My book defines the power of lens using this way:

The power P of a lens is defined as the tangent of the angle by which it converges or diverges a beam of light falling at unit distant from the optical centre $tan(x)=h/f$ and for small $x$ , $x=h/f$ , $h=1, x=1/f$. Thus power p of a lens is $p=1/f$.

The SI unit for power of a lens is dioptre (D): $1D = 1m^{–1}$. The power of a lens of focal length of 1 metre is one dioptre. Power of a lens is positive for a converging lens and negative for a diverging lens. Thus, when an optician prescribes a corrective lens of power $+ 2.5 D$, the required lens is a convex lens of focal length $+ 40 cm$. A lens of power of $– 4.0 D$ means a concave lens of focal length $– 25 cm$.

If power P of a lens is defined as a tangent of angle of convergence or divergence , the tangent is a dimensionless quantity! So its unit should not be $m^{-1}$ or $D$. Taking h=1 does not mean we negate its dimension right ?

  • $\begingroup$ Please consider using text instead of pictures, pictures get broken or lost in the internet very easily. $\endgroup$
    – Mauricio
    Commented Nov 23, 2023 at 18:45
  • $\begingroup$ Okay , just give me a few minutes @Mauricio $\endgroup$
    – Razz
    Commented Nov 23, 2023 at 18:47

1 Answer 1


Power is not the tangent of an angle. That would indeed be dimensionless.

$P = 1/f$, where $f$ is in meters. Thus the dimension of $p$ is $m^{-1}$.

  • $\begingroup$ Then the definition given in the book shouldn't be taken seriously, have they given it for intuitive purposes ? $\endgroup$
    – Razz
    Commented Nov 24, 2023 at 7:28
  • $\begingroup$ The book is being a little loose, saying $h=1$ and then ignoring $h$. Actually $P = tan(\theta)/h$. $\endgroup$
    – mmesser314
    Commented Nov 24, 2023 at 16:41

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