# Confused regarding sign convention of silvered plano convex lens

I'm totally confused while finding the sign convention of a silvered plano convex lens. I know that equivalent power of such a system can be found by adding the power of the mirror and twice the power of the lens.cBut say in this example in the picture below the equivalent focal length has been considered negative. WHY? Isn't the focal length of a plano convex lens positive if the object distance is taken to be negative? Obviously power of a mirror is 0 so it can't affect the signs!

• although plane mirror has 0 power but that does not mean it won't help signs, since it reverses direction of light. – Anubhav Goel Nov 21 '15 at 5:41
• But how can we justify that using the formula that $P_{equivalent}=2P_{lens}+P_{mirror}$ ? @AnubhavGoel – user74370 Nov 21 '15 at 5:44
• the correct formula actually to be usesd is $|P_{equivalent}|=2|P_{lens}|+|P_{mirror}|$ – Anubhav Goel Nov 21 '15 at 6:35
• this method cannot tell us the signs. For that we need to make either diagram or use some arguments. – Anubhav Goel Nov 21 '15 at 6:43
• @AnubhavGoel can you tell me from where you learnt that formula?Please. – user74370 Nov 21 '15 at 6:58

The effect of the mirrored plane surface is the same as doubling the lens thickness and reversing its sign (making it effectively double-concave). This cuts the focal length in half but makes the focal length negative (pretty much the way @mmesser314 described). So: $$1/f_{eq} = -2/f_{pc} = 1/D_o + 1/D_i$$, where $f_{pc}$ is the focal length of the original plano-convex lens without silver, $1/D_o$ is the distance to the object, and $1/D_i$ is the distance to the image.