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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!

enter image description here

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    $\begingroup$ although plane mirror has 0 power but that does not mean it won't help signs, since it reverses direction of light. $\endgroup$ Commented Nov 21, 2015 at 5:41
  • $\begingroup$ But how can we justify that using the formula that $P_{equivalent}=2P_{lens}+P_{mirror}$ ? @AnubhavGoel $\endgroup$
    – user74370
    Commented Nov 21, 2015 at 5:44
  • $\begingroup$ the correct formula actually to be usesd is $|P_{equivalent}|=2|P_{lens}|+|P_{mirror}|$ $\endgroup$ Commented Nov 21, 2015 at 6:35
  • $\begingroup$ this method cannot tell us the signs. For that we need to make either diagram or use some arguments. $\endgroup$ Commented Nov 21, 2015 at 6:43
  • $\begingroup$ @AnubhavGoel can you tell me from where you learnt that formula?Please. $\endgroup$
    – user74370
    Commented Nov 21, 2015 at 6:58

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This plano concave lens has the same object and image positions as your silvered plano convex lens. Hence the same focal length. The sign convention comes from the sign convention of the u and f distances.

enter image description here

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  • $\begingroup$ But without drawing the ray diagram how to tell whether focal length will be positive or negative? $\endgroup$
    – user74370
    Commented Nov 21, 2015 at 4:47
  • $\begingroup$ hyperphysics.phy-astr.gsu.edu/hbase/geoopt/lenseq.html See the link on the Cartesian sign convention. $\endgroup$
    – mmesser314
    Commented Nov 21, 2015 at 5:12
  • $\begingroup$ @sanchyan: because of mirror you know light would come back in front and because of a converging lens, light from infinity would be focused to a point forming a real image in front of mirror.Because final point sized image thus formed must be in front of mirror toward left side , its focal length would be negative. $\endgroup$ Commented Nov 21, 2015 at 5:38
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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.

The big difference is that the silvered lens produces a real image on the left-hand side of the lens, while a double-concave lens will produce a virtual image on the left-hand side of the lens.

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I don't know why nobody pointed about this, but a silvered plano-convex lens acts as a concave mirror. Silvering a lens makes it a mirror, then it depends which lens is mirrored to find what kind of mirror is formed.

Also, Focal length is half the radius of curvature of a sphere whose mirror is made, so whichever side the rays are meeting (actually or apparently) the focal length on that side is considered.

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