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Originally posted by RamnedI think mtthw gave all the answers required, and they were accurate. Regardless of the position of the object, the focal length is positive for a concave mirror and the object and image distance are positive if in front of the mirror. So for (A) and (B), the focal length and object distance are both positive. These are simply conventions, but it seems more important to communicate the meaning, such as "in front of the mirror" or "behind the mirror." Source: http://www.phys.ufl.edu/~phy3054/light/mirror/raydiag/handout1/Welcome.html
What I mean is - Positive or Negative for focal length and object distance...
Edit: You posted while I was writing! So we're already past this.
Originally posted by RamnedYes -- thickness of about 0.75 cm.
wave optics
[b]24. Would it be possible to place a nonreflective coating on an airplane to cancel radar waves of wavelength 3 cm?[/b]
I had to search around even to find a beginning point to understanding this problem (so I guess I'm already out of the running to be a master of physics). Crudely, a radar absorbent material whose thickness is 1/4 of the wavelength of the radar waves expected causes waves to be reflected back from the inner and outer surfaces of the material. I'm not sure about this, but 1/4 is probably the magic fraction, because the resulting waves that are reflected back would be 1/2 wavelength out of phase, not 1/4 wavelength. The waves that get reflected back from the inner surface would traverse the 1/4 wavelength distance twice before they got together again with the waves reflecting off the outer surface. So the answer would be a qualified "Yes"; the thickness would have to be about 0.75 cm. The answer is qualified because there are so many other factors, such as the variability of the radar frequency, the shape of the aircraft, the type of material, etc.
You're on the right track. What steps would you take to do this? I.E. what would you measure? What kind of material would you choose and based on what? Yes, you place a coating 1/4 of the 3 cm wavelength but there's another thing you need to do in the thickness of the coating: Hint - divide the .75 by the ________...
I'm not picking up on your hint, and my additional research on this isn't giving me any clues. In the crude form of wave cancellation (interference) that we're talking about, all we need to worry about are (1) that the material itself is non-reflective and (2) that there will be a reflection off the inside and outside of this layer. Anyone else? I'll have to give up if there are no more hints coming.
Originally posted by HolyTI think the blank is "refractive index". You want quarter of the wavelength inside the layer to get cancellation. This will be less than the wavelength in air.
I'm not picking up on your hint, and my additional research on this isn't giving me any clues. In the crude form of wave cancellation (interference) that we're talking about, all we need to worry about are (1) that the material itself is non-reflective and (2) that there will be a reflection off the inside and outside of this layer. Anyone else? I'll have to give up if there are no more hints coming.
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Originally posted by Ramned(A) According to Wikipedia, the standard value given for the near point is 0.25 m, so this person is farsighted.
Optical Instruments
[b]A patient has a near point of 1.25 m. (A) Is she near sighted or farsighted? (B) Should the corrective lenses be converging or diverging?[/b]
http://en.wikipedia.org/wiki/Near_point
(B) Corrective lenses for farsighted people are converging lenses (biconvex).