Recent content by Callix

Ah okay, thank you!

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.

And since the incidence is normal, cos(0)=1. Well, approximately 0, so cos(~0) = ~1

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...

n=\frac{550}{2(104)}=2.64

Well, it would be destructive since it needs to cancel light. m should be at a minimum as well, which would be 1, right?

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...

Will I have to each of the 4 joins since so far I only have 2 equations and 4 unknowns? Won't I end up with some more unknowns?

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...

Ok! So I got that the equivalence resistance is 750 ohms. So now would I take 30V/750ohms to find the current?

Okay, so the two that are in series on the two on the right, correct? And then in parallel would be the two ends?

Okay, so now aren't these in series then? So you would just add them normally?

Is this correct so far?