# How can the focal length for shake reduction be determined when the lens is mounted using a reverse lens adapter?

My camera (Sony A7RII) offers image stabilisation (SteadyShoot).

I shoot macros using a 50mm and a 28mm Olympus lens reverse mounted with anadapter.

What values should I enter in the menu to configure SteadyShoot (8mm-1000mm)?

Is there a formula? If so can this include the additional use of extension tubes (I currently use 16mm + 10mm).

The lens normal mounted is able to reproduce 370mm on a 36mm sensor this should be a magnification of 0.097:1 roughly a tenth.

When reverse mounted things change 50mm on 36mm gives 0.72:1

With the extension tubes (26mm) 22mm on 36mm gives 1.64:1 magnification.

The calculator on http://extreme-macro.co.uk/extreme-macro-reversed-lenses/#calculator can be used to determine the focal length

Entering the values from sample shot 2 (28mm lens reverse mounted, displays 50mm on 36mm gives 0.72:1)

Adding the extension tubes (26mm) shows the measured maginification (1.64:1) and the focal length of 119mm. So the best setting would be 115mm without extension tubes and 120mm with extension tubes.

The formulars used by the calculator:

Given values:

t = extension in mm
l = lens focal length
m = original magnification
f = intended f/stop

Calculated values:
The new magnification (mag:1)
mag = m + t/l

The new focal length in mm
focal = l*(1+mag)*(1+mag) / mag;

The new fstop
fstop = f*(1+mag)


Stacking lenses is no different in principle than stacking close-up adapter (also sometimes called close-up "filters" because they screw on to a lens's filter threads, and look much like a regular filter).

Close-up adapters are usually described by their diopter — their optical power, which is just the reciprocal of the lens's focal length (in meters). Thus a 5 diopter lens has a focal length of 1/5 = 0.2 m = 200 mm.

Diopter measurement is useful because you can just add the diopter power for every stacked lens. So the equivalent focal length of two stacked lenses is just the reciprocal of the sum of the individual lens's diopter powers:

ƒequiv = 1 / (1/ƒnormal_lens(m) + 1/ƒreversed_lens(m))

In your case, the equivalent focal length is 1/(1/0.050 + 1/0.028) = ~18 mm.

• Thanks for you answer, I provided misleading information by writing that I use 28 and 50mm. I don't use them stacked, only one of them but reversed. This answers the question I would have need to ask when I receive my lens coupling adapter (which I plan to use with the 28mm as front lens and a stacked 300mm on the camera side). Commented Nov 12, 2018 at 12:45