magnified hologram

[tonyr]

Is there a way to make a hologram of something that is magnified in size? You can use a lens to magnify an object but as I understand it that will also distort the longitudinal scaling. Is there a way around that to make an undistorted hologram of a magnified object?

[lobaz]

As far as I know: no.
(At least using just conventional optics.)

[tonyr]

I was afraid of that. How about during playback of a hologram? I've heard you can make an image look deeper by illuminating the hologram using a diverging beam instead of a collimating beam (might have it backward). Can you correct for a lot of longitudinal distortion using that method? Or does it happen that if you correct for the distortion, you always end up cancelling out the magnification too?

[tonyr]

Also wondering about solutions using unconventional optics. One way I can imagine is to record the hologram digitally, alter the pattern somehow (is it a simple algorithm?), and render as a CGH. Also, how about an HOE? Is there a fundamental reason it can't be designed to do undistorted magnification somehow even if lenses can't?

[tonyr]

Here's another idea - what about a multi-stage scenario? Say we magnify the object with a lens so that the image is now magnified 2x but due to the axial distortion it is 4x in depth. Then use a second lens to magnify it along a new axis so that height and depth are magnified 2x and width is 4x. Finally a third lens along another new axis for overall 16x magnification in all 3 dimensions. (Let me know if you'd like a picture.)

There's also the field of view of the lenses to consider, which I'm worried makes it imperfect in practice even with lenses of gigantic diameter. But maybe more stages at different angles would help? Also another scenario could be to make intermediate holograms at each stage each time it's magnified, and use the hologram image for magnification in the next stage. Don't know if that would help any with the field of view problem.

[lobaz]

Simple analysis using conventional methods

Suppose the x-z coordinate system.
The hologram is located in z = 0.
The object is a point located at [xP, zP], where zP < 0.
The reference wave comes from the point [xR, zR], where zR < 0.
Wavelength lambda is used in the hologram recording.

Let's examine the hologram at a point [xH, 0].
The angle of the object wave is omega = atan((xP - xH) / zP).
The angle of the reference wave is rho = atan((xR - xH) / zR).
The frequency of the interference pattern at [xH, 0] is given as

f(xH) = (sin(omega) - sin(rho)) / lambda

For the small angle approximation, sin(angle) = tan(angle) = angle, thus

f(xH) = ((xP - xH)/zP - (xR - xH)/zR) / lambda

For reconctruction, we can select the wavelength of the illumination wave Lambda, and its location [xI, zI].
Thus, the angle of the illumination wave is at [xH, 0] is

iota = atan((xI - xH) / zI).

Moreover, we can "stretch" the hologram, i.e., magnify it by factor mu.
Thus, point xH on the hologram becomes mu*xH, and the frequency of the pattern becomes f(xH)/mu.

We will use the grating equation that determines the angle delta_m of the m-th order diffracted wave as

sin(delta_m) = m * Lambda * f + sin(iota)

In the small angle approximation and stretched hologram, we get

delta_m(mu*xH) = m * Lambda * f(xH)/mu + (xI - mu*xH)/zI

i.e.

delta_m(xH) = m * Lambda * f(xH/mu)/mu + (xI - xH)/zI

To get the position of the reconstructed point, we must examine two outgoing rays from the illuminated hologram and calculate their intersection.
We will use points [xH, 0] and [0, 0] on the hologram. For the latter one, we get by substitution

delta_m(0) = m * Lambda * f(0)/mu - xI/zI
= m/mu * Lambda/lambda * (xP/zP - xR/zR) + xI/zI

We are close to the result.
A ray is given by x = slope * z + intercept. Our two rays are given by equations

x = delta_m(0) * z
x = delta_m(mu*xH) * z + mu*xH

Coordinates of their their intersection are given by the solution of this system of linear equation.
It is not short:

x = xP * (Lambda*m*mu*zI*zR) /
(mu^2*zP*zR*lambda + Lambda*m*zI*zR - Lambda*m*zI*zP)
+ (mu^2*xI*zP*zR*lambda - Lambda*m*mu*xR*zI*zP) /
(mu^2*zP*zR*lambda + Lambda*m*zI*zR - Lambda*m*zI*zP)

z = (mu^2*zI*zP*zR*lambda) /
(mu^2*zP*zR*lambda + Lambda*m*zI*zR - Lambda*m*zI*zP)

Just to see if it goes well, let us substitute

• lambda = Lambda (reconstruction and recording wavelengths match),
• m = 1 (first diffraction order)
• xR = xI, zR = zI (locations of the reconstruction and reference source),
• mu = 1 (original size of the hologram, i.e., no stretching).

The result after substitution is

x = xP
z = zP

i.e., the hologram perfectly reconstructed the original point position [xP, zp].

If we want scaled reconstruction, the result should become

x = constant * xP
y = constant * zP

where the constant should not depend on [xP, zP].

In the full equation for x, we see it is in the form

x = xP * const. + term

Obviously, the term should be 0, which is true for

mu^2*xI*zP*zR*lambda-Lambda*m*mu*xR*zI*zP = 0

i.e., the reconstruction wavelength Lambda should be

Lambda = (mu*xI*zR*lambda) / (m*xR*zI)

Substitution to the equation for x reveals that

x = xP * (mu*xI*zR) / (xI*zR + (mu*xR-xI)*zP)

Obviously, xP should be multiplied by a constant that does not depend on zP.
This is true when xI = mu*xR. After substitution, we get a simple result

x = mu*xP

We can substitute the same conditions to the equation for the coordinate z.
We get

z = zP * (zI/zR)

For uniform scaling, the multiplicative constants for z and x should be the same.
Thus, it must hold

zI = mu*zR.

We should substitute these conditions to the reconstruction wavelength Lambda.
We get

Lambda = (mu*lambda)/m

Conclusion

We have obtained this result: magnified image can be obtained, if following conditions hold:

Lambda = (mu*lambda) / m

i.e., the reconstruction wavelength is equal to mu/m * recording wavelength,
where m (the diffraction order) must be integer.
It also must hold

zI = mu*zR
xI = mu*xR

Then,

x = mu*xP
z = mu*zP

i.e., the reconstruction is scaled by the same factor as the "stretch" of the hologram.
Of course, there is a glitch. If we want to use the same wavelength for recording and illumination (lambda = Lambda), then the magnified reconstruction appears in the m-th diffraction order, which is usually weak.

[lobaz]

Indeed, unconventional methods give you whatever you want. Two ways are straightforward:
You can make a holographic stereogram.
You can make digital model of a real scene and calculate its computer generated hologram of arbitrary scale.

[Din]

The transverse magnification of a hologram recorded with reference at x(r), y(r), z(r) with an object at x((o), y(o), z(o), and reconstructed from x9c), y(c), z(c) is*:

MT= dXi/dxo= 1/(1 +/-zo/mu*zc–zo/zr)(Meyer; paraxial)
MT= (cos(alpha)o/cos(alpha)i){1/(1 +/-zo/muzc–zo/zr)} (Champagne; non-paraxial)

The longitudinal magnification is:
ML= dZi/dzo= -(1/mu){1/(1-zo[(1/muzc) + (1/zr)}2 = -(1/mu)M

So while in conventional optics, we have:
MT = -ML²
In holographic optics, it's possible to increase the transverse magnification without altering the longitudinal magnification, provided you changed the reconstruction wavelength, as Petr has noted.

*Details, if you want, is at http://www.triple-take.com/publications ... graphy.pdf

[tonyr]

Thanks lobaz that is extremely helpful, not just the result but how you derived it.

reconstruction wavelength is equal to mu/m * recording wavelength,

So the bottom line is you can only get perfect magnification by stretching the holographic pattern, and even then you need to use either a higher diffractive order or different reconstruction wavelength.

Separate wavelengths aren't that practical, at most ~2x magnification if you record with blue and playback with red.

Not sure higher orders are practical either considering weakness like you said. 10x magnification sounds out of the question.

Not to mention, how would you even implement the stretch? Use a lens to magnify and reimage the holographic pattern onto a new plate? Has that been done?

Tony

[tonyr]

You can make a holographic stereogram.

This hadn't even occured to me. Seems like by far the best way.