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.