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Anonymous users2024-02-12Why are the interference fringes produced by Newton's rings a set of concentric rings?
3 Where are the interference fringes produced by Newton's rings? What are the two rays of light that are coherent?
4 In the Newton's ring experiment, what is the problem if the radius of curvature of the convex surface of the flat lens is directly measured by the dark grain formula?
5 When using a reading microscope, how can I tell if parallax has been eliminated? What are the main precautions when using it?
6 In optics, there is a simple method to judge the concave and convex of the measured lens by using the principle of Newton ring generation: gently press the edge of the measured lens in the Newton ring device by hand, and observe the direction of the center of the interference fringe at the same time. Think about it, what is the truth of this?
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Anonymous users2024-02-113.Coherent light is the reflected light from the front and back surfaces of the ring.
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Anonymous users2024-02-10On the schematic diagram 1 of Newton's ring, the lower part is a plano-convex lens (flat crystal), a is a plano-convex lens with an o-center of curvature, and around the contact point in the middle of the two is the air gap between the flat glass and the convex lens. When parallel monochromatic light is incident perpendicular to the flat surface of a convex lens. The reflected light caused by the upper and lower surfaces of the air gap forms coherent light.
Light is reflected on the upper and lower surfaces of the air gap (one is reflected on the photophobic media surface, and the other is reflected on the photodense media surface).
The distance of the spatial film between the plano-convex lens and the planar lens is e, and the radius of curvature of the plano-convex lens is r.
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Anonymous users2024-02-091) The interference fringes are all concentric rings, and the more the edge of the fringes are denser 2) The difference is equal inclination interference: h is fixed, =0 is the ** fringe, and the optical path difference and interference order are the largest; When the radius of the ring increases, the corresponding increases, the d loss decreases stupidly, and the interference order decreases
Equal thickness interference: When the radius of the ring increases, the interference order and optical path difference are increasing 3) In the experimental area, if the method of balancing is different, the method of changing the h value can be used (reducing H by hand pressure, and increasing H on the contrary).
Isotilt interference: When h becomes smaller, the circumference shrinks toward the center
Equal thickness interference: when h transforms the rubber with the hour the ring expands outward
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Anonymous users2024-02-08The interference fringes of Newton's rings are produced by the interference of two rays of light sandwiched between the upper and lower surfaces of the air layer between the lens and the smooth plane.
These two beams of light are coherent lights of the same frequency and constant phase difference reflected by a single ray of light at two interfaces.
When the thickness of the air layer is an integer multiple of the half-wavelength of the light in the air, the vibration strengthens and bright streaks appear; When the thickness of the air layer is several times the odd number of times the wavelength of a quarter of the light in the air, the vibration weakens and dark streaks appear.
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Anonymous users2024-02-07Yes, the interference rings of transmitted light and reflected light are complementary, and the image can be determined according to the optical path difference formula, but the reason is that there is a half-wave loss in the reflected light, and there is no half-wave loss in the transmitted light, so that their optical path difference is a difference of 2 (that is, the phase difference is ). So it's complementary.
The interference fringes formed by the reflected light and the transmitted light are complementary, that is, the reflected interference fringes are the bright lines, and the corresponding projected interference fringes are dark lines, because the light is emitted from the photophobic medium to the light-dense medium, and the reflection occurs on the interface, which is a half-wave loss. Transmission does not.
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Anonymous users2024-02-06Yes, but it is more blurry, and its light and dark position is opposite to the light and dark position of the interference fringes formed by the reflected light.
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