GRIN Lens - Gradient Coefficient Expression

[Schreiber__]

Homework Statement

Homework Statement: You have a rod of GRIN material and would like to make a lens that has a pitch of 0.23. The rod had a radius of 1.0 mm with a quadratic radial change in index. The maximum refractive index is 1.6 and the fractional change in index of refraction, Δ, is 0.05. To what length should you cut the rod to get a single lens of 0.23 pitch? Express your answer in mm to two decimal points.

Relevant Equations

Z=(2*pi/g)*P

z= the length of the lens, P is the pitch and g is the gradient constant. I attempted to solve for g using the radius and delta, but I think I am missing a key function. The units of g should be 1/mm.

P=0.23
delta = 0.05
nmax = 1.6
nmin = 1.6 - 0.05 = 1.55
r = 1.0 mm

z = (2*pi/g) * p

Attempt at g, g = delta/r = 0.05/1mm = 0.05/mm, too low gave a length z = 28.9 mm which is incorrect and too long here.

Through some research I found this relationship, P = 2*pi/sqrt(g) where g is the gradient. Using the values above I calculated g = (2*pi/P)^2
Using the values above I calculated g = 746.28, but there are no units? This value is too high and a very small z (~2x10^-3).

The other equations in the lecture were focal length (dependent on z), NA (dependent on the index of refraction, n), and working distance (again dependence on z)

I appreciate the help here!

 

[haruspex]

For the benefit of others unfamiliar with the topic:
"A ray incident on the front surface of a GRIN lens follows a sinusoidal path along the rod. The "pitch" of the lens is the fraction of a full sinusoidal period that the ray traverses in the lens (i.e., a lens with a pitch of 0.25 has a length equal to 1/4 of a sine wave, which would collimate a point source at the surface of the lens)."

Wrt gradient, I gather that the refractive index at radius r is given by $n(r)=n(0)(1-\frac 12g^2r^2)$.
https://www.thorlabs.com/NewGroupPage9_PF.cfm

 

[haruspex]

the fractional change in index of refraction, Δ, is 0.05.

nmin = 1.6 - 0.05 = 1.55

Fractional change suggests to me nmin = 1.6(1 - 0.05). If the radius of the lens is R that gives $\frac 12 g^2R^2=\Delta=0.05$.

 

[Schreiber__]

For the benefit of others unfamiliar with the topic:
"A ray incident on the front surface of a GRIN lens follows a sinusoidal path along the rod. The "pitch" of the lens is the fraction of a full sinusoidal period that the ray traverses in the lens (i.e., a lens with a pitch of 0.25 has a length equal to 1/4 of a sine wave, which would collimate a point source at the surface of the lens)."

Wrt gradient, I gather that the refractive index at radius r is given by $n(r)=n(0)(1-\frac 12g^2r^2)$.
https://www.thorlabs.com/NewGroupPage9_PF.cfm

Thank you, I'll look at that for the solution of the gradient, then solve for the length. There is a second step where we change the radius of the lens (keeping other parameters the same), then evaluate the change in the two calculated length,

The link you provided just hits their general website (at least for me), can you direct me to the page? I tried looking, the reference might be helpful.

 

[Schreiber__]

Fractional change suggests to me nmin = 1.6(1 - 0.05). If the radius of the lens is R that gives $\frac 12 g^2R^2=\Delta=0.05$.

This is a good approach.

 

[haruspex]

The link you provided just hits their general website

https://www.thorlabs.com/NewGroupPage9_PF.cfm?ObjectGroup_ID=1209#