Constant thickness lens varying refractive index

[timetraveller123]

Homework Statement

Homework Equations

The Attempt at a Solution

my initial approach was to just imagine the light rays leaving the focal point and emerging parrallel but then the result had no dependence on thickness how to tackle this problem how does the thickness even factor in

 

[Baluncore]

Focus on wavefronts, not rays.
Consider a plane wavefront entering the lens. The time delay while passing through the lens should convert the plane wavefront to a spherical wavefront. What time delay( radius) is needed. What RI will give that delay?

 

[timetraveller123]

i am really not getting something if by looking at it as wavefronts if the wavefront is normal to the lens why would it refract
and would a time delay make it into spherical wavefront sorry i have not really learned spherical wavefronts before

 

[Baluncore]

https://en.wikipedia.org/wiki/Wavefront
The time delay of rays propagating through a shaped lens is determined by the speed of light in the material and the thickness of the glass.

https://en.wikipedia.org/wiki/Refractive_index
In a flat plate of material with a varying RI, the speed of light can vary with RI and radius from axis, so the time delay can be made the same as in a shaped lens.

https://en.wikipedia.org/wiki/Refractive_index#Inhomogeneity

 

[timetraveller123]

ok now i have a better understanding of what will happen but i am not really able to relate the time delay to the focal length any help on that

 

[Baluncore]

Rays that pass through the lens far from the axis have further to travel to the focus than rays nearer the axis. Assume the lens is very thin, work out the difference in distance and so the time delay required at different radii from the axis for all rays to arrive at the focus at the same time.

You know that the RI, η = c / ν, so you can work out the thickness of glass with a particular RI needed to delay the axial rays more than the outer rays.

You can then work out the RI distribution needed for a flat plate to get the same time delay distribution.

 

[rude man]

I would play with ray optics.
Remember this is a thick lens, not a thin one.
What kind of lens (convergent or divergent) can you make from the stated process?
Then work the geometry with the given r, d and F.
Ray optics give you a better picture of what is going on.
EDIT - see my later post. I don't think this is a lens at all.

 

[timetraveller123]

Rays that pass through the lens far from the axis have further to travel to the focus than rays nearer the axis. Assume the lens is very thin, work out the difference in distance and so the time delay required at different radii from the axis for all rays to arrive at the focus at the same time.

You know that the RI, η = c / ν, so you can work out the thickness of glass with a particular RI needed to delay the axial rays more than the outer rays.

You can then work out the RI distribution needed for a flat plate to get the same time delay distribution.

ok i am getting
##
d_{extra} = \sqrt{F^2 + r^2} - F \\
t_{difference} = \frac{dn_0 - dn_r}{c}\\
hence ,\\
n_r = n_0 - \frac{\sqrt{F^2 + r^2} - F}{d}
##
is this correct

 

[haruspex]

ok i am getting
##
d_{extra} = \sqrt{F^2 + r^2} - F \\
t_{difference} = \frac{dn_0 - dn_r}{c}\\
hence ,\\
n_r = n_0 - \frac{\sqrt{F^2 + r^2} - F}{d}
##
is this correct

Looks right to me.

 

[rude man]

In a positive lens, parallel rays converge to a focus.
In a negative lens, parallel rays diverge (forming a virtual focus on the object side).
Neither happens with this lens. Parallel rays perpendicular to the lens surface just go thru without bending at all.
Strange lens!

 

[Baluncore]

Parallel rays perpendicular to the lens surface just go thru without bending at all.
Strange lens!

That is where the concept of the wavefront comes in. The sum of all rays constructively form a wavefront, with the spherical curvature required to bring the energy to the focus.

Considering a single one dimensional ray is too simplistic a model for diffraction.

 

[rude man]

I forgot about graded-index (GRIN) lenses.

But there is nothing wrong with applying ray optics to the GRIN lens. Outer beams are bent more strongly by the transverse n gradient than ones closer to the lens center, all converging and forming a positive lens, as shown below.

"Geometrical optics does not account for certain optical effects such as diffraction and interference. This simplification is useful in practice; it is an excellent approximation when the wavelength is small compared to the size of structures with which the light interacts." (wikipedia, italics mine).

This is thoroughly covered in http://homepage.tudelft.nl/q1d90/fbweb/diss.pdf

Having been clued in I will try to verify the OP's n(r) derivation. Offhand it looks right.
EDIT: It's right. (It ignores the slight droop in each ray as it passes thru the lens but that is a very reasonable simplification).