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Consider a point source of light placed $40$ cm to the right of a thin converging lens of focal length $15$ cm. We wish to find the distance of its image from the lens.

We can use the lens formula:

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

Putting $f=+15$ cm and $u=-40$ cm,

$$v=\frac{-15\cdot 40}{-25}\text{ cm}=+24 \text{ cm}$$

This implies that the image will be formed $24$ cm to the left of the lens.

My question: we have defined a coordinate system where the center of the lens is the origin, and the object is on the negative x-axis. What happens if we take its distance from the lens to be positive, i.e., $+40$ cm?

My intuition is that the focal length will remain positive, i.e., $+15$ cm, for this reason: there is no change in the value of the focal length derived from the lens maker's formula, which is:

$$\frac{1}{f}=(\mu-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$$

When we take the object to be on the negative x-axis, $R_2>0$ and $R_1<0$. When we take the object to be on the positive x-axis, $R_2<0$ and $R_1>0$. So it is obvious, from the formula at least, that the focal length doesn't change whether we take the object distance to be positive or negative.

But how can this be? Taking the positive object distance and the same focal length,

$$v=\frac{15\cdot 40}{55} \text{ cm}=10.9\text{ cm}$$

This, contrary to our previous calculation, implies that the image will form $10.9$ cm to the right of the lens.

My textbook says that only the former value is correct, which is to say, you have to take the object distance to be negative. Why is this so?

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  • $\begingroup$ Where did you get the first equation? What are the sign conventions for using it correctly? Equations and sign conventions in simple geometrical optics are NOT universal. You must read the book which states them very carefully and apply them very carefully. If you get a sign wrong, you're doomed! $\endgroup$
    – Bill N
    Commented Nov 2, 2020 at 20:16
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    $\begingroup$ If 𝑅2>0 and R1<0, wouldn't your equation for focal length lead to f<0? $\endgroup$
    – Not_Einstein
    Commented Nov 2, 2020 at 22:18

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Is this a problem involving a so-called "virtual object?" Usually the light is considered going from left to right. If the light is converging to a point that would be to the right of the lens if it weren't there, then it's considered a virtual object and you use a negative object distance (negative of the distance from the lens to where the light would converge).

This would be a different situation from having an actual object the same distance to the left of the lens, in which case the object distance is positive, and we would expect a different result.

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  • $\begingroup$ No, it's a real object. Even so, why can't we take its distance from the lens to be positive? $\endgroup$
    – Polemos
    Commented Nov 2, 2020 at 16:36
  • $\begingroup$ You describe a "different" situation where the actual object's distance from the lens is positive (I presume you meant that the object is to the right of the lens, not left). What if we take its distance to be negative instead? How would we have to adjust the focal length so that we get the same image distance as we did when the object distance was positive? $\endgroup$
    – Polemos
    Commented Nov 2, 2020 at 16:39
  • $\begingroup$ If you haven't arrived at an answer to your question yet, perhaps you can tell us the sign convention used in your text with the two equations for 1/f in your original post. $\endgroup$
    – Not_Einstein
    Commented Nov 3, 2020 at 18:57

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