Geometrical optics, lens and image

[fluidistic]

Homework Statement

What should be the focal distance of a negative thin lens such that there's a virtual image situated at 50 cm from the lens, of an ant at 100 cm from the lens? Given that the ant is at the right side of the lens, localize and describe the image.

Homework Equations

1/f=1/s_o+1/s_i.

The Attempt at a Solution

Using the equation above I found that the focal distance is 33.33 cm.
Using my knowledge rather than math, the image is at the right of the lens (same side as the ant), it's virtual as they said and it's erect.
However I don't really like my description since I treated the ant as a point rather than a 2 dimensional object, hence the dislike of using the word "erect". It does not really make sense to me.

Is what I've done good?

 

[ehild]

The image distance is negative if the image is virtual. The focal length of a "negative" (concave ) lens is negative. So your result is wrong.
That ant can be small, and still has a head above its legs. Draw the special rays arising from the head. If they intersect in a point above the image of the legs, the image is erect.

ehild

 

[fluidistic]

The image distance is negative if the image is virtual. The focal length of a "negative" (concave ) lens is negative. So your result is wrong.
That ant can be small, and still has a head above its legs. Draw the special rays arising from the head. If they intersect in a point above the image of the legs, the image is erect.

ehild

Thanks a lot for the information.
Apparently I wasn't aware of the sign conventions. I've done the sketch you suggested. The 2 special rays leaving the head of the ant "cross" in the right side of the lens and the image is erect. Well actually the prolongation of the diverged ray cross the other ray over the ground.

But the formula used is still the same? I mean 1/f=1/s_o+1/s_i.
I take both s_0 and s_i negative because they are in the same side with respect to the lens?
I'm very confused on this.
I get f=(-1/3)m if both are negative.

 

[ehild]

But the formula used is still the same? I mean 1/f=1/s_o+1/s_i.
I take both s_0 and s_i negative because they are in the same side with respect to the lens?
I'm very confused on this.
I get f=(-1/3)m if both are negative.

The object distance is positive every time when the object is real. (the opposite can occur only when you have two or more lenses.) So

1/f (cm) =-1/50+1/100.

ehild

 

[fluidistic]

Ok thanks a lot.
So http://library.thinkquest.org/C003776/ingles/course/lenses.htm is plainly wrong? They say "

# The distance from the center of the lens to the object will be referred to as do.
# The distance from the center of the lens to the image will be referred to as di.

since a distance is always positive, regardless of the characteristics of the image.
I'll have acess to Hecht's book soon, I hope I'll understand why your affirmation stands

The object distance is positive every time when the object is real. (the opposite can occur only when you have two or more lenses.)

.

 

[ehild]

Well, you are right, di and do are not distances but positions with respect to the lens, and they have sign. But they are called "distance". Usually the distance of the image is given as the real distance from the lens, (a positive number ) but you need to use it with a negative sign if the image is virtual. But there can be different sign conventions.
The same for the "focal length". It is negative for a concave lens and positive for the convex ones.

See wikipedia for example:
http://en.wikipedia.org/wiki/Lens_(optics)

ehild