|MadSci Network: Physics|
Hi, First of all, you should check the equations, that you used them to convert the luminance to the light intensity. After that you can continue; Several measures of light are commonly known as light intensity: --Radiant intensity is a radiometric quantity, measured in watts per steradian (W/sr). --Luminous intensity is a photometric quantity, measured in lumens per steradian (lm/sr), or candela (cd). --Irradiance is a radiometric quantity, measured in watts per meter squared (W/m2). The equivalent quantity in other branches of physics is intensity. --Radiance is commonly called "intensity" in astronomy and astrophysics. Luminance is a photometric measure of the density of luminous intensity in a given direction. It describes the amount of light that passes through or is emitted from a particular area, and falls within a given solid angle. The SI unit for luminance is candela per square metre (cd/m2). The CGS unit of luminance is the stilb, which is equal to one candela per square centimetre or 10 kcd/m2. Luminance is often used to characterize emission or reflection from flat, diffuse surfaces. The luminance indicates how much luminous power will be perceived by an eye looking at the surface from a particular angle of view. Luminance is thus an indicator of how bright the surface will appear. In this case, the solid angle of interest is the solid angle subtended by the eye's pupil. Luminance is used in the video industry to characterize the brightness of displays. In this industry, one candela per square metre is commonly called a "nit". A typical computer display emits between 50 and 300 nits. Luminance is invariant in geometric optics. This means that for an ideal optical system, the luminance at the output is the same as the input luminance. For real, passive, optical systems, the output luminance is at most equal to the input. As an example, if you form a demagnified image with a lens, the luminous power is concentrated into a smaller area, meaning that the illuminance is higher at the image. The light at the image plane, however, fills a larger solid angle so the luminance comes out to be the same assuming there is no loss at the lens. The image can never be "brighter" than the source. Luminance is defined by Lv= d^2(F)/(dA.dB.Cos(theta)) (1) where Lv, is the luminance (cd/m2), F, is the luminous flux or luminous power (lm), theta, is the angle between the surface normal and the specified direction, A, is the area of the source (m2), and B,is the solid angle (sr). The luminous flux (or visible energy) in a light source is defined by the photopic luminosity function. The following equation calculates the total luminous flux in a source of light. F=683. integral y(lambda).J(lambda).dlambda (2) Where This integral is obtained in [0,infinite] F is the luminous flux (lm) y(lambda), (also known asV(ë)) is the standard luminosity function (which is dimensionless). J(lambda), is the power spectral density of the radiation, in watts per unit wavelength. For J(lambda), the integral is obtained in [-pi,pi] interval J(lambda)= integral (Iv(w)).e^(-ikw).dw (3) The luminous intensity for monochromatic light of a particular wavelength lambda is given by Iv=683.I.y(lambda) (4) where Iv is the luminous intensity in candelas, I is the radiant intensity in W/sr, If more than one wavelength is present (as is usually the case), one must sum or integrate over the spectrum of wavelengths present to get the luminous intensity: Iv=683 integral I.y(lambda).dlambda (5) Where The integral is obtained in [0,infinite] As a result, The luminous intensity can be written due to luminous flux as follow Power spectral density is calculated by substituting the equation (5) to the equation (3). After that the luminous intensity can be found substituting the equation (3) to the equation (2). Lastly equation (2) is placed to the equation(1)) and the luminance is obtained due to luminous intensity(light intensity). So, we can convert the luminance to the light intensity I hope, this answer can solve your problem. Best Wishes,
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