Recent content by rogue_system

##D## is the diameter of the 3rd newton ring, which for a given radius of curvature, and a given lambda is always going to be the same? Assuming the relationship of the refractive indicies holds and a 180 degree phase shift exists. So as the wavelength gets longer or shorter (depending on if...

$$n_{lens} λ_{lens} = n_{water} λ_{water}$$ Which gives us $$λ_{water} = \frac {n_{lens} λ_{lens}} {n_{water}}$$ The part I was missing was how the lens behaved though, the approximation of $$t = \frac {r^2} {2R}$$ Also I have the equations for constructive and destructive swapped above, so...

I've gone over the textbook, and ultimately had to look elsewhere for the equation relating the radius of the ring to the radius of curvature of the planoconvex lens. I'm also assuming that ## n_air = 1 ##. I've searched high, low, and even my professor's lecture notes and slides, and it's...

The smaller cylinder with curved ends matching the radius of the curvature is more or less how I pictured it (basically a camera lens without the metal barrel around it). Switching the sign got me to the correct distance for S' and thus ##S_{2}##. I had switched S to be negative on the left...

Thre is not, and that information is not given. What I posted in the original problem statement is all it gives.

Not sure how accurate my scale is, but I drew it larger to better visualize. Since the lens is convex I would assume that the image would be inside, but I got a positive value for S' so I calculated it as though it was on the same side as S. It's entirely probable that I have misunderstood the...

Just saw that I multiplied when I should've divided, so I might just need a nap. Simplifying should be: [(0.60/6)-(1/24)]/1.60=1/s'=17/192 Which gives: s' = 192/17 or about 11.29cm Adding to L to find S2 would give about 50.294cm, but I'm still not sure I've done that right? Corrected...

n1=1.00 (air), n2=1.60 (glass), R=6.00cm, s=24cm (1/24) + (1.60/s') = (1.60-1.00)/6 Simplifying I get [(0.60/6)-(1/24)]*1.60=1/s'=7/75 s'=75/7 This puts s' to the left hand (same side as incoming light) of the lens on the left hand side, so the distance to the lens on the right hand side...

Figured it out, I'm supposed to assume the value for 'n' with regard to light trafeling from point B to the interface is just 1 (I know it says it's a vacuum, but I have trust issues with Pearson). From there Snell's law makes short work of it.

Part A was simple enough, if the distance traveled in the new medium takes place over the same amount of time in the old medium, then the distance traveled will be the new velocity divided by t1, which makes d'= (c/n)*t1. Part B is t1/cos(55) = tb Part C is (c/n) * tb Where I'm getting a bit...

The auto-message I got when I signed up here suggested making an into thread, so this is an intro thread. I'm currently an undergraduate student at GMU, who is projected to graduate with a B.S. in physics (no concentration) in spring of 2028. I'm going to school using my G.I. bill, so my...