Difficulty: Intermediate | Prerequisites: Wave optics basics (interference, wavelength), trigonometry, algebra.
Geometric optics treats light as rays that travel in straight lines, reflect off surfaces, and bend (refract) at boundaries between materials. This is the framework for understanding mirrors, lenses, cameras, eyeglasses, and the human eye. It builds directly on Snell's law and the law of reflection, then layers in the thin-lens and mirror equations. If wave optics explains colour and fringe patterns, geometric optics explains where images form and how big they are.
Reflection doubles any mirror rotation. Snell's law lets you track a ray through any number of flat boundaries. The thin-lens and mirror equations (1/f = 1/d_o + 1/d_i) locate images and give magnification. Corrective lenses for the eye are just thin-lens problems where the object and image distances match the eye's defective near or far point.
Law of reflection
The angle of incidence equals the angle of reflection, both measured from the normal to the surface. In simple terms, a ray bounces off a mirror at the same angle it arrived.
Snell's law (law of refraction)
n_1 sin(theta_1) = n_2 sin(theta_2). Describes how a ray bends when crossing a boundary between two media with different refractive indices. Think of it as light changing direction because it changes speed.
Converging (convex) lens
A lens that bends parallel rays inward to meet at a focal point on the far side. Has a positive focal length. The standard magnifying glass.
Diverging (concave) lens
A lens that spreads parallel rays apart so they appear to come from a focal point on the same side as the incoming light. Has a negative focal length. Used in corrective glasses for short-sightedness.
Concave mirror
A mirror whose reflecting surface curves inward (like the inside of a spoon). Converges light, positive focal length. Used in shaving mirrors and telescopes.
Convex mirror
A mirror whose reflecting surface curves outward. Diverges light, negative focal length. Always produces a virtual, upright, reduced image. Used in car side mirrors.
Focal length (f)
The distance from a lens or mirror to its focal point. Positive for converging elements, negative for diverging ones.
Real image
An image formed where rays actually converge. Can be projected onto a screen. Always on the opposite side of a lens from the object (or the same side as the object for a concave mirror when the object is beyond f).
Virtual image
An image formed where rays appear to diverge from but do not actually pass through. Cannot be projected onto a screen. Appears behind a mirror or on the same side of a lens as the object.
Magnification (M)
The ratio of image height to object height: M = h_i / h_o = -d_i / d_o. Positive M means the image is upright; negative M means inverted.
Diopter (D)
The unit of lens power: P = 1/f, where f is in metres. A 2.0 D lens has a focal length of 0.50 m. This is how optometrists describe lens strength.
Near point
The closest distance at which the eye can focus comfortably. Normal near point is about 25 cm.
Far point
The farthest distance at which the eye can focus. For a normal eye, the far point is at infinity. A near-sighted person has a finite far point.
Angle of incidence = angle of reflection, both measured from the normal.
Key result: if the mirror is rotated by an angle theta, the reflected ray rotates by 2 * theta. This follows directly from the geometry of the normal rotating with the mirror.
At each boundary, n_1 sin(theta_1) = n_2 sin(theta_2).
For multiple parallel layers (e.g. air, oil, water), you can skip intermediate layers entirely: n_air sin(theta_air) = n_water sin(theta_water). The oil's refractive index affects the angle inside the oil but not the final angle in the water, because Snell's law chains through every boundary and the intermediate terms cancel.
This is a common exam shortcut. If you are asked only for the angle in the final medium, apply Snell's law directly from the first medium to the last.
1/f = 1/d_o + 1/d_i, where d_o is the object distance and d_i is the image distance.
Sign conventions (lenses):
d_o is positive when the object is on the incoming-light side.
d_i is positive when the image is on the opposite side from the object (real image), negative when on the same side (virtual image).
f is positive for converging lenses, negative for diverging lenses.
Magnification: M = -d_i / d_o. Positive M = upright image; negative M = inverted image. |M| > 1 = enlarged; |M| < 1 = reduced.
When the object is inside the focal length of a converging lens (d_o < f), the image is virtual, upright, and enlarged. This is the magnifying-glass configuration.
Same equation: 1/f = 1/d_o + 1/d_i.
Sign conventions (mirrors):
d_o is positive when the object is in front of the mirror.
d_i is positive when the image is in front of the mirror (real), negative when behind (virtual).
f is positive for concave mirrors, negative for convex mirrors.
A concave mirror producing a virtual image means d_i is negative and the image appears behind the mirror.
If a virtual image is three times the height of the object, then M = +3 (upright), so -d_i/d_o = 3, giving d_i = -3 * d_o.
Solve lens 1 first using the thin-lens equation to find its image.
That image becomes the object for lens 2. The object distance for lens 2 is the separation between the lenses minus the image distance from lens 1 (if the image falls between the lenses).
If the image from lens 1 falls beyond lens 2, the object distance for lens 2 is negative (virtual object).
The total magnification is M_total = M_1 x M_2.
Myopia (short-sightedness): The far point is too close. The corrective lens must take an object at infinity and form a virtual image at the person's far point. P = 1/f = -1/(far point in metres). The power is always negative (diverging lens).
Hyperopia (far-sightedness): The near point is too far. The corrective lens must take an object at the desired near point (usually 25 cm) and form a virtual image at the person's actual near point. Use 1/f = 1/d_o + 1/d_i with d_o = +25 cm and d_i = -(actual near point). The power is positive (converging lens).
With a corrective lens in place, the new near point can be found by setting d_i = -(original near point) and solving for d_o using the lens's known focal length.
n_1 \sin\theta_1 = n_2 \sin\theta_2 \quad (\text{Snell's law})\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \quad (\text{thin lens and mirror equation})M = \frac{h_i}{h_o} = -\frac{d_i}{d_o} \quad (\text{magnification})P = \frac{1}{f} \quad (\text{lens power in diopters, } f \text{ in metres})M_{\text{total}} = M_1 \times M_2 \quad (\text{two-lens system magnification})Snell's law governs fibre-optic cables (total internal reflection keeps light trapped inside the fibre). The thin-lens equation is the basis for camera design, projector optics, and corrective eyewear. Concave mirrors are used in reflecting telescopes, satellite dishes, and car headlamps. Your optometrist's prescription in diopters is just 1/f.
Students often mix up the sign conventions between lenses and mirrors. For lenses, a positive d_i is on the opposite side from the object; for mirrors, a positive d_i is on the same side as the object (in front of the mirror). The equations are the same, but "positive image distance" means different physical locations.
When a converging lens has the object inside the focal point, the image is virtual and on the same side as the object. Students frequently force d_i to be positive and get a nonsensical answer.
In Snell's law through multiple layers, students sometimes apply the law at each boundary sequentially. That works, but the shortcut (first medium directly to last medium) is faster and less error-prone for parallel layers.
For corrective-lens problems, the virtual image must be at the eye's defective point. Students sometimes put the wrong sign on d_i or forget that the image needs to be on the same side as the object (virtual).
Mirror rotation problems (reflected ray rotates by 2 * theta) are quick conceptual marks.
Multi-layer refraction problems test whether you realise the intermediate layers cancel. Do not waste time computing the angle in each layer unless the question specifically asks for it.
Two-lens problems and concave-mirror problems are the most calculation-heavy questions on this exam. Get the sign conventions right first, then the algebra is straightforward.
Corrective-lens problems always appear. Know the setup for myopia (diverging lens, far point) and hyperopia (converging lens, near point) cold.
True or False: If a plane mirror rotates by 10 degrees, the reflected ray rotates by 10 degrees.
False. The reflected ray rotates by 20 degrees (2 * theta).
Fill in the blank: For a converging lens, an object placed inside the focal length produces a ______, ______, ______ image.
Virtual, upright, enlarged.
True or False: A diverging lens can produce a real image.
False. A diverging lens always produces a virtual image (for a real object).
Fill in the blank: A near-sighted person needs a ______ (converging/diverging) corrective lens.
Diverging.
True or False: When light passes through air, then oil, then water (all flat layers), you can find the angle in water using n_air sin(theta_air) = n_water sin(theta_water) directly.
True.
Q: A ray reflects from a plane mirror at 27 degrees incidence. If the mirror is rotated by angle theta, through what angle does the reflected ray rotate?
A: The reflected ray rotates by 2 * theta. The initial angle of incidence is irrelevant to this relationship. Answer: 2 * theta (B).
Q: Oil (n = 1.48) floats on water (n = 1.33). Light hits the oil at 60.0 degrees to the vertical from air (n = 1). Find the angle in the water.
A: Apply Snell's law directly from air to water: (1) sin(60.0) = (1.33) sin(theta). sin(theta) = sin(60.0)/1.33 = 0.8660/1.33 = 0.6511. theta = arcsin(0.6511) = 40.6 degrees. Answer: 40.6 degrees (D).
Q: A 2.50 cm tall object is 10.0 cm to the left of a converging lens with f = 30.0 cm. Describe the image.
A: 1/d_i = 1/f - 1/d_o = 1/30 - 1/10 = (1 - 3)/30 = -2/30. So d_i = -15 cm (negative means same side as object, virtual). M = -d_i/d_o = -(-15)/10 = +1.5. Image height = 1.5 x 2.50 = 3.75 cm, upright. The image is 15 cm to the left of the lens, 3.75 cm tall, and upright. Answer: (D).
Q: Two lenses (f_1 = +20.0 cm, f_2 = +30.0 cm) are separated by x = 115 cm. Object is 50.0 cm left of lens 1. Find the total magnification.
A: Lens 1: 1/d_i1 = 1/20 - 1/50 = (5 - 2)/100 = 3/100, so d_i1 = 33.3 cm. M_1 = -33.3/50 = -0.667. Object distance for lens 2: d_o2 = 115 - 33.3 = 81.7 cm. Lens 2: 1/d_i2 = 1/30 - 1/81.7 = (81.7 - 30)/(30 x 81.7) = 51.7/2451 = 0.02109, so d_i2 = 47.4 cm. M_2 = -47.4/81.7 = -0.580. M_total = (-0.667)(-0.580) = 0.387. Answer: approximately 0.38 (B).
Q: A concave mirror projects an image of a lightbulb onto a wall. Mirror is 1.8 m from bulb and 3.5 m from wall. What is the focal length?
A: d_o = 1.8 m, d_i = 3.5 m (real image, projected on wall). 1/f = 1/1.8 + 1/3.5 = 0.556 + 0.286 = 0.841. f = 1.19 m, approximately 1.2 m. Answer: 1.2 m (B).
Q: A concave mirror makes a virtual image three times as tall as the object. Object is 28 cm away. Find f.
A: M = +3 (virtual, upright, enlarged), so -d_i/d_o = 3, giving d_i = -3(28) = -84 cm. 1/f = 1/28 + 1/(-84) = (3 - 1)/84 = 2/84. f = 42 cm. Answer: 42 cm (E).
Q: A near-sighted person has a near point of 12 cm and a far point of 40 cm. What corrective lens power is needed for distant vision, and what is the new near point?
A: For distant vision: the lens must image infinity at the far point. P = -1/0.40 = -2.5 D. New near point: 1/d_o = 1/f + 1/d_i. With the lens (f = -0.40 m) and the eye's near point as d_i = -0.12 m: 1/d_o = 1/(-0.40) + 1/(-0.12) = -2.5 + (-8.33) = -10.83. That gives d_o negative, which means we need to reconsider: 1/d_o = 1/f - 1/(near point) is not quite right. The correct setup: the lens must form a virtual image at the eye's near point (12 cm from the eye, so d_i = -12 cm = -0.12 m). 1/d_o = 1/f + 1/|d_i| ... Let us use the standard approach: 1/f = 1/d_o + 1/d_i with d_i = -0.12 m and f = -0.40 m. 1/d_o = 1/f - 1/d_i = -2.5 - (-8.33) = -2.5 + 8.33 = 5.83. d_o = 0.171 m = 17.1 cm. Answer: -2.5 D, 17 cm (D).
Q: A farsighted person has a near point of 50.0 cm. What lens strength brings the near point to 25.0 cm?
A: The lens must take an object at 25.0 cm and form a virtual image at the person's near point of 50.0 cm. d_o = 25.0 cm = 0.25 m, d_i = -50.0 cm = -0.50 m. 1/f = 1/0.25 + 1/(-0.50) = 4.0 - 2.0 = 2.0 D. Answer: 2.0 D (C).
Snell's law and total internal reflection connect to fibre optics and waveguide physics. The thin-lens equation is the foundation for compound optical instruments (microscopes, telescopes) covered later. Vision correction ties into biology and the anatomy of the eye. Two-lens systems are a stepping stone to understanding multi-element camera lenses and projectors.
Snell's law, law of reflection, refraction, converging lens, diverging lens, concave mirror, convex mirror, focal length, thin lens equation, mirror equation, magnification, real image, virtual image, diopter, lens power, myopia, hyperopia, near point, far point, corrective lens, two-lens system, sign convention, object distance, image distance, upright, inverted, enlarged, reduced