00:01
So we're told that a 3 millimeter thick pane of glass transmits 90 % of light that has a wavelength in the range of 0 .3 millimeters to 3 millimeters.
00:17
That's my best interpretation of the question wording.
00:20
And as we'll see, this kind of list of problems with the question that you'd have to resolve by maybe re -evaluating the wording.
00:27
But i'll give you the idea of how to solve it, nonetheless.
00:30
And so if we have two objects, one of them has a temperature of 5 ,800 kelvin, and then object b has a temperature of a thousand kelvin.
00:42
We want to know what rate of energy is transmitted through this paint of glass for each of these.
00:47
So to first figure out, we need to figure out if these wavelengths are from these two objects emitted are between this range, because we're told that outside of this range, this glass is opaque.
00:59
It does not transmit any radiation.
01:02
So the wavelength associated with either of these is going to be 0 .003 meters times kelvin divided by the temperature.
01:12
This is wines law.
01:14
So if we plug it in for 5800 kelvin, the wavelength 0 .003 meters times kelvin, divided by 5800.
01:24
This is like about 517 nanometers, which is out.
01:29
Outside of this range.
01:30
It's way too small.
01:32
And the wavelength for object b is about three times 10 to the negative 6 meters...