# Tag Info

124

Your iPhone is a pretty good grating. I just did a simple experiment with an iPhone, a green laser pointer and a sheet of graph paper. This was the result: The display of the iPhone 6 has a resolution of 326 ppi - meaning we have a "grating spacing" of 25.4/326=0.0779 mm. Different models have different resolutions - make sure you find out what your phone ...

115

Something special about the visible range is that water has low absorption in this range. It’s a rather sharp dip near the visible region. Since we know that life began in water, the beings that were receptive to these wavelengths had a significant advantage over the others. Thus natural selection would have favoured these life forms over the others. This ...

84

This is a really interesting question. It turns out that your body is reasonably conductive (think salt water, more on that in the answer to this question), and that it can couple to RF sources capacitively. Referring to the Wikipedia article on keyless entry systems; they typically operate at an RF frequency of $315\text{ MHz}$, the wavelength of which is ...

61

Some areas of physics are counter-intuitive. For them, your everyday experience is a poor guide to how the universe really works. This is one of those areas. Photons have no mass. They all have the same speed. Yet they have energy and momentum, and it isn't the same for all photons. If you are used to $p = mv$, this doesn't make sense. The explanation is ...

55

Why will a blue ray bend lesser than a red ray through a slit of the size a little bigger than the wavelength of the blue ray? Don't think of bending. Think of diffraction like this: if you have a plane wave incident on a slit, then you can think about the space in the slit as being a line of infinitely many point sources that radiate in phase. If you are ...

48

Most computer monitors aren't capable of displaying any spectral color. Some of the RGB monitors could display at most three of them: some red wavelength, some green and some blue. This is because the gamut of the human vision is not triangular, instead it's curved and resembles a horseshoe: In the image above, the black curve represents the spectral colors,...

46

Why not calculate it? Consider a string of length $L$, with its ends fixed at $x=\pm\frac{L}{2}$. Let's assume for convenience that at time $t=0$ the string is "plucked" at $x = 0$, so that the string displacement relative to its equilibrium position is given by $$f(x)=A\left|1-\frac{2x}{L}\right|.$$ The standing wave solutions to the wave equation ...

37

As promised in the comments to my answer, I went out and measured the effect in a number of different configurations (a couple of days later than promised :-)). For those of you who just want the conclusions, here they are: The remote seems to work better when held to the head though the improvement isn't as marked as one might have expected from a google ...

37

In liquids and solids the difference in energy between energy levels becomes very small, due to the electron clouds of several atoms bein in very close proximity of one another. These similar energy levels will form 'bands' of indistinguishable spectral lines. In gases however, atoms will be spaced loosely enough such that the interaction between atoms ...

36

Yes, but not with equal amounts of each. In order to answer this, we need to understand the CIE 1931 color space, and think about its algebraic properties. Essentially what the CIE specification says is that, while light comes to us as a spectrum filled with varying amounts of photons in the wavelength range 380-700nm, our eyes are engineered in such a way ...

35

The range of visible light wavelengths has a special property that makes it the commonly used range for all life forms on the Earth: It is the range of electromagnetic wavelengths that are short enough to be conveniently handled by cell sized detectors and that can pass through the atmosphere. The Earth's atmosphere is not transparent at all wavelengths, ...

32

Color is a double valued variable.For physics there is a one to one correspondence between frequency of light and the color assigned to visible frequencies. As far as the spectrum of colors (rainbow) ultraviolet frequencies are invisible to our eye. The eye is a biological entity, the retina of the eye has color receptors, and these receptors do see the ...

30

What Vasily Mitch says is true (+1). But some objects are colorful because of interactions that take place over a larger region than a single atom. Metals reflect light because electrons spread out through the metal. They can easily move, which makes them conductive. Classically, the oscillating electric field in light vibrates the electrons, and vibrating ...

28

There are two different mechanisms at work here. It's not the case that humans are "ultraviolet colorblind" or something like that. 1) There is the spectrum that the flower petal reflects or absorbs. This spectrum is continuous and includes ultraviolet and everything at lower wavelengths, visible light, and infrared and everything at higher wavelengths. 2) ...

28

It of course depends on what you define as colour. If it is defined as the change in the visible spectrum of a light, then a single atom can definitely absorb a photon of a preferable wavelength and thus slightly change the spectrum of the passing light. Many atoms have excitation energies falling into the visible spectrum when atoms absorb the photons, ...

25

Yes, there are an uncountable infinity of possible wavelengths of light. In general the frequency spectrum for Electromagnetic (e.g light, radio, etc) is continuous and thus between any two frequencies there are an uncountable infinity of possible frequencies (just as there are an uncountable number of numbers between 1 and 2). Two things to consider in ...

25

The answer is actually very dependent on how you pluck the string. If you pluck it closer to the center, you put more energy into the lower modes. Pluck it near either end, and you have more higher harmonics. And then there's the overtone techniques, which intentionally squelch lower harmonics, leaving only higher harmonics.

23

It can be a little confusing because there are two conventions. The modern convention is to distinguish x-rays from gamma rays by how they are produced. X-rays are produced by electron energy transitions, typically inner orbital transitions, whereas gamma rays are produced by electromagnetic transitions in the nucleus. Usually, gamma rays have shorter ...

22

What you have there isn't actually de Broglie's equation for wavelength. The equation you should be using is $$\lambda = \frac{h}{p}$$ And although photons have zero mass, they do have nonzero momentum $p = E/c$. So the wavelength relation works for photons too, you just have to use their momentum. As a side effect you can derive that $\lambda = hc/E$ for ...

22

is there anything special about visible light other than the fact that we use it to see colors? We can see light with wavelengths from $390$ to $650$ nm because in our eyes we have photoreceptor cells which are sensitive only for these wavelengths. If the photoreceptor cells were sensitive to other wavelengths, then we would be able to see those. Does ...

22

Snell's law tells us that the angle of refraction depends on the index of refraction, $n_1 \sin{\alpha_1} = n_2 \sin{\alpha_2}$. However, the question remains, why $n_{\text{blue}} > n_{\text{red}}$. In order to address this, we need a model for the refractive index. The refractive index $n$ of a material is related to the atomic transitions of the ...

22

Wien's displacement is qualitatively quite easy to understand. Consider a black body with temperature $T$. Its atoms are moving around chaotically with an average kinetic energy of $$\bar{E}_\text{atom}\approx kT \tag{1}$$ where $k$ is Boltzmann's constant. On the other hand, you have the black-body radiation. Because the radiation is in thermal equilibrium ...

21

It is an ångström, a unit of length commonly used in chemistry to measure things like atomic radii and bond lengths. Although not an official SI unit, it has a simple relationship to the metric units of length: $$1\:\mathrm{ångström} = 1\:\mathrm{Å} = 10^{−10}\:\mathrm{m} = 0.1\:\mathrm{nm} = 100\:\mathrm{pm}.$$

21

You see line spectra usually only in gases because there the interaction between the atoms can be neglected. In gases with high pressures you get the so-called collision broadening of the lines which eventually become bands. Similarly, in liquids and solid the atoms are so close that the interaction between them leads to the discrete spectral lines becoming ...

20

I will assume you familiar with the properties of waves such as interference and diffraction. Consider an electron orbiting the nucleus. By de Broglie's hypothesis, we would consider it to be a wave orbiting around the nucleus. Now, once the electron wave orbits once, the second time it would interfere with the first wave. For the system to be stable, that ...

19

Formally there are an infinite number of different wavelenghts. However, any given physical system can only be found in a finite number of distinct physical states. To create a light source with a wavelength $\lambda$ that is well defined up to some resolution $\delta\lambda$, requires observing it within a system of size of the order of $\lambda^2/\delta\... 19 Every resonator amplifies just certain frequencies while it inhibits all others. This is true only for very simple resonators. The shape of the guitar body is such that it has a different size at different angles. This corresponds to different resonant frequencies. In addition, the top has a supporting bracing that is very different on different models and ... 19 A sine wave doesn't necessarily have an intrinsic "starting point", you usually can draw its curve starting at any phase and call the corresponding point the beginning of the cycle: Quoting Wikipedia: Wavelength λ, can be measured between any two corresponding points on a waveform and Wavelength of a sine wave, λ, can be measured between any two ... 19 Like electromagnetic waves, gravitational waves can in principle have any wavelength. They obey the usual relationship$v = \lambda f$between the wave's speed$v$, wavelength$\lambda$, and frequency$f$. Einstein's theory predicts (and recent observations have confirmed) that the speed of gravitational waves is$v = c\$, regardless of their frequency. ...

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