I am calculating Responsivity of a pn junction photodiode (a.k.a the target) by irradiating radiation from LED sources. For this purpose, i have two LEDs, one UV and another green LED. Note that LEDs are placed close to the target.UV LED : Manufacturer has given total radiant power to be 20mW...
My lux-meter can be used in two modes, candela and lux. How do I use it to accurately measure a display's luminance in nits?
The screen isn't Lambertian, I have to assume this as an unknown. I'm interested in the luminance as seen by the eye looking at the screen perpendicular to it.
Now, first...
The ##I_i## are the intensity of the rays, in other words energy per surface units per radians by seconds.
The d##\Omega## are the solid angles
The equation p75 isis what I don't understand. I suppose that each side represent the energy going and out of the surface dS but I don't understand...
I'm looking for the distribution of all wavelengths (or frequencies) of light that a stationary observer would receive at his location (at ##r = 0## and time ##t_0##), from all light sources emitting a single wavelength ##\lambda_{\text{e}}## (or angular frequency ##\omega_{\text{e}}##). The...
Homework Statement
I use a probe for converting optical power to an electrical current. I need to convert Volt to candela/m2.
The sensor, a photodiode, measure the brightness of the monitor leaning the probe on the glass (so there is no distance between the monitor and the sensor).
The sensor...
Hello,
My name is Chris and I am trying to understand why UVA radiation, with a longer wavelength, penetrates deeper into the skin than UVB which has a shorter wavelength.
Since E=hv=hc/ λ and ,
then how does a lower energy and lower intensity UVA wave penetrate deeper into the skin than a...
[Mentor's Note: Thread moved from New Member Introductions]
Hello,
My name is Chris and I am trying to understand why UVA radiation, with a longer wavelength, penetrates deeper into the skin than UVB which has a shorter wavelength.
Since E=hv=hc/ λ and
then how does a lower energy...
From what I understand, in the equation P=\sigma AT^4, P is the power output of the star which is the energy radiated per second in EM radiation of all frequencies, and I think luminosity is also defined as the energy radiated per second in EM radiation of all frequencies. Therefore luminosity...
Homework Statement
A point source emits visible light isotropically. Its luminous flux is 0.11 lumen. Find the flux whithin the cone that has half angle of 30 degree from the light source.
Homework Equations
luminous flux = luminous intensity * solid anlge
The Attempt at a Solution
I tried...
It seems to me that luminous intensity should really be put in terms of energy, not a special unit (which itself is based on some arbitrary specification of energy.) The other 5 units and Avogadro's number should be the only fundamental units.
Does the luminous intensity due to an isotropic point source of light at a point on a surface depend on the angle it makes with the normal to the surface?
# The intensity of direct sunlight on a surface normal to the rays is Io. What is the intensity of direct sunlight on a surface whose normal makes an angle of 60 degrees with the rays of the sun?
I think the answer is Io itself because for an isotropic point source of light the luminous...
This is the SI unit for luminous intensity. The definition relates to blackbody radiation emitted at a certain temperature for a certain material (so I guess it ISN'T blackbody!), platinum, I think.
Except... I don't understand the necessity for the introduction of this unit.
Isn't...