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Your assertion If something was not reflecting photons in any direction, you could then see through it. is not right; you haven't thought of the possibility where the light is absorbed and not re-emitted. An object that absorbs all incident light is not seethrough, but black. It would block one's view of objects behind it, and therefore be very ...


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Light allows us to see an object in several ways. You've named some but not others. If the photons are absorbed, then we can't see the object. You got that right. But suppose they aren't absorbed, then what? Suppose the photons travel straight through the object unaffected. Then we wouldn't see it either. For example, a clean sheet of glass and ordinary ...


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Is there actually any evidence that being in water increases your risk of getting sunburn? I suspect that this inference confuses correlation with cause. On brilliant sunny days you are more likely to strip off and get in the pool or go to the beach to cool down. If swimming or diving you are unlikely to wear a sunhat. More of your skin (especially the ...


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Yes. It's is not so much the water is the beach sand reflecting light back to you like a parabolic mirror. The droplets of water on your skin can form more surface area to catch light creating a magnifying effect focusing light on your skin as well. The random texture in the beach sand will also give you even tan. Most sand is white in color even if not the ...


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Most people believe that you need mass to transmit a force, or even to be able make one, but the more fundamental concept happens not the force but the momentum. The momentum is the capacity to interact with another entity and transfer it some change in speed. In the case of the photon, it is massless but however has momentum and energy. The energy and ...


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This is newton second law: $$ \mathbf F = \frac{\Delta\mathbf p}{\Delta t} $$ As you can see, its variation of momentum that brings force. So, its the transfer of momentum. Granted, photons are massless, but they do have momentum. There are two ways of a photon transfer momentum to a solar sail: The photon is absorbed (also heats up the sail), or, the ...


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dL/dx means the rate at which the length (L) is changing as the position of the point of reflection (x) changes. To apply Fermat's principle you want to minimise L (Since the speed of light is constant, to minimise time you must minimise distance.) Now imagine drawing a graph. Put x on the horizontal axis and L on the vertical axis. For any value of x you ...


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Let's first of all negelect the differences in absorption between the miror and the red brick wall. The main difference between the brick wall and a mirror is the flatness of the surface. The morphology of the surface will determine whether you will observe specular or diffuse reflection. I've depicted both situations in the figure below. You can easily ...


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The translucent sheets and optical fibers are being lit from below. Light enters one end with a relative small angle incidence. Once inside the sheet/fiber the light experiences total internal reflections multiple times because it hits the sides of the sheet/fiber with a large angle incidence. It's not until the light gets to the far end of the sheet/...


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According to huygens principle each point in the wave front acts as the source of secondary wavelet.By the time, the secondary wavelets from B travel a distance BC, the secondary wavelets from A on the reflecting surface would travel the same distance BC after reflection.Tking A as centre and BC as radius an arc is drawn.From C a tangent CD is drawn to ...


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As John stated, the relation is not constant. Where does the need to "show that $\frac{i}{r}$ is constant come from? You know that if $i=0\rightarrow r=0$. You know that $$n_i\sin{i}=n_r\sin{r}$$ From this you should be able to see clearly that there's no linear relation except when $i$ approaches $0$, or $n_r$ approaches $n_i$. You should never go into ...


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Light from the focus, when reflected to a parabolic mirror, will all be reflected in parallel rays.


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The best intuitive explanation I can come up with is that the Fresnel equations, which perfectly describe the reflexion of light from a plane, optical quality interface, are an expression of the requirement that the field vectors should be continuous across the interface. This requirement is itself an expression of the absence of either static or flowing ...


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Total Internal Reflection is an example of reflection. In TIR and other forms of reflection (e.g. reflection off of a mirror or other barrier) the angle of incidence will be equal to the angle or reflection. You wrote "TIR reflects with the angle of incidence=angle of refraction." I'm not sure if this is a typo or if this is what you intended but "...



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