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### 1.3 homework

Posted: **Wed Oct 11, 2017 1:26 pm**

by **Alyssa Pelak 1J**

Hello, I have read the solution guide and cannot understand the answers for question 1.3.

Which of the following happens when the frequency of electromagnetic radiation decreases? Explain your reasoning.

A. The speed of radiation decreases.

B. The wavelength of the radiation decreases.

C. The extent of the change in the electrical field at a given point decreases.

D. The energy of the radiation increases.

If someone could please explain the right and wrong answers that would be great!

### Re: 1.3 homework [ENDORSED]

Posted: **Wed Oct 11, 2017 2:30 pm**

by **Hannah Chew 2A**

Hello,

a) No, speed is constant at 3.00 x 10^8 m/s

b) No, wavelength increases. Frequency and wavelength are inversely proportional since c = (lambda)(freq)

c) Yes, the electric field oscillates in strength and direction. If the frequency decreases, then the field oscillates less.

d) No, energy is directly proportional to frequency because E = hv

Hope that helps:)

### Re: 1.3 homework

Posted: **Fri Oct 13, 2017 12:28 am**

by **Brigita1D**

If the frequency goes down, then wavelength increases. This means that the gap between each wave gets bigger and the change in the electric field is not as rapid with the decrease in oscillation. This leads to a decrease in "the extent of the change in the electrical field."

### Re: 1.3 homework

Posted: **Fri Oct 13, 2017 9:31 am**

by **Sarah Brauer**

So "the extent of the change in the electrical field" is referring to oscillations?

### Re: 1.3 homework

Posted: **Fri Oct 13, 2017 12:46 pm**

by **Paula Dowdell 1F**

It is important to note that all of these instances (a)-(d) can be answered by using E=hv and v(lambda)=c

(a) speed does not decrease because it is constant

(b) the wavelength does not decrease because frequency is proportional to (lambda)-1

(c) Yes! as frequency decreases the waves get wider and the slope of the wave decrease.

(d) No, the energy of radiation does not increase because E is proportional to frequency

### Re: 1.3 homework

Posted: **Fri Oct 13, 2017 12:47 pm**

by **Paula Dowdell 1F**

the extent of change in the electrical field is referring to the amplitude!

### Re: 1.3 homework

Posted: **Fri Oct 13, 2017 7:31 pm**

by **Helen Shi 1J**

How does the amplitude of the waves correspond with the lower frequency?

### Re: 1.3 homework

Posted: **Sat Oct 14, 2017 1:58 pm**

by **Nathan Tu 2C**

As to why c is the correct answer, there is a mathematical explanation. Lets say you have a light wave that behaves as the function sin (bx). In this function, b affects the frequency of the wave and is a constant. If you play around with values for b and graphed the resulting equation, you'll notice that as b increases, frequency increases. Lets say you took the derivative of this function. You will get f'(x) = b cos(bx) by the chain rule. Naturally, the larger the value of b, the larger f'(x) will be. By definition, a derivative gives the rate of change at any one point in a function. In the case of light, who's properties can be represented by sin or cos function, the derivative will give you the rate of change in an electric field at a given point. Therefore, the higher the frequency, the higher the magnitude of change at any point on the wave will be.

Long story short, by using derivatives, one can see how frequency affects rate of change. Hope I didn't make it more confusing.

### Re: 1.3 homework

Posted: **Sat Oct 14, 2017 2:19 pm**

by **manasa933**

A) no, speed of radiation, c, is always 3 x 10(to the power 8)

B) no, Frequency is inversely proportional to wavelength, and so, if the frequency decreases, wavelength should increase

C) yes, as the wave gets wider when frequency decreases, the wave is less susceptible to change in electric field.

D) no, E=h x (freq), hence energy would decrease

### Re: 1.3 homework

Posted: **Sat Oct 14, 2017 3:51 pm**

by **Michelle Lee 2E**

I had problems with the wording of this question as well; to understand it, I had to look at it theoretically.

Say the frequency decreases; we know that the wavelength increases. [from the equation c=(lambda)(nu)] As the wavelength increases, the oscillation itself, the wave model, extends. [you can think of this like pulling on a slinky toy] As the wave extends, the amplitude doesn't change and the overall shape doesn't change as well. But now when we look at the electric field, which is described to us on page 4 of our textbook as an arrow from the middle of the graph to the peak or trough, we have to look at how the arrows moves with the extension. As the model extends, the arrows extend bit by bit; the overall change is pretty low because it's over a longer period.

This is opposed to when the model shrinks because when the wavelength is decreased, the arrows have to change more in a more rapid matter to compensate for less distance.

Either way, the electric field is still present so as the frequency decreases and the wave extends, the change in electric field becomes gradual. [not rapid like when frequency increases]

### Re: 1.3 homework

Posted: **Sat Oct 14, 2017 8:51 pm**

by **AtreyiMitra2L**

What actually helped me understand this question was kind of drawing 2 graphs: one with the normal frequency and one with a lower frequency. The one with a lower frequency should have a smaller amplitude. C is saying that "the extent of change at a given point decreases". Take the same random point on both graphs and draw the slope at that point. In the graph with a lower frequency, you will notice that the slope is not as steep compared to the one with normal frequency. From this itself without the process of elimination, you should get the right answer!

### Re: 1.3 homework

Posted: **Sat Oct 14, 2017 8:51 pm**

by **AtreyiMitra2L**

What actually helped me understand this question was kind of drawing 2 graphs: one with the normal frequency and one with a lower frequency. The one with a lower frequency should have a smaller amplitude. C is saying that "the extent of change at a given point decreases". Take the same random point on both graphs and draw the slope at that point. In the graph with a lower frequency, you will notice that the slope is not as steep compared to the one with normal frequency. From this itself without the process of elimination, you should get the right answer!