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8

A polariser is quite good at blocking EM radiation, but not perfect. No polariser is perfect, and does not block 100% of the radiation. This is typically specifed in terms of the extinction ratio. The best laboratory grade polarisers have extincition ratios of the order 100000:1 [1]. Such good polarisers are only possible over certain wavelengths, and for ...


5

The sun is an extended source. This means that it occupies a definite solid angle in the sky $\omega = 6.8\times 10^{-5} Sr$. To visualise this (not to scale), let say that the black area in the following diagram is the angular extend of the sun as seen from the surface of the Earth (ignore the other labels), What happens when we concentrate sunlight ...


4

Have a look at my answer to Slit screen and wave-particle duality because this covers a lot of topics relevant to your question. You're correct that if we imagine the photon as a little ball then if the arms of the interferometer are different lengths the two "halves" of the little ball cannot arrive at the detector at the same time. But this is not how the ...


4

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: ...


4

I don't know the "official" answer but here is what I might try. I am hoping that others will contribute to make this a "good" answer. First - we were not told whether the wavelength of the laser is transmitted at all by the blue foil; but since blue foil typically absorbs red light, and most laser pointers are red (I have a blue one but they are ...


4

Wikipedia has a good article on hybrid photovoltaic-thermal systems. As you proposed they consist of a solar cell with a thermal collector at the rear. Solar energy conversion is a fascinating topic from a thermodynamic perspective and has been summarised beautifully by the work of De Vos, The Thermodynamics of Solar Energy Conversion, ISBN: ...


3

I don't think it's very likely, but one other thing I can think of: when the sign before the $\omega$ is a minus then the wave represents a wave travelling to the "right" - positive x direction - and maybe your TA wants only waves travelling to the right. ^^ In case you don't know why the minus sign represents a wave travelling in the positive x direction: ...


2

Qualitatively, the thing that happens under water (when you wear a diving mask) looks like this: The green lines represent the path the light would have taken without the water, and therefore the "apparent size" of the bubble. But as you can see, the refraction of the light away from the normal (transitioning to a medium of lower refractive index) causes ...


2

You will see transmission peaks when integer multiples of the half-wavelength fit into the cavity length $L$. So, this means the condition (n is an integer): $L = n \lambda /2$, or: $\lambda = 2 L / n$. This is often expressed using "wavenumber" $1/\lambda$ (proportional to frequency), so that the spacing is even: $\frac{1}{\lambda} = \frac{n}{2 L}$. ...


2

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 ...


2

You may have lost marks for leaving out the imaginary unit $i$. If not, then understand that either $e^{\pm i\,\omega\,t}$ can be used to represent the real signal with positive frequency $\omega$ - it's wholly a question of convention. But once you have made the choice you must stick with it and the choice has implications throughout all the equations of ...


2

The author confused things in section 3.1. In macroscopic EM theory of radiation (radiometry), technically irradiance at a given point of a plane is defined as time average of the normal component of the Poynting vector (normal to the plane): $$ I = \overline{\mathbf S \cdot \mathbf n} $$ As you have shown, this is function of $|\mathbf E|^2$ only in some ...


2

You will still "see" the object but it will appear blurred in the shape of the so-called Airy diffraction pattern with a size larger than the actual object (dependent on the numerical aperture of the objective lens and the wavelength of the light used to observe it in the microscope). For VIS light and a 100x objective lens with an N.A. of 1.4 this is ...


2

When one says "photon" one is in the quantum mechanical frame. Quantum mechanics does not follow the rules of classical mechanics if one tries to consider the photon one classical entity, like a bag of energy flowing. The photon is a point like elementary particle in the standard model, it has no extent and when it hits the detector it registers at a ...


2

Polarization maintaining fibers are (as far as I know always) single-mode fibers. It is inherently difficult to efficiently couple light into single-mode fibers because, by definition, they will only support one specific optical mode which you have to carefully match with your input beam. So this mode matching will most likely be your main challenge. Getting ...


2

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 ...


1

Does a spherical wave-front, as you describe it, require at minimum 2 spherical bursts from a single atom? If spherical wave-fronts form as energy bursts that are independent in nature, with time greater than 0 (t>0) between each burst, and each burst being equal in energy, then in order to adhere to E=hv, might determining the frequency and energy of a ...


1

You can actually do this a bit more simply (or at least without integrations). Luminance is invariant in geometrical optics. That is, the brightness of an image cannot be brighter than the source. The radius of the sun is 0.6958 x 10^6 meters. The radius of the earth's orbit is a mean of 149.6 x 10^6. Then the brightness at the surface of the sun is the ...


1

The classic example of when the correct path should maximize the time is inside of a mirrored ellipse. There are four possible paths for a light ray which begins and ends at the center (shown below). Two of those paths are maxima and two are minima. The fact that the original statement of Fermat's principle does not account for this is probably what Hecht ...


1

Dunno what book you're quoting, but you should realize that the index of refraction of air is $n = 1+ \epsilon $ (where I'm using the mathematics standard of $\epsilon$ being a tiny number). Thus the power in air is $1/F$


1

The equation you are quoting gives the power of a lens in terms of its geometry and refractive index. Simply rearranging the terms (dividing by $n$) gives you an expression for $\frac{1}{f}$ which is known as the power of the lens and is expressed in diopters. For the usual situation of a lens in air, we can put $n=1$ which leaves you with an even simpler ...


1

It might be clearer if you read it as "detect THE single particle passing through the slits" . If you use a strong light then there are lots of photons passing through the slits and the interference could be completely classical with different photons going through each slit and interfering at the screen - just like water waves. This was the picture for ...


1

This system is actually a little more complicated that I first thought because the path length to both eyes must be the same to "clone" the light source.


1

We first consider the relation: $$n\delta{\lambda} = d\delta{\theta}\cos{\theta}$$ It's content is that the $n^{th}$ order maximum of a wavelength $\lambda + \delta{\lambda}$ is displaced from the corresponding maximum for a wavelength $\lambda$ by the angle $\delta{\theta}$, related to $\delta{\lambda}$ by the above equation. Now, we can ask the question, ...


1

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 $L$ ...


1

Using your diagram as a reference, the rays from the source are parallel to the (vertical) median of the prism (C). The surface(s) they are incident on are inclined by angles $a$ and $a'$ to this median. So the incident ray makes an angle of $a$ and $a'$ with the respective surfaces (note: not with the normal). From the law of reflection (equality of angles ...


1

there are other parameters like the number of free electrons in the atoms of the material, atomic size etc. Close. While density of particles does matter, it also depends on the material property. More precisely, it is closely related to how the electrons react when situated under electromagnetic oscillation. Each bound electrons has its natural ...


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Every point of a wave front may be considered the source of secondary wavelets that spread out in all directions with a speed equal to the speed of propagation of the waves.


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There are two separate processes going on. Radio waves provide a non-zero electric field throughout the gas. At any time in a gas, some small percentage of atoms are ionized due to thermal collisions. The resulting free electrons are accelerated by the electric field due to the radio waves. These accelerating electrons smash into other atoms, causing ...


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See, first of all our eyes are not a good device to determine the main color of a group of photons. the main color that we see is actually the intensity of a specific range of frequency in the light wave. it means there are all kind of photons coming out of the sun, but the amount of "Yellow" photons are much much more. that's because we see the sun ...



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