# Tag Info

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Observations can be made using different parts of the electromagnetic spectrum. Many of them not within the range of the spectrum that can be made with optical telescopes. Other messengers than photons (e.g gravitational waves, neutrinos,...) can also be observed. Optical counterparts are observations made in the spectrum observable using optical ...

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The CCD has no way of recording the direction, the point in the sky, from which a photon is coming. Say you point your mirror-less telescope at the Moon. Every point on the moon's surface would be reflecting photons onto every part of the CCD at the same time. You've just created an expensive, sensitive, ambient light meter. There would be no image ...

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"ugriz" is short for U-band, G-band, R-band, I-band, and Z-band, detailed on the Wikipedia article on photometric systems: $$\begin{array}{|c|c|} \hline \text{Band}&\text{Effective Wavelength Midpoint }(\lambda_{\text{eff}})\\ \hline \text{U} & \text{365 nm}\\ \hline \text{G} & \sim\text{475 nm}\\ \hline \text{R} & \text{658 nm}\\ \hline \... 9 To answer your question, we need to first show the job each mirror is doing. First up, the Newtonian (lovingly called the "Newt", and invented by Sir Ike Newton): https://en.m.wikipedia.org/wiki/Reflecting_telescope#/media/File%3ANewtonian_telescope2.svg Two mirrors in this design, not surprisingly labeled as primary and secondary. The job of the primary ... 6 TL;DR: There was apparent 11% increase of Neptune brightness during 1980 and 2000. This could be due to multiple reasons. Recent observation suggested the reason to be change in the amount and brightness of the banded cloud mostly in the planet's southern hemisphere due to seasonal variations and thus affect overall brightness of the planet. Neptune's ... 6 The magnitudes don't sum and subtract that way: consider two 0.00 magnitude stars closely together. Is their combined magnitude 0.00? Actually, it is -0.75. This paper might help you with the derivation of the formula for addition of magnitudes, but you are interested in subtraction. You just need to rearrange the formula in the paper to$$m_a=-2.5\log(10^{-...

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Since neither the word "phase" nor "interference" is mentioned in any other answer here, I'll approach it from that direction. In this answer I said In an imaging optical telescope (or any imaging system including eyes) every pixel is illuminated simultaneously and directly by all areas of the aperture. From a given point in the distance a telescope will ...

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Your feeling is right: You shouldn't convolve the spectrum and the filter, you should only multiply so that flux outside the bandpass is suppressed. Subsequently you integrate the resulting function over wavelength, so that flux density (in energy/time/area/wavelength) becomes flux (in energy/time/area). Simply setting the flux to 0 outside $\lambda_1$ and $... 5 I don't know about the telescope, but I know that you will need the telescope that has CCD technology to be able to convert from counts/second (as brightness) that CCD detects to quantitative values easily. For the observational procedure, here is the standard regardless of what kind telescope you use: Take a picture of your object (let's name it ImObj). ... 5 In general, the CCDs used to capture images do not register the energy (therefore, colour) of the incident photons on them - they just count the number of photons observed by each pixel (or a value proportional to the number of photons, as they are not 100% efficient). So, they essentially just show overall brightness variations across the image. If you want ... 5 The optical pulsations of the Crab pulsar have been studied closely since 1969. The observations are actually not that difficult (I did some myself with a photoelectric photometer as a student) and have been achieved with a variety of technologies. A paper by Fordham et al. (2002) slices and dices the Crab pulsar's pulse shape into fine time and spectral ... 5 What matters is how many photons you collect versus how many you would expect to see if the object wasn't there. Photons would be present, without a source, for a variety of astrophysical (e.g. diffuse background) or non-astrophysical (night sky brightness, dark current) reasons, so you need to be able to rule out the null hypothesis that what you have seen ... 4 The quantity you want is basically the extinction law, and is usually called$k(\lambda)$. An extinction law is a fit to several measurements of the extinction$A_\lambda\$ in some direction (or an average of several directions). Cardelli et al. (1989) provides different functional forms for the mean extinction law, parametrized in their Eq. 1 as $$\frac{A_\... 4 I think there is a missing piece of information. The BAT is a coded mask telescope. The imaging is done by photons passing through a mask and falling onto an array of 32768 detectors. http://swift.gsfc.nasa.gov/about_swift/bat_desc.html The "mask-weighted" light curve is produced after a complex ray tracing exercise using an estimate of the position of the ... 4 The brightness of a diffuse area is it's luminance, and can be quoted in units of cd\,m^{-2}. Measuring it would require knowledge of the properties of the camera: the sensitivity (at various wavelengths) of the ccd, the aperture, length of exposure and so on. This could probably be found using calibration images, in which a known luminance is imaged and ... 4 Well "if no known astrophysical model can explain it" then nobody told Wright & Sigurdsson (2016) who, cognisant of Montet & Simon's results, explore a number of astrophysical models. They conclude by saying that the most "plausible model" is that of small scale intervening material between us and the star that may be responsible for the short-term ... 4 As WDC pointed out in his comment, without filters, you simply get a recording of the received irradiance as a function of the sensor's spectral response function. In other words, a normal CCD that detects the light in a camera isn't capable of picking up every wavelength of light perfectly and the response function tells you how good that CCD is at picking ... 4 The main reason is that the intrinsic spectra of galaxies are complex and therefore a redshift of their spectrum, whilst leading to a redder spectrum overall, does not necessarily lead to reddening in all colours. For instance if there is apeak in the intrinsic spectrum, then as that peak moves redward, then colours formed from bands on the same side or ... 4 Actually, the very first binary asteroid was discovered by stellar occultation in 1980. The evidence was rejected by the IAU, and wikipedia credits the first binary asteroid discovery to the Galileo spacecraft finding the moon of 243 Ida in 1993. The first moon of the well known asteroid 216 Kleopatra was observed through stellar occultation in 1980. From ... 4 To calculate a "luminosity light curve" from a time series of V-band photoetry, you need two things. You need to know the distance. The distance to Betelgeuse is uncertain and that means the absolute value of the luminosity you get will also be shifted systematically up or down by whatever distance you adopt. The V-band only contains a small ... 3 Surface brightness of galaxies and nebulae is typically expressed as visual magnitude per square arcsecond. Skyglow is often quantified in the same way. Given the angular dimensions of the frame and the magnitudes of a few stars in it, photometry software could compute sky brightness by excluding all visible stars and dividing the total brightness of the ... 3 I do not know whether complimentary observations of the light curve are being done, but I will try to answer the rest of the question. Is it easy to observe the light curve with ground-based telescopes? No, it's quite difficult. From the Kepler homepage: Since transits only last a fraction of a day, Kepler must monitor all target stars continuously. ... 3 I actually found a concept of 2D easy-to-scale telescope some time ago (here's the link). I guess we will slowly abandon refractor telescopes, for, as I understand, we are pushing them to their limits right now and it is getting really hard to make a bigger one (because of how hard it is to make a sufficiently sized mirror of the quality needed). BUT it ... 3 www.skymaponline.net might be the one you want. Main goal when taking images in terms of timing and exposition: 1. Choose time with good sky condition (e.g., clear sky, no wind) 2. Expose long enough for good S/N ratio, but not too long that CCD is saturated. 3. If you want to measure the length of the eclipse, you can follow since it is coming in until ... 3 The AAVSO eclipsing binary section is a good place to start reading. Their how-to articles address eclipsing binary specific issues such as predicting times of minimum. To get a list of observable Algol-type stars, try their observation planner tool with variable type EA. When you've selected a target, you can use their variable star plotter to make finder ... 3 Your question is a bit unclear. Compare in what sense? The problem boils down to that you have a fixed number of photons in a wavelength band and you can choose, with the appropriate filters or dispersion elements, how many "bins" to divide these photons into. To accumulate the same total number of photons with the same telescope will take the same amount ... 3 Unless you know all the stars in the diagram suffer the same extinction, then you should de-redden them individually. If you do know they have the same extinction (e.g. they are all nearby stars and it is zero; or they are all stars well above the Galactic plane in a similar direction; or stars in a cluster at the same distance) then you can plot a diagram ... 3 They are mostly empirical. Found by measuring the B-V for stars of known T_{\rm eff} (which are in turn measured by knowing the luminosity and radius of a star, and this is only known for a small number of stars). The relationships also depend on stellar surface gravity and composition. An alternative approach is to derive "synthetic" relationships by ... 3 Finding the best-fitting isochrone, a.k.a. isochrone fitting, is a standard approach to determine the age of globular clusters. This problem can be solved with a least-square method, where the data to be fitted are the points on the color-magnitude diagram and the fitting curve is the isochrone. Since there are many ways to apply this method, I will first ... 3 The absolute magnitude of an object is defined as the brightness of the object observed at a distance of d = 10\,\mathrm{pc}. With this distance, you can convert the luminosity density L_\nu in \mathrm{erg}\,\mathrm{s}^{-1}\,\mathrm{Hz}^{-1} to a flux density f_\nu in \mathrm{erg}\,\mathrm{s}^{-1}\,\mathrm{cm}^{-2}\,\mathrm{Hz}^{-1}:$$ f_\nu = \...

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