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 is formed after refracting twice

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 Pisanty Jan 3 '18 at 19:37

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Correct sign convention IS NOT exactly Cartesian.

The right way to think of directions in such cases is following:

  • All the distances in the direction of travelling ray are taken as positive.

  • All the distances against of the ray propagation are negative.

In your question, the ray which is entering the lens after after reflecting is travelling to the left. So the distances from the Pole of the lens to the left are positive. That's why you need to put the value as positive. When you do that you get as +6cm as per 'this' ray propagation direction. But if you go as per the original ray, this distance is to the left of the lens and thus negative regardless of the direction of the ray propagation.


Yes, indeed you can use u=+6cm since the lens is now considering the virtual image formed by the plane mirror as it's object. So you can assume it to be the same and there will be no violation of sign convention.


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