Ah okay, so is my approach correct? And the textbook also notes that if the speed of the first medium is greater than the speed of the second medium, there will also be a phase shift of pi.
Hmm.. I was doing some other research and I found some other equations for thin-film interference..
\huge \phi=\frac{4\pi n_b t \cos(\theta_i)}{\lambda}+\phi_{r2}-\phi_{r1}
Could I make the assumption that since the speed will be greater in air than in the glass that it will experience a pi...
Homework Statement
You are designing a thin transparent reflective coating for the front surface of a sheet of glass. The index of refraction of the glass is 1.52 and when it is in use, the coated glass has air on both sides. Because the coating is expensive, you want to use a layer that has...
Or perhaps since..
# of fringes = 2x+1, where x=m, then 2m+1
a). if m halves, then # of fringes = x+1
b). if m doubles, then # of fringes = 4x + 1
Or maybe I'm just overthinking it?
Homework Statement
A diffracting grating casts a pattern on a screen located a distance L from the grating. The central bright fringe falls directly in the center of the screen. For the highest-order bright fringe that hits the screen, m=x, and this fringe hits exactly on the screen edge. This...
Okay, so if I look at the joint above the voltage source.. there is a current entering from the bottom and current leaving on both the left and right sides. So
I1-I2-I3=0
At join 2 (to the right of the resistor of the top), there is current entering from the left, and leaving out the right and...
Would it just be 1500 then? I'm a bit confused then about how to calculate it.. The two on the far right are in series so you add them and then next is the two that are on the right in parallel, so you would add the reciprocals. Then..? =/
Homework Statement
Assume the electric potential is zero at the negative terminal of the battery. Calculate:
a). the equivalent resistance of the circuit
b). the electric potential at position a
c). the magnitude and direction of the current through each resistor
Homework Equations
The...
I calculated the conductivity by using the information given in the initial problem. The reciprocal gives me the resistivity. I used that value in the resistance equation and that gave me R.
Okay, so I calculated that both copper wires have voltage drops of 0.006V and the Nichrome has a voltage...
I absolutely hate circuits :) lol
So the first part, the copper wire on top. V=IR, so the voltage at the end of the first resistor is V=531.42A*(1.13E-5) and that voltage continues until we hit the next resistor, yeah? 1.13E-5 is the R of copper
Okay, so I found the new total resistances. Copper stayed the same, but Nichrome's value did indeed change!
So then the total current is I=V/R, so 1.5/.002823, which gives me 531.42A. So now I can apply V=IR to each piece?
Ohh, interesting. I did not see that in my search!
So then the reciprocal would give me resistivity which would help me find resistance. So how would I find the new voltage in each piece of wire?
I'm not really sure how...
I know that \sigma A V= I L where sigma is the conductivity. After solving for \sigma, I could take the reciprocal to get resistivity. From there plug in resistivity into the equation for resistance..? Not sure where the mobility comes into to play though.. I'm really...
True. But either way the total resistance rounds to 0.0040A anyway.
So I = 373.94 A. That seems very high doesn't it? I know that the resistivities are correct. As are the lengths and areas. So resistances is also correct..
Okay, I got that the resistance in the two copper wires are 1.13 \times 10^{-5} and the resistance in the Nichrome wire is 0.004.
So V=IR
I=V/R
I=1.5/.0040113?
So let's say I want to find the voltage at the end of the copper wire on the left aka the new voltage for the start of the Nichrome wire on the left.
The current through the copper wire is:
I=V/R
=\frac{1.5V}{\frac{2 \rho L}{A}}=\frac{1.5V}{\frac{2 (1.69 \times 10^{-8}) (0.17)}{2.54 \times...
Speaking of which, I have another question that seems like you would understand:
R= \frac{\rho L}{A}
So if I have half the area, that would mean that the resistance would double correct?
It's a very strange part of circuits that we haven't gone over, yet was included in the assignments, so it's kind of scaring me at the moment. I'm really not sure if I'm doing it correctly.
Well, aren't resistivity and conductivity inverses of each other? so \sigma = \frac{1}{\rho}
I guess I...