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1

That's a good question. Without realising it you have stumbled across the Huygens-Fresnel principle. The starting point it that a single silver atom is far smaller than the wavelength of light, so any scattering from it will be isotropic i.e. it will scatter the light equally in all directions. This type of scattering is known as Rayleigh scattering. But ...


7

I'm imagining a box made completely out of anti-matter so that your situation is realistic. Positrons are the antiparticle of the electron (i.e., the anti-matter equivalent). Meaning, in this case, they're identical to electrons except for charge. Photons, though, have no charge. So don't give a hoot whether a charged particle it's interacting with is ...


0

You'll find a number of images on the internet such as the one below that claim to be a 360° rainbow. Most of these supposed rainbows are not rainbows. They are instead "glories" (wikipedia article). The image below is from that wikipedia article. Rainbows are much bigger than glories. A rainbow, like a glory, naturally is a 360° optical effect. You don't ...


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Take a hose that sprays water in fine droplets and direct it towards a bright object like the sun or a lamp... instant 360 degree rainbow!


2

For the conditions to see a part of a rainbow, I would say you need the following: sufficient moisture to form numerous micro spheres of water, which creates the refraction pattern the right angle between the observer and the sun sufficient mass-thickness with this angle and with this composition to build up a visible amount of color Since the sun is ...


2

Its all upto the relative position of the observer and the sun. The imaginary line connecting the centre of the rainbow and eye of the observer called the line of vision makes the apparent struture of a rainbow. If the line of vision is a straight horizontal line with respect to the surface of the earth. You will definitely see a 360 degree rainbow. I ...


0

Most commercial spectrometers use gratings rather than prisms, but that's a side issue. To answer your main question: the optics inside a spectrometer re-image the entrance slit onto the detector elements. In the case of your 1x128 detector, the entrance "slit" is really a small hole which is optically matched to a single pixel in the image plane. The ...


1

Simply because two light rays intersect at a point it does not mean that an image is formed. You need millions (not necessarily, but a lot) of light rays to intersect at a point to form an image. The reason is that the intensity of light emerging from a two-ray intersection is too less for any human eye to detect. For an image formed due to a concave ...


1

Assuming you want to notice that there is a mirror on the moon, the question comes down to the resolving power of the human eye. Let's assume a healthy person with 20/20 vision. When looking at the moon, their pupil is dilated - say 6 mm diameter. The angular resolution of such a pupil is given by $$\alpha = 1.22 \frac{\lambda}{d} = 1.22 \frac{500\cdot ...


1

The prism doesn't reflect light, it refracts the light. Different wavelengths will be refracted (bent) by different amounts. In the visible spectrum, this will produce the familiar rainbow pattern we're all familiar with. If you direct the refracted light at a linear detector, each pixel of the detector will measure the optical power at a different ...


0

yes, your friend is right you will see the walls of periscope. here's why? . there are infinite number of rays reflecting from first mirror making all possible angles. you will see walls by those light rays which strikes with angle other than 90 degree, and obviously you will not see the walls by rays reflecting at 90 degree. only rays other than 90 degree ...


2

The answer depends on how the periscope is made (the relative size of the mirrors, the shape of the box , and where do you put your eye. You do not need a periscope to see this, just look through a pipe. The mirrors do not matter they only redirect the light, so your problem is equivalent to that of making holes of the size of the mirrors at both ends of a ...


1

Your friend is right, there will be lots of rays of light entering the periscope, a few of which will reflect directly to your eye. Most of the light will bounce off the inside of the periscope and allow you to see the walls of the device. Best thing to do is to try it out!


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I guess this would be more of philosophical question rather than scientific. Mirror would take color of any object being reflected on it so it would have variable color. If you remove the reflectivity from the mirror to see the color of the material, then it is not a mirror anymore.


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In theory there is no in-principle reason why one cannot have a near to perfectly reflecting ring resonator as you wish, simply by making the ring of bigger and bigger radius, so that a repeated reflexion around the edge of a ring keeps the light confined to a circular path. But you will always lose some light, even if the resonator's material is perfectly ...


2

Detecting reflected light from a "past Earth" is an interesting thought. Even a spectrum might tell you something about the composition, temperature of the atmosphere etc, even if as been correctly said in other answers/comments, the spatial resolution to actually image the Earth would need to be several orders of magnitude better than the Hubble Space ...


1

As you know the standard textbook answer to this question would be, for total internal reflection (TIR) the angle of incidence, $\theta$, must be greater than $\theta_c$, which is of course given by $sin^{-1} ({1 \over n}$) so for TIR at $\theta=0$ we need to have $n= \infty$. In the case of a dielectric mirror, this is a multilayered device a combination ...


0

Two potential reasons among others: the wind is not uniform, if it blows larger on one sector it will move the water surface stronger. Second, different smal surface waves that come from different directions can interfere either negatively or positively, depending on their relative phases, over some small region, which again wouls result in inhomoheneities ...



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