Difficulty: Intermediate to Advanced | Prerequisites: Snell's law, law of reflection, basic algebra, comfort with sign conventions.
This is the largest topic block on Midterm 2. Lenses and mirrors take the ray-bending rules from Snell's law and reflection and package them into a single equation (the thin lens or mirror equation) that tells you where an image forms, how large it is, and whether it is real or virtual. Vision correction then applies the same lens equation to the human eye, treating the eye as a lens system with a fixed image distance. If you can handle the sign conventions and the thin lens equation, every problem in this section follows the same pattern.
The thin lens equation (1/f = 1/s + 1/s') and the mirror equation (identical in form) let you find image location and size for any single lens or mirror. Sign conventions decide everything: positive or negative focal length, image distance, and magnification each tell you whether the image is real or virtual, upright or inverted. Vision correction problems are the same equation applied to the eye, with corrective lens power added to the eye's own power.
Thin lens equation
1/f = 1/s + 1/s', where f is the focal length, s is the object distance, and s' is the image distance. This single equation, paired with sign conventions, handles every thin-lens problem. In simple terms, it tells you where the image lands given where the object sits and what the lens does.
Mirror equation
Identical in form to the thin lens equation: 1/f = 1/s + 1/s'. The sign conventions differ slightly (image distance is positive in front of the mirror for real images).
Magnification (m)
m = -s'/s = h'/h. A positive m means the image is upright; a negative m means inverted. |m| > 1 means the image is larger than the object; |m| < 1 means smaller. In simple terms, magnification tells you how much bigger or smaller the image is and which way up it stands.
Focal length (f)
The distance from the lens (or mirror) to the point where parallel incoming rays converge (converging) or appear to diverge from (diverging). Positive for converging lenses and concave mirrors; negative for diverging lenses and convex mirrors.
Real image
An image formed where light rays actually converge. It can be projected onto a screen. The image distance s' is positive for a real image (in the standard lens convention).
Virtual image
An image formed where light rays appear to diverge from, but do not actually pass through that point. It cannot be projected onto a screen. The image distance s' is negative.
Myopia (nearsightedness)
A vision defect where the far point is at a finite distance rather than infinity. The eye's lens is too powerful at rest, so distant objects focus in front of the retina. Corrected with a diverging (negative power) lens.
Presbyopia
Age-related loss of the eye's ability to accommodate (change focal length). The near point recedes, making close-up work difficult. Corrected with a converging (positive power) reading lens.
Dioptre (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. In simple terms, higher dioptres mean stronger bending of light.
Near point and far point
The near point is the closest distance at which the eye can focus. The far point is the farthest. For a normal eye, the near point is about 25 cm and the far point is infinity.
1/f = 1/s + 1/s'. Rearrange to find any one of f, s, or s' given the other two.
s' = (1/f - 1/s)⁻¹. This is the form you will use most often.
For a concave (diverging) lens, f is negative. For a convex (converging) lens, f is positive.
For a concave (converging) mirror, f is positive. For a convex (diverging) mirror, f is negative.
A negative s' means a virtual image (same side as the object for a lens, behind the mirror for a mirror).
m = -s'/s. Also equal to h'/h.
Positive m: upright image. Negative m: inverted image.
|m| > 1: image is enlarged. |m| < 1: image is reduced. |m| = 1: same size.
For a concave lens: the image is always virtual, upright, and reduced. So m is always positive and less than 1.
Solve for the image of lens 1 first using the thin lens equation.
The image of lens 1 becomes the object of lens 2. The object distance for lens 2 is the separation x minus the image distance from lens 1 (if the image is on the same side as lens 2) or x plus s₁' (if the image is beyond lens 2).
Apply the thin lens equation again with lens 2's focal length and the new object distance.
The total magnification is m₁ × m₂.
A convex mirror always produces a virtual, upright, reduced image. The image forms behind the mirror, so s' is negative.
A concave mirror can produce real or virtual images depending on where the object is placed relative to f and 2f (the centre of curvature).
Object beyond 2f: real, inverted, reduced image between f and 2f.
Object at 2f: real, inverted, same-size image at 2f.
Object between f and 2f: real, inverted, enlarged image beyond 2f.
Object inside f: virtual, upright, enlarged image behind the mirror.
Ray 1 (parallel ray): travels parallel to the axis, then through the focal point (or appears to come from it for diverging optics).
Ray 2 (focal ray): passes through the focal point (or aims toward it), then exits parallel to the axis.
Ray 3 (centre ray): for a lens, passes straight through the centre. For a mirror, strikes the centre of curvature and reflects back on itself.
The image forms where these rays converge (real) or where their extensions converge behind the optic (virtual).
An image with |m| > 1 (enlarged) can be formed in two regions:
Region B (between f and 2f): produces a real, inverted, enlarged image.
Region C (inside f): produces a virtual, upright, enlarged image.
At 2f the magnification is exactly 1. Beyond 2f the image is always reduced.
The eye's total optical power = 1/s + 1/s', where s is the object distance and s' is the fixed distance from the lens to the retina.
The eye adjusts its power by changing the shape of the crystalline lens (accommodation). Maximum power focuses at the near point; minimum power focuses at the far point.
The far point is at a finite distance (e.g. 1.50 m) instead of infinity. The eye is too powerful at rest.
Correction: a diverging lens whose power equals -1/(far point distance). For a far point of 1.50 m, the corrective lens has power -0.67 D.
With the corrective lens, the combined system can focus at infinity, but the near point shifts outward.
To find the new near point: P_combined = P_eye,max + P_lens. The new near point is 1 / (P_eye,max + P_lens - 1/s').
Worked example: naked near point 14.0 cm, far point 1.50 m. P_lens = -0.67 D. P_eye,max = 1/0.14 + 1/s' = 7.14 + 1/s'. Combined max power = 7.14 + 1/s' - 0.67 = 6.48 + 1/s'. New near point = 1/6.48 = 15.4 cm.
The near point recedes (e.g. to 0.60 m) while distant vision remains normal.
Correction: converging (positive power) reading glasses.
With reading glasses, the near point moves closer but a new far point appears (no longer infinity).
Worked example: near point 0.60 m, normal far point (infinity). Wearing +2.0 D glasses: new near point = 1/(1/0.60 + 2.0) = 1/3.67 = 0.27 m. New far point = 1/(0 + 2.0) = 0.50 m. Range with glasses: 0.27 m to 0.50 m.
P_combined = P_eye + P_lens at all times.
The lens does not change. The eye changes its power between P_min and P_max.
At minimum power (looking far): s_max = 1 / (P_min,eye + P_lens - 1/s').
At maximum power (looking close): s_min = 1 / (P_max,eye + P_lens - 1/s').
Since 1/s' is the same constant in both, it cancels out in the working, and you can treat the problem in terms of power alone.
\frac{1}{f} = \frac{1}{s} + \frac{1}{s'}Thin lens equation and mirror equation.
s' = \left(\frac{1}{f} - \frac{1}{s}\right)^{-1}Rearranged to solve for image distance.
m = -\frac{s'}{s} = \frac{h'}{h}Magnification: ratio of image to object size. Sign gives orientation.
P = \frac{1}{f}Lens power in dioptres, where f is in metres.
P_{\text{combined}} = P_{\text{eye}} + P_{\text{lens}}Total optical power of eye plus corrective lens.
s = 4fObject distance for a concave mirror when the image distance is s/3 (i.e. |m| = 1/3). Derived from combining the mirror equation with m = -s'/s.
Converging lenses are the basis of cameras, projectors, and magnifying glasses. Concave mirrors are used in telescopes and satellite dishes to focus incoming parallel rays. Vision correction with spectacles, contact lenses, and refractive surgery all rely on adjusting the combined optical power of the eye-lens system to move the focal point onto the retina.
Students often forget that a concave lens always produces a virtual, upright, reduced image. If your answer for a concave lens gives a real or inverted image, check your signs.
A frequent error in mirror problems is getting the sign of s' wrong for a convex mirror. The image is always behind the mirror, so s' is always negative.
In two-lens problems, students sometimes forget that the object distance for lens 2 depends on the separation x. Draw the geometry carefully.
For vision correction, students confuse the sign of the corrective lens. Myopia needs a negative (diverging) lens; presbyopia needs a positive (converging) lens.
⚠️ Concave mirror problems that give you a magnification and ask for the object distance require combining the mirror equation with m = -s'/s. Set up s' in terms of s from the magnification first.
⚠️ Vision problems frequently ask for the new near point or far point after putting on corrective lenses. You need to work with powers (dioptres), not focal lengths directly.
⚠️ Ray diagram problems on exams require three principal rays drawn accurately. Practise these by hand.
⚠️ Two-lens system problems almost always appear on the exam. The key is methodical bookkeeping of which image becomes which object.
True or false: A diverging lens has a positive focal length. (False, it is negative)
Fill in the blank: If m = -0.5, the image is ______ and ______. (inverted and reduced)
True or false: A convex mirror can produce a real image. (False, always virtual)
Fill in the blank: To correct myopia, you need a ______ lens. (diverging / negative power)
True or false: The image of lens 1 becomes the object of lens 2 in a two-lens system. (True)
Q: An upright object of height 2.50 cm is 45.0 cm to the left of a concave lens with f = -20.0 cm. Describe the image.
A: s' = (1/(-20) - 1/45)⁻¹ = -13.8 cm. m = -(-13.8)/45 = +0.307. Image is virtual (s' negative), upright (m positive), and 0.769 cm tall.
Q: A convex mirror has f = -40.0 cm. The image forms 25.0 cm behind the mirror. Find the object distance.
A: Image behind the mirror means s' = -25.0 cm. s = (1/(-40) - 1/(-25))⁻¹ = (1/(-40) + 1/25)⁻¹ = 66.7 cm.
Q: A concave mirror with f = 36 cm produces an image whose distance from the mirror is one-third the object distance. Find s.
A: s' = s/3 (the image is reduced, so the object is beyond the centre of curvature). m = -s'/ s = -1/3. From 1/f = 1/s + 3/s = 4/s, so s = 4f = 144 cm.
Q: Two convex lenses (f₁ = 35 cm, f₂ = 15 cm) are separated by distance x. An object is 50 cm left of lens 1. The final image is 23 cm right of lens 2. Find x.
A: s₁' = (1/35 - 1/50)⁻¹ = 116.7 cm. s₂ = (1/15 - 1/23)⁻¹ = 43.1 cm. x = s₁' + s₂ = 160 cm.
Q: A student with myopia has a far point of 1.50 m and a near point of 14.0 cm. With corrective lenses for distance vision, what is the new near point?
A: P_lens = -1/1.50 = -0.67 D. P_max,eye = 1/0.14 + 1/s' = 7.14 + 1/s'. Combined max power = 6.48 + 1/s'. New near point = 1/6.48 = 15.4 cm.
Q: Rachel has presbyopia with a near point of 0.60 m and normal far vision. She wears +2.0 D reading glasses. What is her range of clear vision with the glasses?
A: New near point = 1/(1/0.60 + 2.0) = 1/3.67 = 0.27 m. New far point = 1/(0 + 2.0) = 0.50 m. Range: 0.27 m to 0.50 m.
The thin lens equation is built on Snell's law applied to two curved surfaces, so everything from geometric optics feeds into this section. Vision correction connects to biology and medicine. Two-lens systems are the basis of compound microscopes and telescopes, which may appear later in the course. The sign conventions here are the same ones used in more advanced optics courses.
thin lens equation, mirror equation, magnification, focal length, real image, virtual image, converging lens, diverging lens, concave mirror, convex mirror, sign conventions, ray diagram, principal rays, two-lens system, object distance, image distance, myopia, nearsightedness, presbyopia, farsightedness, dioptre, lens power, corrective lens, near point, far point, centre of curvature, PHYS 116 midterm 2, optics