Recent content by Axis001

Homework Statement Find the n+1 and n-1 order expansion of \stackrel{df}{dy}Homework Equations (n+1)Pn+1 + nPn-1 = (2n+1)xPn ƒn = \sum CnPn(x) Cn = \int f(x)*Pn(u)The Attempt at a Solution I know you can use the recursion relation for Legendre Polynomials once you combine Cn with the...

:eek: I need to go to sleep. Thank you so much I really appreciate it.

Yes the nsin(...) terms should be there and it is what is giving me such a head ache since I just simply cannot get it out of the equation. A form of the deflection equation that I know is θd = θ - α + arcsin(nsin(α - arcsin((1/n)sin(θ))). It is easily derivable from applying Snell's Law at both...

Ok from "start" to current step with the corrections you helped me find this is what I have. θd = θ - α + arcsin(nsin(α - arcsin((1/n)sin(θ))) θd = θ - α + arcsin(nsin[sin(α)cos[arcsin((1/n)sin(θ))] - cos(α)sin[arcsin((1/n)sin(θ))]) θd = θ - α + arcsin(nsin[sin(α)sqrt[1 - (sin2(θ))/n2] -...

Firstly thank you for your reply I greatly appreciate it. (2) there’s a place where you took no/n1 to be n instead of 1/n (going from the third to last line to the next to last line) Ah yes I'm not sure how I missed. (1) the cosine factor is not quite simplified correctly (going from...

The work in the attachment is mine and your assumption of origin of angles of refraction is correct. I have simplified the equation for the displacement angle I originally found θd = θ - α + arcsin(nsin(α - arcsin((1/n)sin(θ))) into terms of simply θ, n, and α. The last step shown in the...

My apologies not sure how I overlooked defining my variables. α is the apex angle of the prism and θ is the angle of incidence on the prism.

Homework Statement The problem I am having troubles with is proving the deflection of a ray by a prism can be represented by a specific equation : θd = θ - α + arcsin(sin(α)√(n2 - sin2(α)) - sin(α)cos(α)). I have derived another form of the deflection equation for a prism and I am attempting...

That is what is baffling me is there is no provided value for wavelength but my professor insists that a numerical value is possible. Since the two areas should be equivalent I set up a polynomial equation with them and got a L of 0.1135 m. But when I solve for the areas I get 4.1077 x 10^-3 m...

Homework Statement Determine the effective area (Aeff) for a short dipole with L = λ/60 and λ/2 dipole. If the wires used for dipoles has radii a = 1 cm compare Aeff with the physical area. G(short dipole) = 1.5 G(half wave dipole) = 1.64 Homework Equations Aeff = G* λ2/4*pi The...