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course Phy 232
7-30-11 about 10:00pm
labfocusingimages#$&*
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.
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?
2. For the index of refraction you obtained, what therefore will be the angle of refraction for this ray?
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?
4. How far from the 'front' of the lens did the sharpest image form?
5. Should the answer to #3 be greater than, equal to or less than the answer to #4 and why?
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?
7. At what angle with the central ray does the refracted ray therefore emerge from the 'back' of the lens?
8. How far from the 'back' of the lens will the refracted ray therefore be when it crosses the central ray?
Answers:
1. The distance from the cylinder to the screen was 2.9cm (focused) and the radius of the pop bottle is about 5cm. The calculated index of refraction is 1.463. The angle of incidence for this question is 14.48deg.
2. Angle of refraction of about 9.84deg
3. 15.4cm=f, using f=dx/(tan(thetaI - thetaR)), dx=1.25cm
4. 2.9cm + 2(5cm)= 12.9cm
5. Less than because the calculations for the refracted ray are not taking into account all of the environmental factors on the experiment. In a perfect world I suppose they would be equal.
6. How do you calculate this? The dx=1.25cm when entering, but how do you calculate this upon leaving the back side?
7. ?
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@& For a first approximation you can assume that the distance from entering to leaving the container is 10 cm, i.e., equal to the diameter of the container. This isn't quite accurate because of the curvature, and could be corrected, but let's not worry about that right now.
The angle of refraction is 9.8 degrees, approx.. This is therefore the angle of the refracted ray with respect to the radial ray (i.e., the ray from the center to the point where the ray enters the bottle, 1.25 cm from the axis).
What therefore is the angle of the refracted ray with the axis?
&&&&14.48deg - 9.84deg= 4.64deg&&&&
In traveling 10 cm to the 'back' of the bottle, how much closer does the refracted ray therefore get to the axis?
&&&&I still don’t understand how to calculate this, maybe I’m missing the obvious.&&&&
That ray started at 1.25 cm from the axis, and it has gotten closer by an amount you have hopefully just calculated. So as it exits the 'back' of the container, how far is it from the axis?*@
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@& You can form a right triangle as follows:
The hypotenuse extends from the point where the ray enters the bottle to the point where it leaves.
One leg is parallel to the axis through the point where the ray enters the bottle.
The other leg originates at the point where the ray exits the bottle and extends to the first leg, making a right angle.
The second leg is opposite your 4.64 degree angle, while the first leg is adjacent to it. The length of the first leg is very nearly equal to the diameter of the bottle (draw the figure in order to see this).
Alternatively, a vector from the entry point to the exit point coincides with the hypotenuse. The length of the hypotenuse is very nearly equal to the diameter of the bottle (and to the length of the first leg of the above triangle). If the axis is in the x direction, what therefore are the x and y components of this vector?
How far below the axis is the entry point?
How far below the axis, therefore, is the exit point?*@
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&#Please see my notes and, unless my notes indicate that revision is optional, submit a copy of this document with revisions and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).
Be sure to include the entire document, including my notes.
If my notes indicate that revision is optional, use your own judgement as to whether a revision will benefit you. &#
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