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The sun makes very sharp shadows as the photons are essentially parallel. An LED lamp can also make shadows appear very sharp, a florescent lamp, not so much. Your experiment ignores the observational limitations (as pointed out by other answers). It's an important point even in theoretical physics.


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How about a simple test to see if the rings are due to thin film interference? I doubt it, as that implies very high quality manufacturing, with precise rings of very uniform and varying thicknesses, but a test is easy. Try using monochromatic light. Even just the green (or red) 'charging' LED light on a phone/laptop charger should change the pattern if ...


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Judging by the smoothness and regularity of the spectrum, this appears to be simple refraction through the plastic. Thin-film interference would generate a much less regular pattern. In this instance, the plastic disk is acting as a sort of prism. If you want a thin-film interference type of effect, you can look at the disk (from either side, if I recall ...


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Yes, this is precisely how several elements that aren't found on earth were synthesized. Look up the history of elements like Berklium and Californium, for example.


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You need to learn about the Eikonal equation and the equivalent ray path equation, which I talk about in my answer to the question Physics SE question "Ray tracing in a inhomogeneous media", and, if you need to know how it comes as the *slowly varying envelope approximation" from Maxwell's Equations, I talk about this in my answer to the question , "Optics: ...


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A partially reflective Dyson sphere is equivalent to asking what happens if we artificially increase the opacity of the photosphere - akin to covering the star with large starspots - because by reflecting energy back, you are limiting how much (net) flux can actually escape from the photosphere The global effects, depend on the structure of a star and ...


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An analysis would have to look at the effects over time. With my limited understanding of physics and intuition, I see the outer layers of the star reabsorbing the rays. Where the outer portions of the star until now have experienced large amounts of energy flowing in one direction, now has a net outward energy flow of maybe 1/2 to 1/10 of what it used to. ...


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Randall Munroe answered this question in this article Let me (try) to quote: How fast would you have to go in your car to run a red light claiming that it appeared green to you due to the Doppler Effect? —Yitzi Turniansky As expected, quoting and mathjaxing are two things that do not go together well, the rest of this post should be considered a ...


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Imagine you start right next to a star. As you move away from the star, the intensity of the light $I$ (in $W/m^2$) goes down depending on distance $r$ following an inverse-square law: $$I\ \alpha\ \frac{1}{r^2}$$ but the solid angle $\Omega$ (in $sr$) also decreases in the same proportion: $$\Omega\ \alpha\ \frac{1}{r^2}$$ Therefore since radiance $L$ ...


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This is because gravity doesn't exert a "force" on objects, gravity works by bending spacetime. Light doesn't tend to accelerate so it will take a geodesic path through spacetime; this means that it takes a "straight" path through curved space, which is what gravity is. The reason we experience gravity is because when not accelerated we will take a geodesic ...


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Light is quantized so the energy $E = n h \nu$, where $n$ is the number of photons in the optical pulse. (If you don't like to think in terms of pulses, then $P = n h \nu$ denotes the optical power of the beam, with $n$ here being the number of photons per unit time). The polarizing filter absorbs or reflects some of the photons in the light beam, which is ...


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Camera sensitivity is a common topic. In short they mimic the sensitivity of the human eye. Also your statement that we can see violet in pictures of violet objects is not accurate. It refers to violet as a color, not the 'violet wavelengths' of the spectrum. When we see violet objects we do not see 'violet wavelength' light as much as other wavelengths ...


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Yes, it all depends on the refractive index of glass at the wavelengths you are interested in. In the IR (infra red) I think ZnS is a material of choice for lenses and windows - but I know KBr can be used - Potassium Bromide - you have to be careful with KBr as it dissolves in water.... Above the UV it becomes difficult to use windows as all materials are ...


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Rainbows are caused by refraction into, total internal reflection , refraction away/ out of droplets, millions of water droplets on earth. ( They are visible at approx 47 degrees [sun -droplet-Eye or Camera] which we can find out from refractive index of water). No atmosphere on moon, so no rainbow can be seen there by an observer there or recorded by a ...


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Can we see a rainbow on the moon...? Usually no. A rainbow is formed by light refraction in water droplets. On earth we typically see a rainbow during rain while it's sunny. As there is no atmosphere on the moon, there will be no rain and thus no rainbows. However, if you were to spray small water droplets on the moon you might see a rainbow (but the ...


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NO, because moon has no atmosphere. And that's the reason also for that the moon's sky appears black.. Thanks :)


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You are correct in one thing: if an atom in an isotropic medium spontaneously emits a photon, it can do so in any direction at all, and the overall emission will be evenly spread over the unit sphere. However, lasers work using stimulated emission, which is slightly different: if an atom is excited, you can induce it to emit its energy by shining an initial ...


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when we mix colours such as paint together, the colours block out parts of the spectrum when they are mixed. this is called a subtractive colour system. Red, Yellow, and Blue; being primary colours in a subtractive colour system, create black when they are all mixed together because all parts of the spectrum are blocked out (absorbed) by the three colours. ...


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If a single red photon hits your telescope from the direction of a planet in the Andromeda Galaxy, then all you know that the planet emitted a red photon. Was it caused by a fire? A scattering of starlight through its atmosphere? An Andromedan with a laser pointer? A single photon of light from an unknown source has about as much information as a random ...


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Suppose you are using a CCD or a photographic plate to record your image. The interaction with the light occurs when the detector absorbs a photon, and this happens at a point. So the image is built up from a collection of points - one for each photon that is detected. In everyday life, e.g. taking pictures with the CCD in your phone, the intensity of the ...


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Starlight, as emitted by a star, comes in a wide range of colours. For instance see the picture below. Now this is a picture, and pictures can often be tricky with their representation of colour, so you'll have to take my word for it that Betelgeuse does look significantly redder to the naked eye than say Vega until you get a chance to go look yourself on ...


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The colour of stars as observed by an observer on Earth varies just like the colour of our own Sun, depending on where in the sky the source is relative to the observer. However, the light of stars is generally too faint to notice this as clearly with the naked eye, because we cannot perceive colour for weak light sources.


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No, Rayleigh scattering models the probability (and angle) of scattering as a function of wavelength and of the particle sizes. All wavelengths travel a long way but the path followed (scatter or nonscatter) varies. Since space is mostly "empty", there's little scattering. Beyond that, your understanding of stars is quite incomplete. THey do in fact have ...


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EV stands for exposure value. The equation to convert EV to lux is: $$ L = 2.5 \times 2^{EV} $$ Since you're using a Sekonic meter note that Sekonic provide a conversion chart here.


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Since your title starts with "Theoretically", I will give the theoretical answer: no. Intensity does not need to decrease. If you send polarized light into a slab of transparent material at the Brewster angle, then there will be NO reflection, and ALL the light will be transmitted. Theoretically, this means that you might get 100% transmission. In practice, ...


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Light frequency and wavelength are inversely proportional with a constant that is the speed of light (constant in vacuum). Both describe basically the same color within the spectrum, when light traverses a medium with a refractive index, its speed changes and affects the ratio of frequency to wavelength. What really matters is the energy carried by light as ...


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When light hits a barrier, even transparent ones, some light is reflected and some is refracted. This is often described by the transmission coefficient for that material, and at that wavelength. This can happen at the macroscopic barriers and at the smaller barriers between crystals or grains within a material. It is a simple property of waves which does ...


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You may wish to look at the Fresnel formulas (see "Fresnel Equations" Wiki page), which are derived from the Maxwell equations.


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Any energy principle is not being violated since the speed of the photon is never less than $c$ and hence the momentum is unchanging (in the classical sense). Why light travels slower than $c$ in a medium is because of the photons being absorbed and reradiated by atoms in the material. In a sense you can make the analogy of light traveling a longer path in ...


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You are absolutely correct that three-dimensional sections of space-time do not satisfy Euclidean geometry -- they are not flat. However, they are almost flat. On room-sized scales the curvature is very small indeed. I don't know the exact context where you read that "tridimensional space sections of space time continuum (whatever its number of dimensions) ...


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It sometimes occurs. It is called sonochromism. Here is my source.


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I think I get what you are asking. see you are seeing the Light it self. this works like this: Light consists of particles named photons. Atoms, all the time, are shooting photons out. But our eyes are only sensible to some of them. So it works like this, an objects atoms get to "excited state" and that is because they shoot out photons. When we observed ...


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Assuming I've understood your question correctly: The light from a bulb travels outwards in all directions and so hits (almost) all of the room. When it hits the walls etc, it gets reflected off of them (in most directions away from the wall), and then enters your eye. Hence your eye receives light from most of the room, so the room appears light. It's ...


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For example aluminum, even though in the Thorlabs mirror it has been coated to prolong the life of the mirror, the base is the reflective aluminum. Check also Refractiveindex.info for reflectance of materials.


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The LCD panel consists of elements shown in the figure below. The unpolarized light from backlight panel travels through polarizer, after which the light is linearly polarized. TFT panel controls the voltage on the liquid crystal, voltage applied will cause the liquid crystals to "twist" and thus rotate the polarization of the light. Light then passes the ...


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According to Schutz (chapter 11) for the Schwarzschild metric $dr/d\phi$ is given by: $$ \left(\frac{dr}{d\phi}\right)^2 = \frac{E^2}{L^2}r^4 - \left(1 - \frac{r_s}{r}\right)\left(\frac{r^4}{L^2} + r^2\right) $$ where: $$\begin{align} E &= \frac{p_0}{m} \\ L &= \frac{p_\phi}{m} \end{align}$$ Comparing this with the equation in the Wikipedia ...


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The blue shift is actually necessary to conserve energy. The photon has energy, and therefore has potential energy relative to the neutron star. It loses gravitational energy as it approaches the neutron star, and it gains energy in the form of blue shift. Indeed, to the first order in GR, the change in energy of a photon is equal to the change in ...


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Not all light bulbs are thermal emitters. Fluorescent lights do not use incandescence, hence they would not emit an equal spectrum to an incandescent source with an identical maximal light frequency. But in general yes objects do concurrently emit a whole spectrum of waves based on their temperature, regardless of whether their light is visible to us.


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Probably green, since lemons are common objects in daily life we tend to observe them in daylight conditions often. Based on their bright yellow color (relative to daylight brightness of e.g. white paper) they probably reflect a large part of the spectrum that gives a yellow color, i.e. from red(~600nm) to green(~540nm) wavelengths. This means under most ...


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Since both you and your teachers are stumped I will give some pointers. Pointer 1 - light is an electromagnetic wave. The energy flow is given by the Poynting vector. In vacuum (or air) this is $$\vec S = \vec E \times \vec H$$ Conveniently, for plane waves the time averaged Pointing vector is (see wiki ) $$\langle S \rangle = \frac12 \epsilon_0 c E^2$$ ...


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UV filters for photography filter out near-blue uv light. They were commonly used when film was still popular as many films were sensitive to these wavelengths and, without a filter, the result was a bluish tinge to the pictures. Digital sensors don't have this issue so uv filters are far less common nowadays. Hence when they are used with digital cameras, ...


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They certainly do exist, but you might have trouble finding one mounted for compatibility with a camera whose sensitivity is in the visible region. One example is Schott UG5.


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No you are getting it wrong! UV filter , Will "Filter" the Ultra violet light. it means no ultra violet can get trough it. it is very cheap and you can find a UV filter every where. almost every camera has a UV filter to prevent ultra violet light to get through the lens. and also there are filters that will pass only ultra violet lights, as they absorb ...


1

You're looking at solar cells for terrestrial operation. The main efficiency number is not Power_electric/Power_solar, but Power_electric/investment. Capturing the last few bits of blue light just isn't worth it. In space applications, the investment is dominated by the launch costs. Using a more exotic material to capture 1% more energy might shave a ...


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We would expect the screen to display bright images as dark (and vice versa) as one of the polarisers has changed 90degs


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The CIE standard colorimetric tables give $\bar x(\lambda)$, $\bar y(\lambda)$ and $\bar z(\lambda)$ as pure numbers, normalized so that $\int\bar x(\lambda)d\lambda$ is the same for the three components. In the end, it boils down to what units you expect your tristimulus signals $X$, $Y$ and $Z$ to have. For dimensionless $\bar x(\lambda)$, $$X=\int ...


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Due to the Structure of Glass, No Interference. To determine the thickness required to cause thin-film interference, both in light and in oil or soap bubbles, you rely on the following equations: $$2n_{film}d_{film} \cos{\theta} = m\lambda$$ $$2n_{film}d_{film} \cos{\theta} = (m-\frac{1}{2})\lambda$$ These being the equations for constructive and ...


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See,the philosophy of relativity is that all the observers will agree with the natural phenomena, or what is happening; like, if one observer sees two particles to collide or a lightning causing damage then all the observers will agree with the collision and lightning causing damage. But there will be disagreement in positioning and timing (and in interval) ...


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Special Relativity is based on two assumptions: any reference frame (viewpoint) is valid and light moves at the same speed in all reference frames. In the first picture picture, light is trapped between two mirrors. If the distance is 300,000m then it takes 2 seconds to go up and down. As you can see, when the mirrors moves (with the light) it makes a ...


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Perhaps part of the problem is that the statement $2 ( l - l') = n \lambda$ is not correct in general. It applies to the specific situation where both the incident wavefront and the refracted wavefront accumulates a path difference of $l - l'$ between each layer of your crystal. That is true if the diffraction causes a reflection back in the same direction ...



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