Difficulty: Intermediate | Prerequisites: Snell's law, law of reflection, basic algebra, comfort with sign conventions for distances.
This section covers how curved mirrors and lenses form images, the equations that let you predict where images appear and how large they are, and how these elements combine in optical instruments like the eye, glasses, and microscopes. The thin-lens equation and the mirror equation are the same formula, but the sign conventions differ for mirrors versus lenses and for real versus virtual images. Getting comfortable with those conventions is the single biggest step toward solving these problems reliably. You should already understand reflection, refraction, and how light bends at a curved surface.
The thin-lens/mirror equation 1/f = 1/d_o + 1/d_i relates the focal length to the object and image distances. Magnification m = -d_i/d_o tells you the image size and orientation. Diverging lenses (f < 0) always produce virtual, upright, reduced images. Convex mirrors (f < 0) do the same. For optical instruments, know how to chain two lenses (the image of the first becomes the object of the second) and how refractive power in diopters relates to focal length (P = 1/f in metres).
Thin-lens equation / mirror equation
1/f = 1/d_o + 1/d_i. Works for both thin lenses and spherical mirrors with appropriate sign conventions. In simple terms, this one equation lets you find where the image forms for any lens or mirror, given the object distance and focal length.
Focal length (f)
The distance from the lens (or mirror) to the point where parallel incoming rays converge (converging lens/concave mirror, f > 0) or appear to diverge from (diverging lens/convex mirror, f < 0).
Magnification (m)
m = -d_i/d_o = h_i/h_o. A positive m means the image is upright; negative means inverted. |m| > 1 means enlarged; |m| < 1 means reduced.
Real image
Formed where light rays actually converge. Can be projected onto a screen. d_i is positive for lenses, positive for mirrors (in front of the mirror).
Virtual image
Formed where light rays appear to diverge from but do not actually pass through. Cannot be projected onto a screen. d_i is negative for lenses (same side as the object), negative for mirrors (behind the mirror).
Concave (converging) mirror
A mirror that curves inward, with a positive focal length. f = R/2, where R is the radius of curvature.
Convex (diverging) mirror
A mirror that curves outward, with a negative focal length. f = -R/2. Always produces virtual, upright, reduced images.
Concave (diverging) lens
A lens that is thinner in the middle, with a negative focal length. Always produces virtual, upright, reduced images regardless of object position.
Convex (converging) lens
A lens that is thicker in the middle, with a positive focal length. Can produce real or virtual images depending on the object distance.
Diopter (D)
The unit of refractive power. P = 1/f, where f is in metres. A positive diopter value means a converging lens; negative means diverging.
Near point
The closest distance at which the eye can focus on an object. For a normal eye this is about 25 cm. Farsighted people have a near point further away.
Tube length (L) of a microscope
The distance between the second focal point of the objective lens and the first focal point of the eyepiece. Standard value is often 160 mm. This is used to calculate the objective's image distance.
A diverging (concave) lens has f < 0. The image is always virtual (d_i < 0), upright (m > 0), and reduced (|m| < 1).
Magnification: m = -d_i/d_o. For a diverging lens, since d_i is negative, m = -(-|d_i|)/d_o = |d_i|/d_o, which is positive (upright) and less than 1.
To find the object distance for a given magnification, use m = f/(f + d_o) for a diverging lens (derived from combining the lens equation with the magnification formula). Alternatively: from 1/f = 1/d_o + 1/d_i and m = -d_i/d_o → d_i = -m × d_o, substitute into the lens equation.
f = -34 cm. Currently m = 1/3.
From m = -d_i/d_o and 1/f = 1/d_o + 1/d_i: d_i = -m × d_o. Substituting: 1/(-34) = 1/d_o + 1/(-d_o/3) = 1/d_o - 3/d_o = -2/d_o. So d_o = -2 × (-34) = 68 cm.
Now find d_o for m = 1/4: 1/(-34) = 1/d_o - 4/d_o = -3/d_o → d_o = -3 × (-34) = 102 cm.
The object must move from 68 cm to 102 cm from the lens, a shift of 34 cm further to the left.
Answer: A (moved to the left by 34 cm)
Step 1: find the image from lens 1 using 1/f₁ = 1/d_o1 + 1/d_i1.
Step 2: the image from lens 1 becomes the object for lens 2. The object distance for lens 2 is the separation between the lenses minus d_i1 (if the image is between the lenses) or calculated from the image position.
Step 3: use 1/f₂ = 1/d_o2 + 1/d_i2 to find the final image.
f₁ = +20.0 cm, object is 50.0 cm to the left of lens 1.
1/20 = 1/50 + 1/d_i1 → 1/d_i1 = 1/20 - 1/50 = (5 - 2)/100 = 3/100 → d_i1 = 33.33 cm (to the right of lens 1).
Lens 2 is 90.0 cm to the right of lens 1. Object distance for lens 2: d_o2 = 90.0 - 33.33 = 56.67 cm.
Final image is 359.0 cm from the original object. The original object is 50.0 cm left of lens 1, and lens 2 is 90.0 cm right of lens 1, so lens 2 is 140.0 cm from the original object. The final image is at 359.0 - 140.0 = 219.0 cm to the right of lens 2. So d_i2 = 219.0 cm.
1/f₂ = 1/56.67 + 1/219.0 = 0.01765 + 0.004566 = 0.02221 → f₂ = 45.0 cm.
Answer: D (45.0 cm)
A convex mirror has f = -R/2 (negative focal length). All images are virtual (behind the mirror), upright, and reduced.
As the object moves from very close to infinitely far, the image moves from just behind the mirror surface toward the focal point. The image distance d_i approaches -|f| = -R/2 but never reaches it.
The greatest distance behind the mirror that an image can form is R/2, approached as d_o → ∞.
Answer: B (R/2)
For a concave mirror, the mirror equation 1/f = 1/d_o + 1/d_i applies with f = R/2 > 0.
Magnification: m = -d_i/d_o = h_i/h_o.
The inverted image of a tree is 3.8 cm high and located 7.0 cm in front of the mirror. The tree is 23 m = 2300 cm from the mirror.
|m| = d_i/d_o = 7.0/2300 = 0.003043.
h_object = h_image / |m| = 3.8 cm / 0.003043 = 1249 cm ≈ 12.5 m.
Answer: D (12 m)
Q13: The image is upright, on the same side as the object, and larger. From the figure, the object is in front of the mirror and the image appears behind it (virtual). An upright, enlarged, virtual image is produced by a concave mirror when the object is between the mirror and the focal point. Using the distances from the scale drawing, you can find R from the mirror equation.
Q14: The image is always upright, and moving the object further away makes the image smaller. This is the signature of a convex mirror. Convex mirrors always produce virtual, upright, reduced images, and the image gets smaller as the object recedes. A flat mirror also gives upright images but the size stays the same. A concave mirror with the object inside f gives upright enlarged images that get larger, not smaller, as you back away (until you pass f). So the answer is a convex mirror.
A farsighted person cannot focus on close objects. Their near point is further than 25 cm. A converging lens (positive diopters) brings the near point to 25 cm.
Refractive power P = 1/f (f in metres). P = 1.20 D means f = 1/1.20 = 0.8333 m = 83.33 cm.
The glasses take an object at 25.0 cm (the desired near point) and produce a virtual image at the person's actual near point (d_i < 0, since the image is on the same side as the object for a simple magnifier/reading glasses configuration).
Using the lens equation: 1/f = 1/d_o + 1/d_i → 1/83.33 = 1/25.0 + 1/d_i → 1/d_i = 1/83.33 - 1/25.0 = 0.01200 - 0.04000 = -0.02800 → d_i = -35.71 cm.
The actual near point is |d_i| = 35.71 cm. Wait, this does not match the choices. Let me reconsider: the near point is the distance from the eye where the virtual image forms. So the person's actual near point is |d_i| ≈ 36 cm.
Actually checking: the person places the object at 25 cm. The lens creates a virtual image at the person's near point N (which is further away). 1/f = 1/25 + 1/(-N) → 1/83.33 = 1/25 - 1/N → 1/N = 1/25 - 1/83.33 = 0.04 - 0.012 = 0.028 → N = 35.7 cm.
Hmm, that gives about 36 cm. But checking the listed answers (36, 49, 60, 72, 90), the answer is A (36 cm). Let me re-examine with f in cm: P = 1.20 D = 1/(f in m) → f = 83.33 cm. 1/83.33 = 1/25 - 1/N → 1/N = 0.04 - 0.012 = 0.028 → N = 35.7 ≈ 36 cm.
Answer: A (36 cm)
A compound microscope has an objective lens that forms a real, magnified image of the specimen inside the tube. The tube length L = 160 mm is the distance between the objective's back focal point and the eyepiece's front focal point.
The image distance for the objective is d_i = f_obj + L = 8.0 + 160 = 168 mm.
Using the lens equation: 1/f = 1/d_o + 1/d_i → 1/8.0 = 1/d_o + 1/168 → 1/d_o = 1/8.0 - 1/168 = 0.1250 - 0.005952 = 0.11905 → d_o = 8.4 mm.
The objective should be 8.4 mm from the specimen.
Answer: C (8.4 mm)
Q15: To trace rays through a two-lens system, first find the image from lens 1 using the standard three principal rays (parallel ray, focal ray, centre ray). Then treat that image as the object for lens 2 and trace three new principal rays through lens 2.
The final image location depends on where the first image falls relative to lens 2's focal points.
Q16: If the object is moved further to the left (further from lens 1), the image from lens 1 moves closer to lens 1's focal point (toward f₁ on the right side). This changes the object distance for lens 2, which shifts the final image. For a converging first lens, as the object goes to infinity, the first image approaches f₁. Whether the final image moves left, right, or stays put depends on the specific geometry. In many standard configurations, moving the object further left causes the final image to move to the right.
\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \quad \text{(thin-lens / mirror equation)}m = -\frac{d_i}{d_o} = \frac{h_i}{h_o} \quad \text{(magnification)}f = \frac{R}{2} \quad \text{(mirror focal length from radius of curvature; positive for concave, negative for convex)}P = \frac{1}{f} \quad \text{(refractive power in diopters, } f \text{ in metres)}d_i = f_{\text{obj}} + L \quad \text{(microscope objective image distance, } L = \text{tube length)}Every pair of prescription glasses uses the thin-lens equation. The optometrist measures your eye's focusing error and prescribes a lens power in diopters to correct it. Positive diopters for farsightedness, negative for nearsightedness.
Compound microscopes use two lenses in series (objective + eyepiece) to achieve very high magnification. The tube length is standardised so that objectives and eyepieces from different manufacturers are interchangeable.
Students often confuse real and virtual images. A real image forms where rays converge (d_i > 0 for lenses). A virtual image forms where rays appear to diverge from (d_i < 0 for lenses). Virtual images cannot be projected onto a screen, but you can see them by looking through the lens or into the mirror.
Students sometimes think a diverging lens can produce a real image. It cannot. Diverging lenses and convex mirrors always produce virtual images.
Students often get the sign of f wrong. For diverging lenses and convex mirrors, f is negative. Forgetting the negative sign leads to completely wrong image positions.
Students sometimes confuse the near point with the focal length of the eye's lens. The near point is a distance from the eye; the focal length of the corrective lens is a separate quantity determined by the lens equation.
⚠️ Sign conventions are the most common source of errors. For lenses: d_o > 0 (object on the incoming side), d_i > 0 for real images (opposite side), d_i < 0 for virtual images (same side). For mirrors: d_o > 0 (in front), d_i > 0 for real images (in front), d_i < 0 for virtual (behind).
⚠️ Two-lens problems require you to complete the full calculation for lens 1 before starting lens 2. The most common mistake is using the wrong object distance for the second lens.
⚠️ For convex mirrors, know that the image distance is always between 0 and R/2 behind the mirror. This is a common conceptual multiple-choice question.
⚠️ Microscope problems: the tube length L is not the full tube, but the distance between the objective's back focal point and the eyepiece's front focal point.
True or false: A convex mirror can produce a real image. (False. Convex mirrors always produce virtual images.)
Fill in the blank: A lens with a refractive power of +2.00 D has a focal length of ______ cm. (50 cm)
True or false: For a diverging lens, the magnification is always less than 1 but greater than 0. (True.)
Fill in the blank: The mirror equation for a concave mirror with radius of curvature 20 cm gives f = ______ cm. (10 cm)
True or false: In a two-lens system, if the image from lens 1 falls at the focal point of lens 2, the final image is at infinity. (True.)
Q: An object is placed 30 cm from a converging lens with f = 20 cm. Where does the image form, and what is the magnification?
A: 1/20 = 1/30 + 1/d_i → 1/d_i = 1/20 - 1/30 = 1/60 → d_i = 60 cm. m = -60/30 = -2. The image is 60 cm on the far side, inverted, and twice the object height.
Q: An object is placed 15 cm from a diverging lens with f = -30 cm. Where is the image?
A: 1/(-30) = 1/15 + 1/d_i → 1/d_i = -1/30 - 1/15 = -1/30 - 2/30 = -3/30 = -1/10 → d_i = -10 cm. Virtual, upright, on the same side as the object, reduced (m = -(-10)/15 = 2/3).
Q: A concave mirror has f = 12 cm. An object is 8 cm from the mirror. Describe the image.
A: 1/12 = 1/8 + 1/d_i → 1/d_i = 1/12 - 1/8 = (2 - 3)/24 = -1/24 → d_i = -24 cm. The image is virtual (behind the mirror), upright, and enlarged (m = -(-24)/8 = 3).
Q: A nearsighted person has a far point of 2.0 m. What lens power (in diopters) is needed to correct their distance vision?
A: The lens must take an object at infinity and produce a virtual image at 2.0 m (the far point). 1/f = 1/∞ + 1/(-2.0) = -0.50 D. The prescription is -0.50 D.
The thin-lens equation is derived from Snell's law applied at two curved surfaces, so it connects directly back to refraction. Understanding sign conventions here also prepares you for more advanced optics courses.
Vision correction ties into biology and human anatomy. The eye is itself a converging lens system, and corrective lenses modify the total system focal length.
Microscope and telescope optics extend the two-lens system concept. If you understand how to chain two lenses, you can handle any compound optical instrument.
Thin lens equation, mirror equation, 1/f = 1/do + 1/di, magnification, m = -di/do, focal length, converging lens, diverging lens, concave lens, convex lens, concave mirror, convex mirror, real image, virtual image, radius of curvature, f = R/2, diopter, refractive power, near point, far point, farsighted, hyperopia, nearsighted, myopia, corrective lens, vision correction, compound microscope, objective lens, eyepiece, tube length, two-lens system, ray diagram, principal rays, sign convention, PHYS 116, Physics 116, UW, University of Washington, midterm 2, practice exam 2B