this problem). This is about knowing the relationships of reflection/transmission and what happens to the wave's phase. Your professor/book probably referred to wave inversion as a phase shift (two waves become 1/2 a wavelength out of phase of each other). I'm used to calling it a phase shift but we'll use inversion for the purposes of this question's wording.

Transmitted light doesn't experience an inversion/shift. Reflected light off a GREATER index of refraction will experience an inversion. Reflected light off a LESSER index will not experience an inversion. So your reflection off the film will be inverted while the reflected light from the other side of the film (reflects off the lesser index of air on the other side) will NOT be inverted. Answer should be A.

A basic image of this happening: http://labman.phys.utk.edu/phys222/modules/m9/images/ref.gif

 

[Jay2910]

Hi Deadlifts,

Thanks for the help🙂
Couple of quick follow up questions( its just on the first one).

So, the normal is defined as a perpendicular line right? and the closer the ray gets to the normal the higher the refractive index right?
So, if I draw a perpendicular line near the left end of all of those 3 boxes(where those rays start) . . I feel like the ray A, is more closer to the normal than B or C. Why isn't that have a bigger refractive index then?

Are we getting the order of the refractive indexes from the angle?
The book basically says that is the lesser the angle difference from the surface=lower the index of refraction and then it makes the same conclusion( answer being C).

Thanks again for your input!

 

[Deadlifts]

Hi Deadlifts,

Thanks for the help🙂
Couple of quick follow up questions( its just on the first one).

So, the normal is defined as a perpendicular line right? and the closer the ray gets to the normal the higher the refractive index right?
So, if I draw a perpendicular line near the left end of all of those 3 boxes(where those rays start) . . I feel like the ray A, is more closer to the normal than B or C. Why isn't that have a bigger refractive index then?

Are we getting the order of the refractive indexes from the angle?
The book basically says that is the lesser the angle difference from the surface=lower the index of refraction and then it makes the same conclusion( answer being C).

Thanks again for your input!

This page has a pretty good applet to show changes (you can modify the angle of light, index values, etc. Play around with it for a more conceptual understanding.
http://www.eserc.stonybrook.edu/projectjava/snell/

When I say draw a line perpendicular to the surface, I mean intersecting the point where the two rays meet on said surface (see: http://www.eserc.stonybrook.edu/projectjava/snell/23a.gif). It's the angles from that normal line that are used in Snell's law.

Ray A has the greatest angle of incidence (aka entry angle from normal) from our choices. It's drawn at a much larger angle than the other two intentionally. The greater the angle from the normal, the smaller its index has to be. Remember, n1*sin(theta1)=n2*sin(theta2)