Power of a Lens

When you visit an optician and hear them say “your left eye needs a $+2.5$ D lens,” they are talking about the power of the lens. But what exactly does that number mean, and why is it more useful than simply stating the focal length? The answer lies in how strongly the lens bends light, and power gives us a single, clean number to express that bending ability.

What Lens Power Tells You

Every lens bends light that passes through it. A convex lens pulls parallel rays together (converges them), while a concave lens spreads them apart (diverges them). Some lenses bend light sharply, others only gently. The power of a lens is a direct measure of this bending: how much convergence or divergence the lens introduces into a beam of light.

Here is the key insight: a lens with a shorter focal length bends light more. Think about it this way. If a convex lens brings parallel rays to a focus just 10 cm away, it must be bending them quite steeply. Another convex lens that focuses the same rays 50 cm away is bending them much more gently. Similarly, a concave lens that makes rays diverge as though they came from a point only 10 cm behind it is spreading them far more aggressively than one whose virtual focus sits 50 cm back.

So focal length and bending strength are inversely related: shorter focal length means stronger bending.

The Formal Definition

Fig 9.18: Power of a lens, showing the convergence angle $\delta$ for a ray at height $h$ from the optical centre

Consider a ray of light travelling parallel to the principal axis and hitting the lens at a height $h$ above the optical centre O. After passing through the lens, this ray bends and heads toward the focus F, which sits at distance $f$ from O. The angle $\delta$ between the incoming direction and the outgoing ray measures how much the lens has bent the light.

From the geometry of the triangle formed by the ray, the optical centre, and the focus:

$\tan \delta = \frac{h}{f}$

Now, the power $P$ of the lens is defined as the value of $\tan \delta$ when $h = 1$ (that is, at unit distance from the optical centre):

$\text{When } h = 1: \quad \tan \delta = \frac{1}{f}$

For small angles (the paraxial approximation that we have been using throughout ray optics), $\tan \delta \approx \delta$, so:

$P = \frac{1}{f} \qquad \text{(9.25)}$

This is the definition of lens power. It is simply the reciprocal of the focal length.

The Dioptre: SI Unit of Lens Power

Since $P = 1/f$ and focal length is measured in metres, the unit of power comes out to $\text{m}^{-1}$ (inverse metres). This unit has its own name: the dioptre, abbreviated D.

$1\,\text{D} = 1\,\text{m}^{-1}$

A lens whose focal length is exactly 1 metre has a power of exactly 1 dioptre.

Sign Convention: Positive for Converging, Negative for Diverging

The sign of the power tells you what kind of lens you are dealing with:

• Converging (convex) lens: focal length is positive, so power is positive
• Diverging (concave) lens: focal length is negative, so power is negative

This makes practical sense. When an optician prescribes $+2.5$ D, the positive sign immediately signals a convex lens. The focal length is:

$f = \frac{1}{P} = \frac{1}{2.5} = 0.4\,\text{m} = +40\,\text{cm}$

A prescription of $-4.0$ D means a concave lens with:

$f = \frac{1}{-4.0} = -0.25\,\text{m} = -25\,\text{cm}$

The power notation is more convenient than quoting focal lengths because opticians and patients can immediately compare lens strengths without converting units. A $+3$ D lens bends light more than a $+1$ D lens, and a $-5$ D lens diverges more strongly than a $-2$ D lens. The number directly encodes the bending strength.

Worked Examples

Finding Power from Focal Length

Problem: A glass lens has $f = 0.5$ m. What is its power?

Solution:

$P = \frac{1}{f} = \frac{1}{0.5\,\text{m}} = +2\,\text{D}$

The positive value confirms this is a converging lens.

Finding Refractive Index from Curvatures and Focal Length

Problem: A double convex lens has radii of curvature 10 cm and 15 cm for its two faces. Its focal length is 12 cm. What is the refractive index of the glass?

Solution:

First, assign signs using the Cartesian convention. For a double convex lens, the first surface curves away from the incoming light (centre of curvature on the far side), so $R_1 = +10$ cm. The second surface curves toward the incoming light (centre of curvature on the near side), so $R_2 = -15$ cm. The focal length is positive: $f = +12$ cm.

Apply the lens maker’s formula (Equation 9.22):

$\frac{1}{f} = (n - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$

Substitute the values:

$\frac{1}{12} = (n - 1)\left(\frac{1}{10} - \frac{1}{-15}\right)$

Simplify the bracket. Subtracting a negative fraction is the same as adding:

$\frac{1}{10} - \frac{1}{-15} = \frac{1}{10} + \frac{1}{15}$

Find the common denominator (30):

$\frac{1}{10} + \frac{1}{15} = \frac{3}{30} + \frac{2}{30} = \frac{5}{30} = \frac{1}{6}$

Now the equation becomes:

$\frac{1}{12} = (n - 1) \times \frac{1}{6}$

Multiply both sides by 6:

$\frac{6}{12} = n - 1$

$0.5 = n - 1$

$n = 1.5$

The refractive index of the glass is 1.5.

Focal Length of a Lens in a Different Medium

Problem: A convex lens has focal length $+20$ cm in air. What is its focal length when placed in water? Given: refractive index of air-water = 1.33, refractive index of air-glass = 1.5.

Solution:

This problem shows a striking result: the same lens behaves very differently depending on the medium it sits in.

Step 1: Set up the air equation.

In air, the surrounding medium has $n_1 = 1$ (air) and the glass has $n_2 = 1.5$. The lens maker’s formula gives:

$\frac{1}{f_{\text{air}}} = (n_2 - n_1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$

$\frac{1}{20} = (1.5 - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) = 0.5\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$

From this, extract the curvature term:

$\frac{1}{R_1} - \frac{1}{R_2} = \frac{1}{0.5 \times 20} = \frac{1}{10}$

Step 2: Set up the water equation.

In water, $n_1 = 1.33$ and $n_2 = 1.5$ (the glass has not changed). The general form of the lens maker’s formula for a lens in a medium other than air is:

$\frac{n_1}{f_{\text{water}}} = (n_2 - n_1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \qquad \text{(9.26)}$

Substitute the known values:

$\frac{1.33}{f_{\text{water}}} = (1.5 - 1.33) \times \frac{1}{10}$

$\frac{1.33}{f_{\text{water}}} = 0.17 \times \frac{1}{10} = 0.017$

Step 3: Solve for $f_{\text{water}}$.

$f_{\text{water}} = \frac{1.33}{0.017} \approx 78.2\,\text{cm}$

The focal length nearly quadruples, jumping from 20 cm in air to about 78.2 cm in water. This happens because water’s refractive index (1.33) is much closer to that of glass (1.5) than air’s (1.0) is. With a smaller refractive index contrast, the lens bends light far less, producing a much longer focal length. This is the same reason a glass rod becomes nearly invisible when submerged in a liquid whose refractive index matches the glass.