Finding image positions with a biconvex lens using ray diagrams and calculations

  • Thread starter Thread starter questions_uk
  • Start date Start date
  • Tags Tags
    Optics
Join the discussion
Registration is free. Start your own thread to ask a follow-up.
Sign up Log in
33 replies · 6K views
Thanks.
 
Physics news on Phys.org
Hi there. Just a quick question regarding an equation. In order to find the transverse magnification of an object 10 mm high and 5 mm long, it gives the equation:

fo / xo = xi / fi = - yi / yo giving xo x xi / fo x fi

Now which part of that is supposed to be used to calculate the transverse magnification?

Am not sure why some variables are equated to other variables e.g. xo x xi = fo x fi

Thanks.
 
questions_uk said:
Hi there. Just a quick question regarding an equation. In order to find the transverse magnification of an object 10 mm high and 5 mm long, it gives the equation:

fo / xo = xi / fi = - yi / yo giving xo x xi / fo x fi

Now which part of that is supposed to be used to calculate the transverse magnification?

Am not sure why some variables are equated to other variables e.g. xo x xi = fo x fi
The definition of transverse magnification is:
m ≡ yi/yo

It can be shown (using similar triangles) that:
m ≡ yi/yo = -si/so (Where si is the image distance; so is the object distance. Both measured from the lens.)

If you measure distances from the focal points (xi & xo are the image and object distances measured from the focal points), then you can show (using the thin lens equation) that:
xi * xo = f*f

m ≡ yi/yo = -xi/f = -f/xo

Using that last term is the easy way to calculate the transverse magnification.