Formula - Light

Light Properties - shows the relationship between the speed of light, its wavelength and its frequency. A fairly simple, but important relationship.
Formula: c = f where:

This is a very important relationship since it tells you several things - first of all, the speed of light is constant - it never changes (as far as we're concerned in this class). So the left side of the formula always has the same value. That means if you change something on the right side (either the wavelength or the frequency) then the other thing has to also change, but in the opposite sense. So if the wavelength goes down the frequency goes up, and vice versa.

Typical Problems

1. If a light's wavelength is increased by a factor of 10, how does its frequency change?

Solution: Since the speed of light (c) cant change, when you change one part of the formula, the other part mush change in the opposite sense:
c = f
So for the left side to stay the same while becomes 10 times larger, f must become 10 times smaller. Mathematically this would be written as
c = 10 x x (1/10) f
And the 10s cancel out. So f must be 10 times smaller.

2. If a particular type of light has a wavelength of 6430 ┼, what is its frequency?

Solution: In order to use the formula, you should first convert the wavelength from ┼ to meters -
6430 ┼ = 6430 x 10-10 m = 6.430 x 10-7 m
Now you can use it in the formula -
c = f
Rearrange it to solve for f
f = c/
f = 3 x 108/(6.430 x 10-7) = 4.67 x 1014 /seconds.

3. If a particular type of light has a frequency of 1 million /second, what is its wavelength?

Solution: Rearrange the formula to solve for wavelength -
c = f
= c/f
And put in the various values, no conversions are needed, though you need to write frequency properly -
= 3 x 108/ 106 = 300 m.


Energy of a photon - There are two versions of this formula, one using the frequency, the other using the wavelength. The basic upshot is that as frequency goes up, so does energy, however wavelength goes down. So as wavelength goes up frequency and energy go down. This formula is for the energy of an individual photon. In general, only relational values will be needed (no exact values calculated).

Formula: E = h f where:

The second version of the formula is found by using the light properties formula (c = f) and substituting that in for the above relation. The result is how the energy depends upon the wavelength of light.
Formula: E = hc/ where:

Typical Problems

1. If you increase the frequency of a light source by a factor of 30, how much does the energy of the photons change?

Solution: Look at this version of the formula -
E = hf
The only thing that can change on the right side is the frequency (f), which becomes 30 times larger, therefore the Energy of the photons is 30 times larger

2. If a light sources wavelength is 25 times smaller than before how does that change the energy of the photons?

Solution: Look at the second version of the formula -
E = hc/
By changing the wavelength, you are dividing the bottom by a factor of 25, or written mathematically
E = hc/(1/25)
But when you divide by a fraction that is the same as multiplying by its reciprocal, so the end result is
E = hc/ x(25/1) = 25 hc/
So the energy is 25 times greater.
You could also figure this out by looking at how the frequency changes when you change the wavelength. If the wavelength is 25 times smaller than the frequency is 25 times larger. Looking at the first version of this formula -
E = hf
You are basically multiplying the right side by 25 when you change the wavelength, so the result is the same, the energy is 25 times greater.


Wien's Law - This law is used for black bodies, perfect radiation (light) emitters and absorbers and indicates at which wavelength they tend to give off most of their light. This law is used to explain why the colors of objects change as you change their temperatures.

Formula: max=0.0029/T where:

The constant in the formula shows how these two things are related, and could be different if you measured the wavelength or temperature in different units. Basically as the temperature goes up, max goes up and vice versa.

Typical Problems

1. If you increase the temperature of a black body by a factor of 5, how does its value of max change?

Solution: This is a relational change to the formula -
max = 0.0029/T
If T goes up by a factor of 5, that means the right side is being divided by 5, so the end result is that the left side is 5 times smaller than it was before. The value of max is 5 times smaller than what it was before.

2. An object has a temperature of 10,000 K. What is its value of max?

Solution: This is just a straight plug into the formula -
max = 0.0029/T
max = 0.0029/10,000
max = 2.9 x 10-7 m.
This is also the equivalent to a wavelength of 2900 ┼, and corresponds to UV light.

3. If an object has a max of 450 ┼, what is its temperature, assuming that it's a black body?

Solution: You'll have to rearrange the formula and solve it for T
max = 0.0029/T
T = 0.0029/max
And you'll also have to change the wavelength from ┼ to meters. 450 ┼ = 450 x 10-10 m = 4.50 x 10-8 m. Now you can put that into the formula -
T = 0.0029/(4.50 x 10-8) = 64,400 K - which is pretty dang hot!


Stefan-Boltzmann Law - This law is used for black bodies, perfect radiation (light) emitters and absorbers and indicates how much total energy they give off. Interesting thing to note - the energy they give off depends only on their temperatures, nothing else about the object matters (like what it is made of). Also this measure the total energy given off, so all energy at all wavelengths is what this indicates.

Formula: Etotal = T4 where:

Generally this is one of those relational formulas, where I would ask, if you increase/decrease T by a certain amount, how does that change the value of the total energy given off. But you have to be careful, T is taken to the fourth power, so a small change in it results in a very large change in the total amount of energy given off.

Typical Problems

1. If you increase the temperature of a black body by a factor of 5, how does that change the amount of total energy given off?

Solution: This is a relational change to the formula -
Etotal = T4
So if the temperature is 5 times greater, the power on the temperature also works on the 5. The end result is that the total energy goes up by a factor of 5 x 5 x 5 x 5 = 625.

2. One object (object A) is giving of 400 times more energy than another (object B). How do their temperatures compare?

Solution: Even though you might not think it is the case, this is another relational problem, but in this case you need to find the value of T that results in an energy change of a factor of 400. So here's the formula -
Etotal = T4
For E to be 400 times greater, T must be greater by a certain amount (let's call it X). The change in temperature results in a factor of 400, so X to the forth power must equal 400. Mathematically this would be written as
400 = (X)4
To solve for X you need to take each side by the 1/4 power or X = 400(1/4)=4.47.
So the temperature of object A is 4.47 times greater than that of object B. You can check this by going the other direction. If an object's temperature is 4.47 times greater, how does that change the energy output? Well you just take 4.47 to the fourth power, 4.474 = 399 (rounding makes it not give exactly 400).


Doppler Effect - One of the most important formulas in astronomy since it helps us to figure out how fast things are moving towards/away from us, just by looking at how their light is altered. It doesn't show motion that is perpendicular to our line of sight, but just velocity towards or away from us, so it does have its limitations.

Formula: V = c / where:

You can use this formula to get the velocity of an object. The velocity and the speed of light are in the same units (either km/s or m/s), while the change in wavelength and normal wavelength are also in their same units. By change of wavelength, this is how much the light has been altered or "shifted" from what its normal wavelength should be. The normal wavelength is what it has when it isn't moving, sometimes this is called the "lab" wavelength or "zero velocity" wavelength. In any event, is the measure between the observed wavelength and the normal or measured - , and it can be either positive or negative. When the value of is positive, that indicates that the wavelengths that are observed are larger than normal, or "red shifted". This indicates motion away from the observer. When the value of is negative, the wavelength that is observed is smaller than normal, which is called "blue shifted". This means the object is moving towards the observer.

Typical Problems

1. If a spectral feature has a normal wavelength of 4800 ┼ and you observe an object to have this spectral feature appearing at a wavelength of 4820 ┼ what is its velocity?

Solution: First of all you need to figure out what the value of is -
= observed - normal = 4820 - 4800 = 20 ┼
Now put that in the formula, be careful to put in the correct units for the wavelength -
V = c /
V = 300,000 x 20/4800 = 1240 km/s.
Since this is a positive velocity, it indicates motion away from us.

2. A feature of hydrogen normally appears at a wavelength of 912 ┼. You observe such a feature in an object's spectrum, but it is at a wavelength of 900 ┼. What is the objects velocity?

Solution: First of all you need to figure out what the value of is -
= observed - normal = 900 - 912 = -12 ┼
Now put that in the formula, be careful to put in the correct units for the wavelength -
V = c /
V = 300,000 x (-12)/912 = -3950 km/s.
Since this is a negative velocity, it indicates motion towards us. Also noteworthy is the fact that the change in velocity in this problem is only 12 ┼, while in the previous problem the change was larger. The velocity though here is larger due to the fact that the normal wavelength is much smaller here than in the previous problem.

3. If an object is travelling at a velocity of 5000 km/s away from us, where would the hydrogen feature that is normally observed at 912 ┼ appear at in this objects spectrum?

Solution: This one is a bit tricky. You are trying to figure out the value of , and from that the value of the observed wavelength. So let's rearrange the formula to solve for
V = c /
= V x /c
Not very pretty, but now you can plug stuff in -
= 5000 x 912/300,000 = 15.2 ┼.
So the value of is 15.2 ┼. This again is the value of the observed wavelength - normal wavelength, so
15.2 = observed -
15.2 = observed - 912
observed = 912 + 15.2 = 927 ┼.
The object would have the feature that normally appears at 912 ┼ appear at a longer wavelength, about 927 ┼ when it is moving at that speed.

4. Do the same as in problem 3, but now have the object moving at a velocity of 5000 km/s towards us.

Solution: Again, you are trying to figure out the value of , and from that the value of the observed wavelength. So let's rearrange the formula to solve for
V = c /
= V x /c
Which is pretty much what you had before, but now the velocity is towards us, so we put a negative sign on it:
= -5000 x 912/300,000 = -15.2 ┼.
So the value of is -15.2 ┼. This again is the value of the observed wavelength - normal wavelength, so
-15.2 = observed -
-15.2 = observed - 912
observed = 912 - 15.2 = 897 ┼.
The object would have the feature that normally appears at 912 ┼ appear at a shorter wavelength, about 897 ┼ when it is moving at that speed towards us.