Mirrors lenses and image construction

An image point is the point where rays intersect after reflection or refraction: if the rays themselves intersect, the image is real, and if only their backward extensions intersect, the image is virtual.

Two principal rays are enough for a construction. A ray parallel to the principal optical axis passes through the focal point $F$ after reflection or refraction; a ray through $F$ leaves parallel to the axis. A ray through the optical centre of a thin lens keeps its direction. For a mirror, the third principal ray is directed to the vertex (pole): there it reflects symmetrically about the principal optical axis according to the law of reflection, so its direction changes; only the ray travelling along the principal optical axis itself is undeviated.

The image position follows from the mirror or thin-lens equation

$$\frac{1}{f}=\frac{1}{d_o}+\frac{1}{d_i},$$

where for a spherical mirror $f=R/2$ and $R$ is the radius of curvature.

The sign of $d_i$ encodes the nature of the image: $d_i>0$ means real (in front of the mirror, behind the lens), $d_i<0$ means virtual (behind the mirror, in front of the lens). The linear magnification is

$$m=-\frac{d_i}{d_o},$$

so $|m|>1$ means enlarged and $m<0$ means inverted.

Choose a sign convention first and keep it to the end: diverging elements have $f<0$, while the object distance is always $d_o>0$. Comparing the signs of $d_i$ and $m$ with your ray diagram is a fast check of the result.

A concave spherical mirror has a radius of curvature $R=40$ cm. An object of height $h_o=2$ cm is placed on the principal optical axis at a distance $d_o=60$ cm from the mirror vertex. Find the focal length, the image position, and the magnification.

A converging lens has focal length $f=10$ cm. An object of height $h_o=2$ cm is placed at a distance $d_o=15$ cm from the lens. Find the image distance, the magnification, and the image height; state where a screen must be placed to show a sharp image.

A convex (diverging) mirror with $|f|=15$ cm is used as a rear-view mirror. An object (a car) is at a distance $d_o=10$ cm from the mirror. Find $d_i$ and $m$, determine the nature and orientation of the image, and explain why such a mirror is convenient for a driver.

Self-study plan with a simulation (no hazardous light sources).

1. Open the PhET "Geometric Optics" simulation or a similar ray-tracing program.
2. Choose a concave mirror, place the object at $d_o>2f$, and lock the screen at the ray intersection; record $d_i$ from the simulation and compare it with $1/f=1/d_o+1/d_i$.
3. Repeat for $f<d_o<2f$ and for $d_o<f$; in the third case the rays do not intersect — note that the image is virtual and locate it using the backward extensions of the rays.
4. Switch to a converging and then a diverging lens and repeat the same three measurements; fill in a table: element type, $d_o$, $f$, $d_i$, sign of $d_i$, sign of $m$.
5. Compare the signs in the table with your ray diagrams and state a conclusion: how the signs of $f$, $d_i$, and $m$ predict the nature and orientation of the image.

Safety: do not send a laser pointer beam or sunlight through a lens — a concentrated beam is dangerous to the eyes and can start a fire. Perform all observations in the simulation only.

Show that the image position is set by the intersection of the principal rays, and link the geometry of the construction to the signs in the mirror or lens equation.

A sketch of a concave mirror on the principal optical axis: the mirror arc on the left, points $C$ and $F$ marked on the axis; the object as a vertical arrow at $d_o>2f$; three principal rays: parallel to the axis (reflected through $F$), through $F$ (reflected parallel to the axis), and into the mirror vertex (reflected symmetrically); their intersection gives an inverted image arrow between $F$ and $C$. Beside it, a smaller version of the same diagram for lenses: converging and diverging lenses drawn as the two-arrow symbol, focal points $F$ on both sides, the same principal rays, and dashed backward extensions of the rays for a virtual image.

Driving at night, you glance at a convex rear-view mirror with radius of curvature $R=1200\ \text{mm}$, so its focal length is $f=-R/2=-600\ \text{mm}$ (the focus is virtual, behind the mirror). A car behind you is at $d_o=20\ \text{m}$ from the mirror.

The mirror equation $\frac{1}{f}=\frac{1}{d_o}+\frac{1}{d_i}$ with $f=-0.60\ \text{m}$ and $d_o=20\ \text{m}$ gives $\frac{1}{d_i}=\frac{1}{-0.60}-\frac{1}{20}=-1.72\ \text{m}^{-1}$, so $d_i\approx -0.58\ \text{m}$: the image is virtual, behind the mirror. The magnification $m=-\frac{d_i}{d_o}=\frac{0.58}{20}\approx 0.029$ makes a $1.5\ \text{m}$ car appear about $4.4\ \text{cm}$ tall and upright — this uses the sign convention and the principal-ray rule that a ray parallel to the axis reflects as if it came from the virtual focus.

A smartphone camera uses a converging lens with $f=4.0\ \text{mm}$. You photograph a friend's face $d_o=0.50\ \text{m}=500\ \text{mm}$ from the lens; the sensor must sit where the image forms.

The thin-lens equation gives $\frac{1}{d_i}=\frac{1}{4.0\ \text{mm}}-\frac{1}{500\ \text{mm}}=0.248\ \text{mm}^{-1}$, so $d_i\approx 4.03\ \text{mm}$ — the sensor sits essentially at the focal plane. The magnification $m=-\frac{d_i}{d_o}\approx -0.0081$ makes the image real, inverted, and about $0.8\%$ of the object size; the field of view at that distance is about $62\ \text{cm}$ wide. This is exactly the ray construction: the ray parallel to the axis goes through the focal point, and the ray through the lens centre passes straight.

A concave makeup mirror in the bathroom has radius of curvature $R=500\ \text{mm}$, so $f=+250\ \text{mm}$. You hold your face at $d_o=150\ \text{mm}$ from the mirror, inside the focal length.

Using $\frac{1}{f}=\frac{1}{d_o}+\frac{1}{d_i}$: $\frac{1}{d_i}=\frac{1}{250}-\frac{1}{150}=\frac{6-10}{1500}=-0.00267\ \text{mm}^{-1}$, so $d_i=-375\ \text{mm}$ behind the mirror and $m=-\frac{d_i}{d_o}=+2.5$. The image is virtual, upright and $2.5$ times larger — an $8\ \text{mm}$ eyelash looks $20\ \text{mm}$ long. The construction uses two principal rays: one parallel to the axis reflects through $F$, one hitting the vertex reflects at equal angles; their backward extensions meet behind the mirror.