...

27-8. Two thin lenses L and L', of focal lengths f and f', are separated by a distance D. Find the equivalent focal length F of the combination, and the distances Δ, Δ' of the principal planes from the respective lenses L and L'.

—  R. B. Leighton , Feynman Lectures on Physics. Exercises

The Solutions give next explanation:

...

27.8. From the definition of the main planes of the optical system follows their important property: if the point is in the main plane at a distance y from the optical axis, then its image is obtained in the other principal plane at the same distance from the optical axis. We use this property of the principal planes to solve the problem. The image of the point located in the main plane is shown in the figure. The following relations are obvious:

${\displaystyle y=f\tan(\varphi )=F\tan(\Phi )}$
${\displaystyle y'=(D-f)\tan(\varphi )=(f'+x')\tan(\Phi )}$.

It follows from the lens equation for ${\displaystyle L'}$ also

${\displaystyle f'^{2}=x'(D-f-f')}$.

Eliminating from this the angles ${\displaystyle \varphi }$ and ${\displaystyle \Phi }$, as well as the distance ${\displaystyle x'}$, we obtain

${\displaystyle F={\tfrac {ff'}{D-f-f'}}}$
${\displaystyle \Delta '=f'+x'+F={\tfrac {f'D}{D-f-f'}}}$ .

From a similar consideration for the lens ${\displaystyle L}$:

${\displaystyle \Delta ={\tfrac {fD}{D-f-f'}}}$ .

—  MEPhI , Solutions (Google Translate)

But I don't understand why does MEPhI use both -- principal planes and lenses -- simultaneously . The principal planes are only for replacing the lens system, not for using simultaneously . Username160611000000 (talk) 06:02, 26 August 2018 (UTC)[reply]

If we draw the rays correctly PNG, we see that the ray will go through the path BNE'S' if the principal planes are used , or the path BEE'S' if the lenses are used. From this correct image I write down:
Consider light path BEE'S'.
for lens L':
${\displaystyle {\tfrac {1}{s'}}+{\tfrac {1}{s''}}={\tfrac {1}{f'}}}$
${\displaystyle {\tfrac {1}{s'}}+{\tfrac {1}{\infty }}={\tfrac {1}{f'}}}$
${\displaystyle {\tfrac {1}{s'}}={\tfrac {1}{f'}}}$
for lens L :
${\displaystyle {\tfrac {1}{F-\Delta }}+{\tfrac {1}{s'}}={\tfrac {1}{f}}}$
${\displaystyle {\tfrac {1}{F-\Delta }}+{\tfrac {1}{f'}}={\tfrac {1}{f}}}$
By analogy consider the light path B'G'GS:
${\displaystyle {\tfrac {1}{F-\Delta '}}+{\tfrac {1}{f}}={\tfrac {1}{f'}}}$.
Now we see
${\displaystyle {\tfrac {1}{F-\Delta '}}=-{\tfrac {1}{F-\Delta }}}$
${\displaystyle 2F=\Delta +\Delta '}$, which absolutely doesn't coincide with The Exercises answer and The Solution answer. Where is my mistake???
Username160611000000 (talk) 07:00, 26 August 2018 (UTC)[reply]

Magbetic flux density is quoted as nunber of lines per unit area coning our of the magnet. So how do these lines actually originate and is there flux between the lines? Is not the fkux equalky distivuted icer the surface of the poke?.80.2.20.210 (talk) 21:18, 26 August 2018 (UTC)[reply]

Here line is the old name for the Maxwell (unit), equal to 10^-8 Wb. Think of the lines, not as discrete curves with no cross-sectional area, but rather as fat bundles packed together and each carrying this much flux. --catslash (talk) 21:46, 26 August 2018 (UTC)[reply]

Wikipedia has an article titled Field line which may be useful here. --Jayron32 17:28, 27 August 2018 (UTC)[reply]

The lines of magnetic flux form closed loops. The path of a 'line' leaves the magnet at one pole, passes through the air to the other pole where it re-enters the magnet and passes through the metal back to the first pole (I assume that originate asks for a starting-point rather than for a reason for existing).

According to classical physics the flux is distributed smoothly (if not quite equally) over the surface of the pole, and not in discrete lines with gaps between them. In quantum physics the flux can be quantized (see magnetic flux quantum), but the quantum of flux is far smaller than one 'line' --catslash (talk) 23:01, 27 August 2018 (UTC)[reply]

Wht they call th leccye field effect flux,? I thot flux id sonthin flowing.. but leccy fields Dolby fliw anywhere. Tbe jyst exist?80.2.20.210 (talk) 21:22, 26 August 2018 (UTC)[reply]

It's just an analogy which -for the reason you mention- is less than perfect. See flux and the section Flux#Flux_as_a_surface_integral. --catslash (talk) 22:11, 26 August 2018 (UTC)[reply]

shoulda be called effective ekectic or magnetic field.as the permirtivity or permeability affwcts the field strength. Yes? 80.2.20.210 (talk) 23:57, 26 August 2018 (UTC)[reply]