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course Phy 232
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?
Cylinder radius = 5.08 cm
Distance = (1/4)*(5.08) = 1.27 cm
The sin(angle of incidence) = dx / R. Therefore, the change in x is 1.27 cm and the radius is 5.08 cm. Therefore, the angle of incidence is 14.48 degrees.
2. For the index of refraction you obtained, what therefore will be the angle of refraction for this ray?
Sin(angle of incidence) / sin(angle of refraction) = n2 / n1. N2 is the index of refraction of water, which is 1.33 and n1 is the refraction of air, which is 1.0003. Setting up this equation including the calculated angle of incidence, we can solve for the angle of refraction. Sin(14.48 degrees) / sin(refraction angle) = 1.33 / 1.0003. Sin(angle of refraction) = .188 which gives us an angle of refraction of 10.84 degrees.
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?
To find the distance from the lens to where the two rays meet can be solved using the angle of refraction and triangles. Knowing the angle of refraction and knowing how to reflect that angle, we can use the formula set up tan(angle of refraction) = change in x / distance. Plugging in the numbers, we get the equation tan(10.84 degrees) = 1.27 cm / distance. Therefore, the distance we are looking for is 6.63 cm.
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10.84 degrees is not the angle between the refracted ray and the central axis, but rather the angle between the refracted ray and the radial line from the center of the lens to the point where the ray enters.
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4. How far from the 'front' of the lens did the sharpest image form?
6.25 cm.
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At this distance the image would be inside the lens.
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5. Should the answer to #3 be greater than, equal to or less than the answer to #4 and why?
The numbers should be equivalent but because of uncertainties such as human error on measuring the distance with the measuring device and the rounding error potential in the calculations.
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?
To get the distance from the back of the lens where the rays meet would require subtracting the thickness of the lens. This would result in a distance from where the rays meet the back of the lens of 6.63 cm - .64 cm = 5.99 cm. The angle of incidence, according the Snell’s Law, is the same in the front as in the back of the lens, 14.48 degrees. The angle of refraction is also the same in the back as in the front. The angle of refraction is 10.68 degrees.
7. At what angle with the central ray does the refracted ray therefore emerge from the 'back' of the lens?
The angle of refraction is also the same in the back as in the front. The angle of refraction is 10.68 degrees. The angle remains the same in front of and before the lens.
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The back of the lens is not parallel to the front, so the angle of incidence will not be the same at the back as at the front.
The angle of incidence will be the angle between the ray incident on the back of the lens, and the radial line from the center of the lens to the point where the ray strikes the back of the lens.
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8. How far from the 'back' of the lens will the refracted ray therefore be when it crosses the central ray?
The distance from where the rays meet to the back of the lens is 5.99 cm. This comes from calculating the distance from the front of the lens and subtracting the thickness of the lens.
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You're doing some good things here, but you're missing some details so revisions
are necessary.
&#Please see my notes and submit a copy of this document with revisions, comments 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.
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