what is the ratio of the intensity of light if their difference is one bell(unit of intensity)?
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I have never heard of bells (bels ) being used to measure light, only sound.\n\nIf two sounds differ by 1 bel, one is 10 times as loud as the other.
437996
Answer is D. Candela\n Candela is the luminous intensity, in the perpendicular direction, of a surface of 1/600 000 square meter of a full radiator at the temperature of freezing platinum under a pressure of 101325 newtons per square meter.  \n\nA is wrong. Angstrom is the unit of wavelength.\n\nB is wrong. Lux is unit of luminous of a surface area which receiving light from a source, emits it.\n\nC is wrong. Lumen is the light emitted per second within a unit solid angle by a uniform POINT source of light.  1 cd = 1 lm sr^ (-1)
750681
A photon with a wavelength of about 650 nm
400641
There's a difference.  It refers to (sin x)^2.  Plug in a few numbers and you can easily tell the difference.
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Dark is the absense of light.\n\nIllumination is measured in foot candles (in the US anyway) which is roughly equivalent to the amount of illumination created from a candle 1 foot away.
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Angle at which light is totally reflected. \n\nThis website has a good diagram.\n\nhttp://www.physicsclassroom.com/Class/refrn/U14L3c.html
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E is normally expressed as joules (kilogram-meters per second^2)\n\nSo the mass is in Kg, and the speed of light is in Meters/Sec
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http://math.ucr.edu/home/baez/physics/Relativity/SpeedOfLight/fast_light.html\n\nhave a look at this website
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u can find out by\n\n   1/wavelength  =  R(1/(n2)^2 - 1/(n1)^2}\n    and substituting n1=3,n2=2,R(rydberg constant)= 10973731.6\n    you get wavelength in meters then find the corresponding colour of light
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The energy contained within a single photon of light can be found using Planck's equations,\nE = hv\nWhere E is the energy, h is Planck's constant, and v is the frequency of the light.\n\nYou can see that the higher the frequency of light, the higher the energy per photon.\nExamining the spectrum of light one can see that green light (roughly a wavelength of 550 nm) has a higher frequency than orange light (roughly a wavelength of 650 nm).\nTherefore, green light photons contain more energy than orange light photons.\n\nPlanck's constant = h = 6.626 E-34 J s
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I'll let you do the calculation, but the equation you need is:\n\nE = hc/lambda \n\nwhere h is Planck's constant, c is the speed of light in m/s and lambda is the wavelength of the light (in meters). You can find the rest of the numbers in your text book, I'm sure...
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One lux is equal to one lumen per square metre, where 4π lumens is the total luminous flux of a light source of one candela of luminous intensity.\n\nThe difference between the lux and the lumen is that the lux takes into account the area over which the luminous flux is spread. 1000 lumens, concentrated into an area of one square metre, lights up that square metre with an illuminance of 1000 lux. The same 1000 lumens, spread out over ten square metres, produces a dimmer illuminance of only 100 lux.\n\nAchieving an illuminance of 500 lux might be possible in a home kitchen with a single fluorescent light fixture with an output of 12000 lumens. To light a factory floor with dozens of times the area of the kitchen would require dozens of such fixtures. Thus, lighting a larger area to the same level of lux requires a greater number of lumens.\nlumens/sq ft      10.76             lumen/sq metre \nlumens/sq ft      10.76391       lux (lx) \n\n\n\nMore at the link below:
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nu = c / lamda\n\nsolve for lamda - you want wavelength.\nc is the speed of light = 3 x 10^8 m/sec\nnu is your frequency = 3.2 e-2\n\nplug into the calculator: 9.375 e9 meters
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You can use the Rydberg formula, which relates the reciprocal of the wavelength of the photon absorbed or emitted to the squares of the reciprocals of the principal quantum numbers.\n\n1/wavelength = R(1/n_l^2 - 1/n_u^2)\n\nSo, for the listed values of n, \n\nA.  1/w = R(1/ 2^2 - 1/4^2) = 0.1875R\nB.  1/w = -0.75R\nC.  1/w = 0.21R\nD.  1/w = -0.04861R\n\nSince energy and frequency are proportional to 1/wavelength,\nlooks like the order is D, A, C, B.
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There are two transitions in the energy levels of mercury that are close to each other in energy difference, therefore the wavelength of the light from these two transitions are very close. See also "fine structure".
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It's 0.00002777778 inches, or approximately 705.55 nanometers.  It's an inch divided into 36000 parts, so it is pretty small.\n\nI believe the wavelength of red light falls in a range of 630 to 760 nanometers or approximately 1/36000 of an inch.
586314
ñ = n - iκ\n\nHere, n is the refractive index indicating the phase velocity as above, while κ is called the extinction coefficient, which indicates the amount of absorption loss when the electromagnetic wave propagates through the material. Both n and κ are dependent on the frequency (wavelength).