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The image in the mirror is a virtual image. For a virtual image, the incoming light rays do not originate from a point from within the mirror as you can clearly see by drawing a ray diagram. However, imagine that you replace the mirror with a piece of glass and the 'image' in the mirror with real objects. Now the two set ups are different, but as far as the ...


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See, it's ironic, you giving us a two-dimensional image of what you are talking about, because for us of course all of those things are in the same plane, the plane of the screen that we are looking at it with. What is happening to you is that you are born into a set of biases which this science training that we get in physics is meant to try and suppress. ...


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The mirror creates a virtual image, which is a way of saying that the rays of light bounce off the object and off of the mirror in exactly the same paths they would if there was an actual object where the virtual image appears to be. Physically, they are indistinguishable. The brain completes this illusion because the physics of this virtual image is ...


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So comparing the image and object, they are perpendicular to each other. But our brain senses the image to be below object. Why? Because from your perspective, the light appears to be coming from point 2. But really this is a reflection. The light from point 2 is really coming from point 1.


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It depends on the medium. For a medium with scalar refractive index, the situation is as described by Roger Wood. Pure s-/p-polarization will keep their polarization. This is in some sense by construction, since they are chosen as the "eigenpolarizations" of the reflection response. Since s-/p-polarization typically have different reflection ...


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You're correct, the diagram would look something like this. A Snellen chart could be L distance from the mirror, with the observer sitting L distance from the mirror too. The virtual image would lie 2L from the observer. For two parallel plane mirrors you'll end up with a ray diagram that looks something like: Where $I_n$ are the $n^{\text{th}}$ virtual ...


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When an electromagnetic wave is reflected from a perfectly conducting mirror,* the electric field ${\bf E}$ is phase shifted by $180^{\circ}$. However, there is also the magnetic field ${\bf B}$, and the magnetic field is not phase shifted. There is a tendency to identify the wave with its electric field, and this is entirely natural, since in most ...


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For an ideal mirror it is shifted by 180 degrees. For imaging this means a shift of $\lambda/2$ along the propagation direction. This is not related to the mirroring effect, which is caused by the change of sign of $k_\perp$.


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This gets a little long to explain the situation. See the bottom for answer to your questions. Many textbooks do a bad job explaining the theory of blackbody radiation. In particular, often they assume perfectly reflecting cavity without explaining why they do so. It seems to be often just copied from textbook to textbook. I don't know who and where ...


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