#$&*
course Phy 232
7/3 5:30 PM
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?
Radius= 5 cm
Distance=5/4 cm=1.25 cm
7 degrees
You don't say how you got 7 degrees. However that looks like about half of the angle of incidence.
2. For the index of refraction you obtained, what therefore will be the angle of refraction for this ray?
Index of refraction of water: 1.33
Index of refraction of air: 1.0003
Snell’s Law: n1sin(theta1)=n2sin(theta2)
(1.0003)sin(7)=(1.33)sin(theta2)
Theta2=5.27degrees
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?
I estimate the angle of incidence of that ray to be about 15 degrees, give or take a degree or two. You can find the accurate angle. The angle of incidence is the angle between the incoming ray and a radial line at the point where the ray enters the cylinder.
Using 15 degrees, the angle of refraction would be about 12 degrees. This is the angle of the refracted ray with the radial line. The radial line is at 15 degrees to the central ray, so the angle of the refracted ray with the central ray is (very approximately) 3 degrees.
If you follow the refracted ray to the point where it crosses the central ray, you form a right triangle with the refracted ray as the hypotenuse, and the shorter leg equal to 1/4 the radius of the cylinder. The angle opposite this leg is 3 degrees, and the sine of that angle is about .05. The hypotenuse therefore has length about 1/4 * radius / (.05) = 5 * radius.
You should find the accurate result.
The figures and the discussion in Class Notes #19 should be helpful.
The refracted light starts from a distance of 1.25 cm away from the lens (1/4 of the diameter) and is parallel to the central ray so it would meet 1.25 cm away from the lens.
4. How far from the 'front' of the lens did the sharpest image form?
The sharpest image formed when the light was 3.5 cm away from the bottle (or lens).
5. Should the answer to #3 be greater than, equal to or less than the answer to #4 and why?
I think that they should be about the same (although they aren’t) because this would be the ideal focal distance.
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?
The angle of incidence was the same as that on the front, 7 degrees and the angle of refraction is 5.27 degrees according to Snell’s Law.
7. At what angle with the central ray does the refracted ray therefore emerge from the 'back' of the lens? 5.27 degrees
8. How far from the 'back' of the lens will the refracted ray therefore be when it crosses the central ray? 1.25 cm
The back will be the same as the front and the distance will be 3 cm . It is 3 cm from the back of the lens where the refracted ray crosses the central ray.
"
&#Please see my notes and 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. &#
#$&*