1
$\begingroup$

I am just totally confused about the rule sign of convex and concave lenses. The general formula: $$1/v-1/u=1/f.$$

Is okay but when solving problem sums sometimes $v$ becomes negative sometimes $u$ and many times $f$. Sometimes both $v$ and $f$ becomes negative. I just can't mug up why this is happening.

Can anyone please explain this to me.

$v$ = image distance $u$ = object distance $f$ = focal length of lens.

$\endgroup$
1
  • $\begingroup$ A more general formula for the "secondary" focus of the lens would be $$\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2}{f_s}$$ where $n_1$ and $n_2$ are the refractive indices of the media on the right and left sides of the lens. The one you posted holds true only for $n_1 = n_2$. $\endgroup$ Aug 15 '20 at 16:20
0
$\begingroup$

Just think in terms of their properties. A concave lens diverges the rays incident on it. So the image will be formed behind the lens. So the image distance becomes negative. (The image distance is measured from the principal point to the point at which the image is formed.)
enter image description here

Now the convex lens converges the incident rays. Hence the rays converge to a single point infront of the lens. So the image distance is positive here.
enter image description here

you can easily verify that the focal length is in the positive x direction for a convex lens while it is along the negative x direction for a concave lens. So f is positive for a convex lens and negative for a concave lens. The reason is the sign altering of v.

$\endgroup$
3
$\begingroup$

I have a great deal of sympathy with your position. As taught in schools the sign convention is somewhat vague and that gets confusing with complicated setups.

If you're dealing with anything complicated I recommend keeping strictly to the Cartesian sign convention. I've linked an article that seems to be a good summary of this, but Googling will find you lots more articles to look at. Basically:

  • the light travels from left to right

  • to the left of the lens is negative

  • to the right of the lens is positive

  • converging lenses have positive $f$, diverging lenses have negative $f$

This convention means $u$ is normally negative because the object is to the left of the lens, while $v$ is normally positive because a (real) image is to the right of the lens. The lens equation becomes:

$$ \frac{1}{v} - \frac{1}{u} = \frac{1}{f} $$

$\endgroup$
2
  • $\begingroup$ Thanks. Cleared a bit of my doubts. But still I'm not confident. Maybe practicing will let me learn some new stuff. I will also look up to the Cartesian sign as you told. Thanks again. :) $\endgroup$ Oct 8 '15 at 11:03
  • $\begingroup$ Not only are some explanations vague, but most US introductory physics texts and older optics texts use a different sign convention based on object/entering light side being positive and image/exiting light side being positive. It would take a severe revolution to switch everyone to the Cartesian system, but in the long run, it might be worth the fight. Anybody up for it? (remember VHS vs Beta?) $\endgroup$
    – Bill N
    Feb 2 '16 at 16:13
0
$\begingroup$

See according to the New Cartesian Sign convention U i.e., the object distance is always in the left hand side of the lens or mirror i.e., to -X side. That is why U is always taken as negative. For focal length, f in lens is always taken as negative for concave and positive for convex. In case of mirror the f is taken as negative if it is in -X side i.e., is in Concave mirror and taken as positive if it is in +X side i.e., is in Convex mirror. And for image distance, V in lens it is taken as positive in Convex lens since image is formed on +X side. It is taken as negative in Concave lens since image is formed in -X side of the Cartesian. In Concave mirror V is taken as positive and negative for convex mirror.

$\endgroup$
0
$\begingroup$

This is a ray tracing convention you can google it.

Basically everything to the left and down is negative, everything to the right and up is positive. Curvatures are measured from the vertex to the center of the curvature.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.