What Power of a Lens Means
The power of a lens is a measure of the convergence or divergence that the lens introduces into light falling on it. A lens of shorter focal length bends incident light more strongly, converging it in the case of a convex lens and diverging it in the case of a concave lens. NCERT defines the power $P$ as the tangent of the angle by which the lens deviates a ray that runs parallel to the principal axis and strikes it at unit distance from the optical centre.
For a ray at height $h$ that emerges to cross the axis at the focus, $\tan\delta = h/f$, and for a ray at unit distance ($h = 1$) the deviation angle for small $\delta$ gives $\tan\delta = 1/f$. This leads directly to the defining relation:
$$ P = \frac{1}{f} $$Power equals the reciprocal of focal length, with $f$ measured in metres.
The figure below shows the same parallel ray meeting two lenses of different focal length: the shorter focal length produces the sharper bend and therefore the larger power.
The Dioptre and Sign Convention
The SI unit of power is the dioptre, written D, where $1\,\text{D} = 1\,\text{m}^{-1}$. A lens whose focal length is exactly one metre has a power of one dioptre. Because power is the reciprocal of focal length, the sign of $P$ follows the sign of $f$ under the Cartesian sign convention: a converging convex lens has positive focal length and so positive power, while a diverging concave lens has negative focal length and so negative power.
This is why an optician's prescription is just a signed number in dioptres. NCERT gives two concrete readings: a corrective lens of power $+2.5\,\text{D}$ is a convex lens of focal length $+40\,\text{cm}$, while a lens of power $-4.0\,\text{D}$ is a concave lens of focal length $-25\,\text{cm}$.
| Quantity | Symbol / relation | Sign and value |
|---|---|---|
| Power of a lens | P = 1/f | $f$ in metres; unit dioptre (D) |
| Dioptre | 1 D = 1 m⁻¹ | power of a 1 m focal length lens |
| Convex (converging) lens | $f > 0$ | $P$ positive, e.g. $f=+40$ cm → $+2.5$ D |
| Concave (diverging) lens | $f < 0$ | $P$ negative, e.g. $f=-25$ cm → $-4.0$ D |
Centimetres sneak into the dioptre formula
The single most common slip is leaving the focal length in centimetres. A convex lens of focal length $20\,\text{cm}$ has power $P = 1/0.20\,\text{m} = +5\,\text{D}$, exactly the answer in NEET 2022 Q.34. Writing $1/20$ gives $0.05$, off by a factor of 100. Equally important: keep the sign. A concave lens never has positive power, so any answer that gives a diverging lens a $+$ power is wrong by inspection.
Convert $f$ to metres first, then take the reciprocal, then attach the sign of $f$: convex $\Rightarrow +P$, concave $\Rightarrow -P$.
Power from the Lens Maker's Formula
Power can also be written directly in terms of the lens geometry. Combining $P = 1/f$ with the lens maker's formula gives a single expression linking power to the refractive index and the two radii of curvature:
$$ P = \frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) $$
For a biconvex lens of equal radii $R_1 = R$ and $R_2 = -R$ this becomes $P = (\mu-1)\,(2/R)$. With $\mu = 1.5$ and $R = 20\,\text{cm}$, the focal length is $20\,\text{cm}$ and the power is $+5\,\text{D}$, which is precisely the route NEET 2022 Q.34 expects. The relation is also the bridge that lets a layered or cemented lens be treated as a sum of powers, the situation in NEET 2023 Q.48.
Power leans entirely on focal length. Revisit how $f$ is derived from index and curvature in Lens Maker's Formula before attempting power problems.
Combination of Thin Lenses in Contact
Consider two thin lenses $A$ and $B$ of focal lengths $f_1$ and $f_2$ held in contact. An object beyond the focus of lens $A$ first forms an image at $I_1$; that image acts as the object for lens $B$, which forms the final image at $I$. Because the lenses are thin, their optical centres are taken to coincide at a single point $P$.
Applying the thin-lens relation $\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$ to each lens in turn and adding the two equations, the intermediate image distance cancels. The pair then behaves like one equivalent lens of focal length $f$ obeying:
$$ \frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2} + \cdots \qquad\Longleftrightarrow\qquad P = P_1 + P_2 + \cdots $$
The derivation extends to any number of thin lenses in contact, so the powers of all the lenses simply add. The sum is algebraic: convex lenses contribute positive terms and concave lenses negative terms. This is what lets a combination be engineered to a desired total power, converging or diverging, and to sharpen the final image.
Powers add only in contact, and only with their signs
$P = P_1 + P_2$ is valid only when the lenses are in contact, so their optical centres coincide. If a question separates the lenses by a distance $d$, the rule changes to $P = P_1 + P_2 - d\,P_1 P_2$ and the separation cannot be ignored, as in NEET 2021 Q.12. The second pitfall is forgetting signs: a convex lens of focal length $f$ in contact with a concave lens of focal length $f$ gives $\tfrac{1}{f}-\tfrac{1}{f}=0$, so the equivalent focal length is infinite, the exact answer to NEET 2023 Q.50.
Add powers directly only for lenses in contact; carry the $+$/$-$ sign of each lens; a separation $d$ forces $P = P_1 + P_2 - d\,P_1 P_2$.
Net Magnification of a Combination
Powers add, but magnifications do not. Since the image formed by the first lens becomes the object for the second, each lens multiplies the size of what reaches it. The net linear magnification of a combination is therefore the product of the individual magnifications:
$$ m = m_1 \cdot m_2 \cdot m_3 \cdots $$
This product rule is exactly what designers exploit in cameras, microscopes and telescopes, where several lenses are stacked to reach a large overall magnification. It also resolves the 2025 camera question directly: four identical lenses each of power $p$ and magnification $m$ give a net power $p+p+p+p = 4p$, but a net magnification $m\cdot m\cdot m\cdot m = m^4$.
| Property | Lenses in contact | Rule type |
|---|---|---|
| Reciprocal focal length | $\dfrac{1}{f} = \dfrac{1}{f_1}+\dfrac{1}{f_2}+\cdots$ | Algebraic sum |
| Power | $P = P_1 + P_2 + \cdots$ | Algebraic sum (with signs) |
| Magnification | $m = m_1 \cdot m_2 \cdots$ | Product |
| $n$ identical lenses (power $p$, mag $m$) | $P = np,\;\; M = m^{\,n}$ | Sum vs product |
Worked Examples
A glass lens has focal length $f = 0.5\,\text{m}$. What is its power?
With $f$ already in metres, $P = \dfrac{1}{f} = \dfrac{1}{0.5} = +2\,\text{D}$. The positive sign marks it as a converging lens.
Two thin convex lenses of focal lengths $20\,\text{cm}$ and $40\,\text{cm}$ are in contact. Find the equivalent focal length and power.
$\dfrac{1}{F} = \dfrac{1}{20} + \dfrac{1}{40} = \dfrac{3}{40}$, so $F = \dfrac{40}{3} = 13.3\,\text{cm} = 0.133\,\text{m}$. The power is $P = \dfrac{1}{F} = \dfrac{1}{0.133} = +7.5\,\text{D}$. Equivalently, $P = P_1 + P_2 = 5 + 2.5 = +7.5\,\text{D}$, the same answer by adding powers.
A combination of three thin lenses (focal lengths $+10\,\text{cm}$, $-10\,\text{cm}$ and $+30\,\text{cm}$) is arranged along an axis with separations between them. Where does the final image form?
Lens by lens, the image of the first lens at $v_1 = 15\,\text{cm}$ becomes the object for the second; the second lens sends the rays to infinity ($v_2 = \infty$); for the third lens an object at infinity gives a final image at $v_3 = 30\,\text{cm}$ to its right. NCERT uses this case to stress that when lenses are separated, the image of one is the object of the next and the simple sum-of-powers shortcut does not apply.
Power and combinations at a glance
- Power $P = 1/f$ with $f$ in metres; SI unit dioptre, $1\,\text{D} = 1\,\text{m}^{-1}$.
- $P$ is positive for a convex (converging) lens and negative for a concave (diverging) lens.
- From the lens maker's formula, $P = (\mu-1)\left(\tfrac{1}{R_1}-\tfrac{1}{R_2}\right)$.
- Thin lenses in contact: $\tfrac{1}{f} = \tfrac{1}{f_1}+\tfrac{1}{f_2}+\cdots$ and $P = P_1 + P_2 + \cdots$ (algebraic sum).
- Net magnification multiplies: $m = m_1 m_2 m_3 \cdots$; $n$ identical lenses give $P = np$ but $M = m^{\,n}$.
- Powers add directly only in contact; with a gap $d$, $P = P_1 + P_2 - d\,P_1 P_2$.