几何光学

A straw in a glass of water looks bent. A shop's round security mirror shows the whole aisle, but small. A magnifying glass 放大镜 makes print larger, yet moved a little further away, the same lens turns the room upside down. One rule for bending light and one equation explain all of these.

Geometric optics 几何光学 treats light as straight-line rays. The law of reflection 反射定律: when light bounces off a surface, the angle of incidence 入射角 equals the angle of reflection 反射角, both measured from the normal 法线 (the line perpendicular to the surface). Smooth surfaces reflect a clear image; rough surfaces scatter the rays.

For rays near the principal axis 主光轴, a spherical mirror 球面镜 makes parallel rays converge at a real focal point 焦点, or appear to diverge from a virtual focus, a distance $f$ from the mirror. The mirror equation 面镜公式 relates object and image distances:

A concave mirror 凹面镜 (converging) can form a real image 实像 (rays actually meet, projectable, inverted 倒立); a convex mirror 凸面镜 always forms a virtual image 虚像 (rays only appear to meet, upright 正立, reduced). The magnification 放大率 $m$ gives the image's size and orientation. Sign convention 符号规则 matters: with the real-is-positive convention, a negative image distance means a virtual image. It lies behind a mirror, or on the object side of a lens.

Light bends when it passes between materials because its speed changes – refraction 折射. Each material has an index of refraction 折射率 $n=\dfrac{c}{v}$ (how much it slows light). Snell's law 斯涅尔定律:

At oblique incidence, light entering a higher-index medium 介质 bends toward the normal. Beyond the critical angle 临界角, light going from a higher to a lower refractive index reflects entirely – total internal reflection 全反射, used in optical fibres 光纤.

Worked example. A ray in air ($n_1=1.00$) strikes water ($n_2=1.33$) at $40^{\circ}$ to the normal. By Snell's law,

The ray bends toward the normal, as expected going into the denser medium.

Worked example (critical angle). For light trying to leave glass ($n=1.50$) for air, total internal reflection begins at the critical angle $\theta_c$ where the refracted ray grazes the surface:

Any ray hitting the inside surface more steeply than $42^{\circ}$ is trapped – the reason optical fibres carry light for kilometres.

What changes, and what does not. Entering glass, light slows to $v=c/n$ and its wavelength 波长 shrinks to $\lambda_n=\lambda/n$. The frequency 频率 stays the same: it is set by the source, and the waves cannot pile up at the boundary. So the colour of light, which depends on frequency, does not change inside the glass.

A lens 透镜 bends light through refraction. The thin-lens equation 薄透镜方程 has the same form as the mirror equation:

A converging 会聚 (convex) lens can form a real, inverted image or, for a close object, a virtual, upright, enlarged image (a magnifying glass). A diverging 发散 (concave) lens always forms a virtual, upright, reduced image. Ray diagrams 光路图 locate the image quickly using two easy rays: one parallel to the axis that bends to pass through the far focus, and one straight through the centre of the lens (undeviated 不偏折).

Worked example. An object sits $30\ \text{cm}$ from a converging lens of focal length 焦距 $f=10\ \text{cm}$. Find the image. From the thin-lens equation,

and the magnification is $m=-d_i/d_o=-15/30=-0.5$. The image is real (positive $d_i$), inverted (negative $m$), and half the object's size – just what a camera lens does.

Every mirror and lens problem uses the same equation, so the signs carry the physics.

Quantity	Positive	Negative
$f$	converging lens, concave mirror	diverging lens, convex mirror
$d_i$	real image (can be projected)	virtual image (cannot be projected)
$m$	upright image	inverted image
$	m	$

Worked example (a magnifying glass). An object sits $6.0\ \text{cm}$ from a converging lens with $f=10\ \text{cm}$, inside the focal length.

The image is virtual (negative $d_i$, on the same side as the object), upright and enlarged 2.5 times: a magnifier. Move the object outside $f$ and $d_i$ turns positive: the image becomes real and inverted, as in the camera example above.

Worked example (two mirrors). A concave mirror has $f=+20\ \text{cm}$ and an object $60\ \text{cm}$ away: $\frac{1}{d_i}=\frac{1}{20}-\frac{1}{60}=\frac{1}{30}$, so $d_i=+30\ \text{cm}$ and $m=-0.5$, a real, inverted, half-size image. A convex security mirror has $f=-15\ \text{cm}$ and a shopper $30\ \text{cm}$ away: $\frac{1}{d_i}=-\frac{1}{15}-\frac{1}{30}=-\frac{1}{10}$, so $d_i=-10\ \text{cm}$ and $m=+0.33$. The image is virtual, upright and small, which is why the whole aisle fits in the mirror.

Worked example (a diverging lens). An object stands $24\ \text{cm}$ from a diverging lens with $f = -12\ \text{cm}$.

The image is virtual, upright and one-third the size, $8.0\ \text{cm}$ from the lens on the object's side. Whatever the object distance, $d_i$ stays negative and $|d_i| < |f|$: a diverging lens always makes a small, upright, virtual image between the lens and its focal point.

The ray diagram checks the algebra: the image sits a third of the way from the lens to the object, and it is a third of the object's height.

Answer each before reading the answers below. Take $c=3.0\times10^8\ \text{m/s}$.

1. Light passes from air into glass ($n=1.5$) at $30°$ to the normal. Find the angle of refraction.
2. Find the speed of light in that glass.
3. Diamond has $n=2.42$. Find its critical angle with air, and explain why diamonds sparkle.
4. An object is $20\ \text{cm}$ from a converging lens with $f=15\ \text{cm}$. Find the image distance and magnification, and describe the image.
5. A student says: "A diverging lens can project an image onto a screen if the object is placed carefully." Correct the student.

1. $\sin\theta_2=\dfrac{1.0}{1.5}\sin30°=0.333$, so $\theta_2=19.5°$, toward the normal.
2. $v=\dfrac{c}{n}=\dfrac{3.0\times10^8}{1.5}=2.0\times10^8\ \text{m/s}$.
3. $\sin\theta_c=\dfrac{1}{2.42}=0.413$, so $\theta_c=24°$. Such a small critical angle means most light that enters is totally internally reflected several times before it escapes, so diamonds flash brightly.
4. $\dfrac{1}{d_i}=\dfrac{1}{15}-\dfrac{1}{20}=\dfrac{1}{60}$, so $d_i=+60\ \text{cm}$ and $m=-3$: real, inverted and three times larger, as in a projector.
5. A diverging lens has negative $f$, so $d_i$ is negative for every real object: the image is always virtual and cannot be formed on a screen.

• Measure all angles from the normal, not the surface.
• Light entering a denser medium (larger $n$) bends toward the normal; use Snell's law $n_1\sin\theta_1=n_2\sin\theta_2$.
• Total internal reflection happens only going into a less dense medium, beyond the critical angle ($\sin\theta_c=1/n$).
• For mirrors and lenses use $\tfrac1{d_o}+\tfrac1{d_i}=\tfrac1f$ and $m=-d_i/d_o$: a positive $d_i$ is a real image, negative is virtual.
• $n=c/v$, so a bigger index means a slower speed of light in the material.

Common mistakes

• Measuring angles from the surface in Snell's law. Use the angle to the normal; if the question gives the angle to the surface, subtract it from $90^{\circ}$.
• Dropping the minus sign of a diverging lens or convex mirror focal length. Put $f < 0$ into the equation, or the image comes out real.
• Forgetting to invert at the end: $\dfrac{1}{d_i} = \dfrac{2}{30}$ means $d_i = 15\ \text{cm}$, not $0.067\ \text{cm}$.
• Saying the frequency (colour) of light changes in glass. Speed and wavelength drop by the factor $n$; frequency does not change.
• Expecting total internal reflection when light goes from air into glass. It needs light travelling toward a lower index, at more than the critical angle.
• Drawing the ray through the centre of a lens with a bend. It goes straight on; only the parallel ray and the focal ray bend.