# Should power of lens be dimensionless quantity?

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 ?

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

$$P = 1/f$$, where $$f$$ is in meters. Thus the dimension of $$p$$ is $$m^{-1}$$.
• The book is being a little loose, saying $h=1$ and then ignoring $h$. Actually $P = tan(\theta)/h$. Commented Nov 24, 2023 at 16:41