# Confusion regarding Cartesian sign convention in optics [closed]

Here's a question I've been trying to solve:

A biconvex lens of focal length $15$ cm is in front of a plane mirror. The distance between the lens and the mirror is $10$ cm. A small object is kept at a distance of $30$ cm from the lens. At what distance from the plane mirror does the final image occur ?

I could solve a major part of the question, but my confusion occurs at the last step of the question. Consider the diagram:

The image which would have been formed at a distance of $30$ cm from the lens (at $I_1$) had the lens been absent, essentially acts as a virtual object for the plane mirror and a real image should be formed at a distance of $20$ cm (at $I_2$) from the plane mirror by the reflection of rays by the plane mirror.

BUT, the returning rays are further refracted by the lens, and it's here my doubt is. Applying the lens formula,

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

Substituting values,

$\frac1v - \frac{1}{-10}=\frac{1}{15}$

However, doing it this way gives me $v=-30$ cm, which is not the answer. The answer is actually $v=-6$ cm, that is the image forms at 6 cm before the lens, and thus at a distance of $10$ cm+$6$ cm=$16$ cm from the mirror.

The only way I see that happening is to take $u=+10$ cm, instead of $-10$ cm, as I have. But isn't that against the convention?

Could someone please clarify with regards to the correct convention to be used? Thank you.

## closed as off-topic by John Rennie, Jon Custer, Gert, Kyle Kanos, Emilio PisantyJan 3 '18 at 19:37

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