Using Thin Lens Equation to find how far 1972 Blue Marble photo was taken

Question

I am having difficulty using the information I found in Optics Eugene Hecht to calculate how far away the 1972 Blue Marble photo (shown here) was taken. The camera used was an 80mm Hasselblad camera. The photograph was taken aboard the Apollo 17 spacecraft. It is viewed from around 28,000 km from Earth's surface.

I first started with the magnification formula where $h_i$ is the image height which seems to be around 56mm, and $h_o$ is the object height and is the diameter of Earth in milimeters.

$$M=\frac{h_{i}}{h_o}=\frac{56mm}{12,742,000,000mm} $$

$$ M=4.39\times 10^{-9}$$

For Object Distance we use: $$d_o=\frac{d_i}{M}$$

Using Thin Lens Equation: $$\frac{1}{f}=\frac{1}{d_o} +\frac{1}{d_i} $$ $$\frac{1}{80mm}=\frac{1}{(\frac{d_i}{M})} +\frac{1}{d_i} $$ $$ \frac{1}{80mm}=\frac{M}{d_i} +\frac{1}{d_i} $$ $$\frac{1}{80mm}=\frac{M+1}{d_i} $$ Solving for $d_i$ we get $$ d_i=(M+1)80 mm$$ $$d_i=(M=4.39\times 10^{-9}+1)80mm$$

$$d_i\approx 80mm$$

now to find $d_o$ we use the following:

$$d_o=\frac{d_i}{M}$$ $$d_o=\frac{80mm}{4.39\times 10^{-9}}$$ $$d_o =18,200 km$$

I am off by 10,000 km, so I am trying to understand what information I am missing or needing to get a more accurate answer. I do know the photo is cropped so I assume that may play a role. Or I am not applying the equations correctly. Any assistance is greatly appreciated.

Answer 1

You made two mistakes: the image is indeed cropped, and you calculated the distance from the center of the earth instead of the surface.

Here is the original image:

The source is this nasa.gov page, near the bottom, which also says that the diameter of the earth on the negative is about 29mm. The height of the camera above the surface is therefore about $\displaystyle \frac{80\text{ mm}}{29\text{ mm}}\cdot 2 r_\text{earth} - r_\text{earth} \approx 29000\text{ km}$.

Answer 2

I could not find the photograph in my fifth edition of Hecht.

$\dots\, h_{\rm i}$ is the image height which seems to be around 56mm but it was 130mm for me.
You need to know the size of the image formed in the focal plane of the camera not the image size in the book.

You can save yourself time by approximating that the visual angle subtended at the lens by the Moon is equal to the visual angle subtended by the image formed in the focal plane of the camera.

Using the NASA webpage found by @benrg where it states
By measurement of the size of Earth's image in these photographs (29mm), they were taken at a distance of about 29,400 kilometres (15,900 nautical miles).
the calculation can be done as follows.

$\dfrac{2r}{r+h}=\dfrac if \Rightarrow h \approx 29,000\,\rm km$

Note that a $29\,\rm mm$ marble would be rated as being large.