CP Light from an ideal spherical black body 15.0 cm in diameter is analyzed by using a diffract

CP Light from an ideal spherical black body 15.0 cm in...

CP Light from an ideal spherical black body $15.0 \mathrm{~cm}$ in diameter is analyzed by using a diffraction grating that has 3850 lines/cm. When you shine this light through the grating. you observe that the peak-intensity wavelength forms a first-order bright fringe at $\pm 14.4^{\circ}$ from the central bright fringe. (a) What is the temperature of the black body? (b) How long will it take this sphere to radiate $12.0 \mathrm{MJ}$ of energy at constant temperature?

Key Concepts

Diffraction Grating

A diffraction grating is an optical component with a regular pattern that diffracts light into several beams traveling in different directions. The directions depend on the spacing of the grating lines and the wavelength of the light, and they can be described by the equation that relates the order of the diffracted beam, the grating spacing, and the angle of diffraction. This concept is key to analyzing the wavelengths present in a source of light.

Stefan-Boltzmann Law

The Stefan-Boltzmann Law states that the total energy radiated per unit surface area of a black body across all wavelengths per unit time is proportional to the fourth power of its absolute temperature. This relationship is crucial for determining the power output of a radiating black body and is widely applied in thermal physics and astrophysics to estimate energy outputs.

Black Body Radiation

Black body radiation refers to the idealized emission of electromagnetic radiation by an object that absorbs all incident radiation, regardless of frequency or angle of incidence. The emitted radiation depends solely on the temperature of the black body, and its spectrum across different wavelengths is well-defined, serving as a fundamental concept in thermal physics and quantum mechanics.

Wien's Displacement Law

Wien's Displacement Law provides a relationship between the temperature of a black body and the wavelength at which its emission spectrum is at maximum intensity. According to this law, the product of the temperature and the peak wavelength remains constant, allowing the temperature of a black body to be determined from its peak emission wavelength.

Transcript

00:01 Okay, in this question we're talking about the diffraction pattern of weight.

00:05 So we have a sphere, which we can treat as a perfect black body, and it has a radius of 7 .5 centimeters, it tells the diameter is 15.

00:18 And it's going through this grading, which has 3 ,850 lines per centimeter.

00:27 And then the pattern it makes on the wall has its first peak or the first area between peaks happen at 14 .4 degrees or negative 14 .4 degrees on this side.

00:49 So we're asked to find what the temperature of the black body must be for question a.

00:55 So we need to find a way to relate temperature to these other things that we know, which are the density of splits in the grading and the angle where this first interference pattern happens.

01:10 So let's find an equation which first relates theta and the density of the lines.

01:18 So an equation we can use that does that is d -sign theta is equal to lambda.

01:28 And lambda is the wavelength of incoming light.

01:32 We might realize this will be useful because we know an equation which relates lambda to temperature, but we'll get to that later.

01:38 So first let's figure out on, let's focus on figuring out what lambda is.

01:44 So we need to figure out what d is, and we can do that using the density of the lines in the greater.

01:50 So if there are 3 ,850 lines per centimeter, then the distance between each line is going to be one over this.

01:57 Let's use si units, so we'll do 385 ,000 lines per meter.

02:07 So d is equal to 1 over 3 ,000 or 385 ,000, which i'll write is 3 .85 times 10 of 5.

02:19 This is going to give us 2 .597 times 10 to minus 6 meters.

02:26 So we have d and we have theta.

02:31 I should have specified the formula is theta 1 and lambda 1, but we're just using lambda 1, so we can forget about the indices and just plug these numbers in.

02:44 So this will give us lambda in terms of these things, which we know.

02:51 So now we need to find a way to connect lambda to temperature.

02:54 So for this we can use vien's law, which says that lambda peak or lambda max, which is what i normally call it is equal to, or times t is equal to 2 .9 times 10 to the negative 3 meters kelvin...

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