A constant part of a photo taken

[marcnn]

Homework Statement

(56th Polish Olympiad in Physics, II stage)

A photographer has a camera with a lens of focal $f$ with can be set to a value from the interval $[f_{min}, f_{max}]$. The diameter of the diaphragm is $d$.

The photographer wants to make a photo of a friend so that the friend's face is sharp and takes up the half of the photo's height. Besides a building, lying the distance $l$ behind the friend should be maximally blurred. Which focal should be used if

- $d = \text{const}$
- $d/f = \mathrm{const}$

Homework Equations

+ the attempt at a solution[/B]

A part of the solution: we can consider only points on the optical axis. Let $x_A < x_B$ be the distances of points $A$ and $B$ from the lens and $y_A, y_B$ the distances from the lens of their images in the optical system.

Now the solution suggests that the problem description (actually the part underlined) implies that the augmentation of the face has to be constant, i.e. $$\frac {y_A}{x_A} = p = \text{const}$$
Why does it imply this?

 

[Aceix]

Can you post the original question or a link to it?

 

[BvU]

So the condition is that the image size is constant (or rather: a given, namely half the picture height).
The object size (the head) is also fixed.
Do you have an expression for the enlargment factor of a lens ? Happens to be y_A/x_A in this exercise,

 

[marcnn]

It's the original problem: http://www.kgof.edu.pl/archiwum/56/of56-2-1-R.pdf
Is it what you meant?

Yes, it was given in the opening post and is $\frac {y_A}{x_A} = p$

 

[Aceix]

It's the original problem: http://www.kgof.edu.pl/archiwum/56/of56-2-1-R.pdf
Is it what you meant?

Yes, it was given in the opening post and is $\frac {y_A}{x_A} = p$

Yes, thanks