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What you are describing is the diffraction pattern from a single slit as described in How can a single slit diffraction produce an interference pattern? The intensity as a function of angle is given by: $$ I(\theta) = I_0 ~ \mathrm{sinc}^2 \left(\frac{d\pi\sin\theta}{\lambda}\right) $$ where $d$ is the slit width and $\lambda$ is the wavelength of the light. ...


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My guess is that the author of the textbook has the common misconception that there are only three colors—red, green, and blue—and yellow is red+green. Therefore, you're expected to answer that the red+yellow filters will let through red light, and the green+yellow filters will let through green light, and the others will let through nothing. That isn't ...


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The results will depend on the filter transmission bandwidth. For near-ideal filters with short bandwidth (transmitting only one color and absorbing/reflecting others), virtually no light will be transmitted in any of these cases. The order of the two filters doesn't matter. When using wide band filters, some light between the peak transmission wavelengths (...


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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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You are looking for a way to uniquely identify a particular diamond. Yes, this is commonly done. Because the facet sizes are slightly different on different cut diamonds, and because every diamond has its own unique imperfections, the pattern of light refracted and scattered by each cut diamond is unique. Some people have made holograms of the scatter ...


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jewelers use a handheld magnifier that can be used as a quick approximate refractometer with some practice; they can tell the difference between glass and diamond this way. Precision refractometers can measure the refractive index accurately enough to identify the composition of transparent materials. Conversely, if the refractive index of a given type of ...


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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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Speed is not the only property of light - there is also frequency, which measures how often the light's electric and magnetic fields oscillate in unit time. It is the frequency that is changing when light is redshifted.


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The following may be useful to consider. It is true that Huygen's principle states that every point on a wavefront acts as a source of secondary waves which travel in all possible directions. However, those individual waves are not seen as separate, but only as a superposition. In other words, say you are looking at a source that is very far away, it will ...


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For us to see an object, our eyes' receptors need to absorb photons coming (or scattering off of) from the object. To see something light must enter the eye and the rods (and cones) must be stimulated sufficiently for the signals to be produced for processing by the brain. Why can we only "see" reflected light? In vacuum, there is nothing for the ...


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For an object (the green arrow) standing in front of the mirror, all the rays that impinge on the mirror send back at the same angle at which they impinge on it. These reflected rays seem to be coming from an object behind the mirror. It only seems further away (if you look in the mirror you see yourself twice the distance as you are standing in front of the ...


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