#$&*
course PHY 202
3/8/15 at 8:30PMI struggled with a few parts of this analysis... Please see my notes inserted with my answers.
This experiment uses a cylindrical container and two lamps or other compact light sources.
Fill a cylindrical container with water. The cylindrical section of a soft-drink bottle will suffice. The larger the bottle the better (e.g., a 2-liter bottle is preferable to a 20-oz bottle) but any size will suffice.
Position two lamps with bare bulbs (i.e., without the lampshades) about a foot apart and 10 feet or more from the container, with the container at the same height as the lamps. The line separating the two bulbs should be perpendicular to the line from one of the bulbs to the cylindrical container. The room should not be brightly lit by anything other than the two bulbs (e.g., don't do this in front of a picture window on a bright day).
The direction of the light from the bulbs changes as it passes into, then out of, the container in such a way that at a certain distance behind the container the light focuses. When the light focuses the images of the two bulbs will appear on a vertical screen behind the cylinder as distinct vertical lines. At the focal point the images will be sharpest and most distinct.
Using a book, a CD case or any flat container measure the distance behind the cylinder at which the sharpest image forms. Measure also the radius of the cylinder.
As explained in Index of Refraction using a Liquid and also in Class Notes #18, find the index of refraction of water.
****
n = (2x + 2)/(2x + 1)
where x = d_i / r_lens
x = 0.051m / 0.050m
x = 1.02
n = [2(1.02) + 2] / [2(1.02) + 1]
n = 4.04 / 3.04
n = 1.33
I have to admit that I’m shocked that my index of refraction for water is this accurate.
****
Then using a ray-tracing analysis, as describe in Class Notes, answer the following:
1. If a ray of light parallel to the central ray strikes the cylinder at a distance equal to 1/4 of the cylinder's radius then what is its angle of incidence on the cylinder?
****
d = 1/4*r
d = 1/4*(0.05m)
d = 0.0125m
sin(theta_i) = d/r
theta_i = arcsin(0.0125/0.05)
theta_i = 14.48 degrees = angle of incidence
****
2. For the index of refraction you obtained, what therefore will be the angle of refraction for this ray?
****
n1*sin(theta_i) = n2*sin(theta_r)
1.0*sin(14.48) = 1.33*sin(theta_r)
Theta_r = arcsin[sin(14.48) / 1.33]
Theta_r = 10.83 degrees = angle of refraction
*****
3. If this refracted ray continued far enough along a straight-line path then how far from the 'front' of the lens would it be when it crossed the central ray?
****
f = d / tan(theta_i - theta_r)
f = 0.0125m / tan(14.48 - 10.83)
f = 0.196m
f is the focal length, which is where the refracted ray crosses the central ray, so this point would occur 0.196m in front of the lens.
@&
The ray would cross the axis after being refracted, so the .196 meters would be on the other side of the surface (in back, not in front).
*@
I’m not sure if this makes sense to me, as this would occur far beyond the lens (since the lens is approximately 10 cm in diameter). This would make sense if the lens was of a negligible diameter, but this is not the case, so I had thought that the situation would require me to do as in Ch 23 figure 23-22.
@&
That is about right.
I don't have the most recent edition of the book at this location so I'm not sure which figure you are looking at, but the analysis with the glass rod would seem to be the relevant one. That rod doesn't have a back surface, which the cylinderical lens does, but the diagram is representative of what would happen in the rod. The distance from the front of the rod to the focal point is clearly greater than the diameter of the sphere, and is consistent with your result that the ray would, if extended, cross the axis at about double the radius of the sphere.
*@
****
4. How far from the 'front' of the lens did the sharpest image form?
****
Diameter of lens + image distance = distance of image from front of lens
10.0cm + 5.1cm = 15.1cm
****
5. Should the answer to #3 be greater than, equal to or less than the answer to #4 and why?
****
I think they should be the same (which would answer my question from #3).
@&
The ray will not actually continue along the line of the first refraction. It will do so while still inside the cylinder, but refraction will occur when it gets to the back of the bottle.
*@
****
6. How far is the actual refracted ray from the central ray when it strikes the 'back' of the lens? What is its angle of incidence at that point? What therefore is its angle of refraction?
****
Considering one large triangle with base 0.0125m (this is the 1/4*r distance of the parallel ray), height 0.196m (this is the focal length), and angle 3.65 degrees (theta_i - theta_r, which is 14.48 - 10.83), I drew a straight line for a new base for a smaller triangle within the larger triangle. I drew this line (of distance x) 0.1m “above” the original base, because this is the diameter of the lens and thus denotes the location of the back of the lens. This gave me a small triangle with base x, height of 0.096m, and angle 3.65 degrees. Thus,
tan(3.65) = x/0.096m
x = 6.12*10^-3 m
So, the refracted ray is 6.12*10^-3 m away from the central ray at the back of the lens.
@&
The incoming ray is 1.25 cm from the axis, and the refracted ray is .61 cm from the axis when it reaches the back of the lens.
The refracted ray makes angle 3.65 degrees with the axis. That's around .06 radian, so the tangent of that (small) angle should be about .06, using a small-angle approximation. So the ray should, over about a 10 cm 'run', get about .06 * 10 cm = .6 cm closer to the axis. This is consistent with your result.
*@
****
7. At what angle with the central ray does the refracted ray therefore emerge from the 'back' of the lens?
****
I think this is related to the question I had in my response to #3. I had expected the ray to refract a second time at the interface between water and air (back of the lens) but in order to calculate the focal length, I had based it only one refraction. Now I’m not sure how to proceed because the refracted ray does not cross the central ray at the back of the lens.
Using triangle geometry, I find that the angle of the refracted ray with regards to the central ray is 86.35 degrees, but I’m not sure if this is what the question is referring to.
A small triangle within the main triangle with height 0.196m, composed of the base and the normal lines. One angle is a right angle, another angle is theta_i, and the third angle remains to be found. This third angle plus theta_r gives the third angle of the main triangle, which would be the angle of the refracted ray with regards to the central.
Theta_i + 90 + angle_3 = 180
Angle_3 = 75.52 degrees
Angle_3 + theta_r = Angle of refracted ray
75.52 + 10.83 = 86.35 degrees
@&
The refracted ray will be at 86.35 degrees relative to the plane perpendicular to the front of the lens at the point where the ray meets it.
However you need to find the angle of the ray with the normal to the back of the lens, at the point where the ray meets the back of the lens.
The radial line (the line from the center to that point) will be the normal for this refraction. That line will make an angle of roughly 6 degrees with the axis.
Can you work out the details, find the angle of incidence of the ray that meets the back of the lens, then calculate its angle of refraction?
*@
****
8. How far from the 'back' of the lens will the refracted ray therefore be when it crosses the central ray?
****
Focal length - lens diameter = distance from back of lens
0.196m - 0.1m = 0.096m
@&
This neglects the refraction at the back of the lens, as I believe you understand.
*@
****
"
@&
Check my notes and see if you can work out that second refraction and arrive at a complete solution.
*@
@&
You can insert modifications, questions, etc. using &&&& before and after each insertion.
*@