Lens formula optical power and optical instruments

A thin lens links the positions of object and image by the lens equation: $\frac{1}{f}=\frac{1}{d_o}+\frac{1}{d_i},$ where $f$ is the focal length, $d_o$ the object distance, and $d_i$ the image distance. For a real object and a real image these quantities are positive; a virtual image gives $d_i<0$, while a diverging lens has $f<0$. Optical power $D=\frac{1}{f}$ is measured in diopters (m$^{-1}$) and, for thin lenses in close contact, powers add: $D=D_1+D_2$. The linear magnification $\Gamma=\left|\frac{d_i}{d_o}\right|=\left|\frac{h_i}{h_o}\right|$ tells how many times the image is larger than the object. Drawing two rays — through the optical centre and parallel to the principal axis — lets you check the position and type of the image without calculation, just as you did for mirrors. Optical instruments are combinations of lenses: a telescope made of two converging lenses with focal lengths $f_1$ (objective) and $f_2$ (eyepiece) has angular magnification $M=\frac{f_1}{f_2}$ and tube length $L=f_1+f_2$, while a magnifying glass enlarges the image because the object is placed closer than the focal length.

A converging lens has focal length $f=10\ \text{cm}$. An object stands at a distance $d_o=30\ \text{cm}$ from the lens. Find the image distance, the optical power, and the magnification.

A lens has optical power $D=+4\ \text{D}$. An object is placed $40\ \text{cm}$ from it. Where is the image, what is the magnification, and is the image real?

In a telescope the distance between objective and eyepiece is $L=105\ \text{cm}$, and the angular magnification is $M=20$. Find the focal lengths and optical powers of both lenses. Where does the objective form the image of a distant object?

Work with a geometric-optics simulation (for example, PhET «Geometric Optics») — it is safer and more precise than experiments with real lenses.

1. Choose a converging lens and record pairs of $d_o$ and $d_i$ for 6–8 object positions.
2. Plot $\frac{1}{d_i}$ against $\frac{1}{d_o}$. The points should lie on a straight line with slope $-1$, and the intercept on the vertical axis equals $\frac{1}{f}$.
3. Obtain $f$ from the graph and compute the optical power $D=\frac{1}{f}$; compare it with the lens value given in the simulation.
4. Repeat with a diverging lens and explain the sign of the resulting $d_i$.

Safety warning: never point a lens at the Sun and never use a laser pointer — focused light irreversibly damages the retina.

Show the geometric meaning of the lens equation: how object distance, focal length, and optical power determine the position, size, and type of the image.

Diagram: a horizontal principal optical axis, a thin lens at the centre, focal points $F$ and $F'$ at distance $f$ on either side; an object arrow at distance $d_o$; two rays — one parallel to the axis that passes through $F'$ after refraction, and one through the optical centre — intersect to give the image arrow at distance $d_i$.

Inset on the right: a graph of $\frac{1}{d_i}$ against $\frac{1}{d_o}$ — a straight line with slope $-1$ and intercept $\frac{1}{f}$ on the vertical axis.

A phone camera has a lens of focal length $f = 4.0\ \text{mm}$, with the sensor at distance $d_i$ from the lens. The phone first focuses on a friend's face at $d_o = 1.0\ \text{m}$, then on a flower at $d_o = 0.20\ \text{m}$. From the lens formula $\frac{1}{d_i} = \frac{1}{f} - \frac{1}{d_o}$ we get $d_i = 4.016\ \text{mm}$ in the first case and $d_i = 4.082\ \text{mm}$ in the second. The lens therefore has to move by only $66\ \mu\text{m}$ — exactly what the tiny autofocus motor does.

This is the thin-lens formula at work: $\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}$ shows that with a fixed $f$ the image distance $d_i$ grows as the object comes closer. The transverse magnification $m = -\frac{d_i}{d_o}$ gives the image size: a face $0.10\ \text{m}$ wide at $1.0\ \text{m}$ becomes a $0.40\ \text{mm}$ strip on the sensor.

From the stands a fan watches a $0.5\ \text{m}$ flag on the far side of the pitch, $100\ \text{m}$ away. The binocular objective has $f_o = 20\ \text{cm}$ and the eyepiece $f_e = 2.0\ \text{cm}$, so the tube is about $f_o + f_e = 22\ \text{cm}$ long. To the naked eye the flag subtends $\frac{0.5}{100} = 0.005\ \text{rad}$; through the binoculars it subtends $0.005 \cdot 10 = 0.05\ \text{rad} \approx 2.9^\circ$.

The lens formula is applied twice. For a distant object $d_o \gg f_o$, so $\frac{1}{d_i} \approx \frac{1}{f_o}$ and the objective forms a real image essentially in its focal plane, $20\ \text{cm}$ away. The eyepiece then acts as a magnifier: its object (that image) sits in its focal plane, and the instrument's angular magnification is $M = \frac{f_o}{f_e} = \frac{20}{2.0} = 10\times$.

In the kitchen the recipe lies on the table $25\ \text{cm}$ from the eyes, but this person can read comfortably only at $50\ \text{cm}$. The optician prescribes glasses of optical power $D = +2.0\ \text{D}$, that is $f = \frac{1}{D} = +0.50\ \text{m}$. Substituting $d_o = 0.25\ \text{m}$: $\frac{1}{d_i} = \frac{1}{0.50} - \frac{1}{0.25} = 2 - 4 = -2\ \text{m}^{-1}$, hence $d_i = -0.50\ \text{m}$.

A negative $d_i$ means a virtual image: it lies $0.50\ \text{m}$ behind the lens, on the same side as the recipe, and the eye focuses on it without strain. This uses both the sign convention inside the lens formula and the link between optical power and focal length, $D = \frac{1}{f}$: a value $D = +2.0\ \text{D}$ means a converging lens with $f = +0.50\ \text{m}$.