Thin lenses — lens equation, lens maker's equation: Worked examples

This page is the "no surprises" drill. We take the two boxed rules from the parent note and hit them with every kind of input: converging and diverging lenses, objects near and far, the special focal-point cases, and even a flat plate. If you can do all of these, no exam version can ambush you.

Before we start, one reminder of the machinery we will reuse. Everything below rides on three earned facts:

The scenario matrix

Think of every thin-lens problem as choosing one row from each of these three dials. The matrix below lists the case-classes; each worked example is tagged with the cell(s) it lands on.

Dial	Cases that behave differently	Covered in
Lens type	converging $f > 0$ · diverging $f < 0$ · flat plate $f = \infty$	Ex 1,2,3,4,6 · Ex 5,7 · Ex 8
Object position (vs converging $f$ )	beyond $2 f$ · at $2 f$ · between $f$ and $2 f$ · at $f$ (degenerate) · inside $f$ (virtual)	Ex 2 · Ex 3 · Ex 4 · Ex 6 · Ex 5(div)/Ex 7
Shape / sign of $R$	biconvex ( $+, -$ ) · plano-convex (one $\infty$ ) · biconcave ( $-, +$ ) · flat ( $\infty, \infty$ )	Ex 1 · Ex 8 · Ex 5 · Ex 8
Extra flavours	real-world word problem · exam-style twist (image height, magnifier)	Ex 6 (camera) · Ex 4,7

The one "degenerate" input everyone fears — object at the focal point — is Ex 6. The one "nothing happens" input — flat plate — is Ex 8. We handle both head-on.

Example 1 — Biconvex shape → focal length

(Cell: converging · biconvex $R_{1} > 0, R_{2} < 0$ )

Worked example Find

$f$ A glass lens ( $n = 1.5$ ) has front surface $R_{1} = + 15 cm$ and back surface $R_{2} = - 30 cm$ . Find its focal length.

Forecast: Both faces bulge outward, so this is converging — expect $f > 0$ , somewhere around a couple of tens of cm.

Step 1 — Fix each radius's sign. Why this step? The maker's equation is meaningless until every $R$ carries its centre-of-curvature sign. Front face's centre is to the right (outgoing) ⇒ $R_{1} = + 15$ . Back face's centre is to the left (incoming) ⇒ $R_{2} = - 30$ .

Step 2 — Plug into the maker's equation. Why this step? This is the only rule that turns shape into $f$ . $\frac{1}{f} = (1.5 - 1) (\frac{1}{15} - \frac{1}{- 30}) = 0.5 (\frac{1}{15} + \frac{1}{30})$

Step 3 — Add the fractions. Why this step? Common denominator 30: $\frac{2}{30} + \frac{1}{30} = \frac{3}{30} = \frac{1}{10}$ . $\frac{1}{f} = 0.5 \cdot \frac{1}{10} = \frac{1}{20} cm^{- 1} \Rightarrow f = + 20 cm$

Verify: $f > 0$ ✔ (converging, as forecast). Units: $[cm^{- 1}]^{- 1} = cm$ ✔. A quick sanity check — a symmetric biconvex $\pm 15$ would give $f = 15$ cm; making the back flatter ( $30$ ) should lengthen $f$ , and indeed $20 > 15$ . ✔

Example 2 — Object beyond $2 f$

(Cell: converging · object beyond $2 f$ )

Now we place objects at every distance and watch the image march. Study the figure once — every remaining converging example is just a point on this line.

Figure — Thin lenses — lens equation, lens maker's equation

Example 3 — Object exactly at $2 f$

(Cell: converging · object at $2 f$ — the symmetry point)

Example 4 — Object between $f$ and $2 f$ (exam twist: find image height)

(Cell: converging · object between $f$ and $2 f$ · exam-style height twist)

Worked example Projector geometry

Same lens $f = + 20$ cm. A $2.0$ cm tall object sits at $u = - 30$ cm (between $f = 20$ and $2 f = 40$ ). Find $v$ , $m$ , and the image height $h^{'}$ .

Forecast: Between $f$ and $2 f$ ⇒ image real, inverted, magnified, forming beyond $2 f$ . This is exactly Magnification and image formation in action.

Step 1 — Lens equation. Why this step? Locate the image first; height needs $m$ , which needs $v$ . $\frac{1}{v} = \frac{1}{20} + \frac{1}{- 30} = \frac{3 - 2}{60} = \frac{1}{60} \Rightarrow v = + 60 cm$

Step 2 — Magnification. $m = \frac{v}{u} = \frac{60}{- 30} = - 2$

Step 3 — Image height. Why this step? $m = h^{'} / h$ , so $h^{'} = m h$ . This is the twist: the exam wants the size, not just position. $h^{'} = m h = (- 2) (2.0) = - 4.0 cm$

Verify: $v = 60 > 2 f = 40$ ✔ (beyond $2 f$ , as forecast). $∣ m ∣ = 2 > 1$ magnified ✔. $h^{'} = - 4.0$ cm: negative sign = inverted, and $∣ h^{'} ∣ = 4.0 = 2 \times 2.0$ cm ✔. Units of height are cm ✔.

Example 5 — Diverging lens (biconcave)

(Cell: diverging $f < 0$ · biconcave $R_{1} < 0, R_{2} > 0$ · virtual image)

Worked example Two rules in one problem

A biconcave glass lens, $n = 1.5$ , $R_{1} = - 25 cm$ , $R_{2} = + 25 cm$ . An object stands at $u = - 20$ cm. Find $f$ , then $v$ and $m$ .

Forecast: Concave ⇒ diverging ⇒ $f < 0$ . A diverging lens always gives a virtual, erect, diminished image ( $v < 0$ , $0 < m < 1$ ), no matter where the object is.

Step 1 — Maker's equation for $f$ . Why this step? We're given shape, not $f$ . Front face's centre is on the incoming (left) side ⇒ $R_{1} = - 25$ ; back face's centre is on the outgoing side ⇒ $R_{2} = + 25$ . $\frac{1}{f} = (0.5) (\frac{1}{- 25} - \frac{1}{+ 25}) = 0.5 (- \frac{2}{25}) = - \frac{1}{25} \Rightarrow f = - 25 cm$

Step 2 — Lens equation for $v$ . $\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{- 25} + \frac{1}{- 20} = \frac{- 4 - 5}{100} = - \frac{9}{100}$ $v = - \frac{100}{9} \approx - 11.11 cm$

Step 3 — Magnification. $m = \frac{v}{u} = \frac{- 100/9}{- 20} = \frac{100}{180} = \frac{5}{9} \approx + 0.556$

Verify: $f < 0$ ✔ (diverging). $v < 0$ ✔ (virtual, incoming side, look at the dashed rays in the figure). $0 < m < 1$ and positive ✔ (erect, diminished). Sanity: virtual image is closer to the lens than the object ( $11.1 < 20$ ) — exactly what your glasses do when you're near-sighted.

Example 6 — Object exactly at the focal point (degenerate) + camera word problem

(Cell: converging · object AT $f$ — degenerate limit · real-world)

Worked example Where does the image go when

$u = - f$ ? (a) Converging lens $f = + 20$ cm, object at $u = - 20$ cm exactly. Find $v$ . (b) Word problem: A camera has a $50$ mm ( $f = + 5.0$ cm) lens. To photograph a distant mountain, how far behind the lens must the sensor sit?

Forecast (a): Object at the focal point means rays leave the lens parallel — they never reconverge, so the image is at infinity ( $v \to \infty$ ). (b): A very distant object is the same limit from the other side ( $u \to - \infty$ ): image should land right at $f$ .

Step 1 — Part (a), push the algebra. Why this step? We must see the degeneracy in the equation, not just assert it. $\frac{1}{v} = \frac{1}{20} + \frac{1}{- 20} = 0 \Rightarrow v = \infty$ The reciprocal is exactly zero, so $v$ is infinite: the image forms nowhere finite. Physically the outgoing rays are parallel.

Step 2 — Part (b), the opposite limit. Why this step? A mountain is effectively $u \to - \infty$ , so $1/ u \to 0$ . $\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{5.0} + 0 = \frac{1}{5.0} \Rightarrow v = + 5.0 cm$

Verify: (a) $1/ v = 0$ confirms the "image at infinity" ✔ — this is why a lit object at $f$ makes a spotlight/collimator. (b) $v = f = 5.0$ cm ✔: a distant object always images at the focal plane, which is why the sensor sits one focal length back. The two parts are the same degenerate case seen from opposite sides. ✔

Example 7 — Magnifying glass (object inside $f$ )

(Cell: converging · object inside $f$ → virtual · exam twist: magnifier)

Worked example Why a converging lens can also give a virtual image

Converging lens $f = + 10$ cm. A stamp sits at $u = - 6$ cm (inside the focal length). Find $v$ and $m$ .

Forecast: Inside $f$ , a converging lens acts as a magnifying glass: virtual, erect, enlarged image ( $v < 0$ , $m > 1$ ).

Step 1 — Lens equation. Why this step? Test whether "inside $f$ " really flips $v$ negative. $\frac{1}{v} = \frac{1}{10} + \frac{1}{- 6} = \frac{3 - 5}{30} = - \frac{2}{30} = - \frac{1}{15} \Rightarrow v = - 15 cm$

Step 2 — Magnification. $m = \frac{v}{u} = \frac{- 15}{- 6} = + 2.5$

Verify: $v = - 15 < 0$ ✔ (virtual, same side as object — you see it through the lens). $m = + 2.5$ : positive ⇒ erect, $> 1$ ⇒ enlarged ✔. Note the contrast with Ex 5: same negative $v$ , but here $∣ m ∣ > 1$ because the lens converges. That is the whole difference between a magnifier and spectacles. See Power of a lens (dioptres).

Example 8 — Flat plate & near-flat limit

(Cell: flat plate $f = \infty$ · limiting behaviour · plano check)

Worked example The "does nothing" lens

(a) A flat slab of glass, $R_{1} = \infty$ , $R_{2} = \infty$ , $n = 1.5$ . Find $1/ f$ . (b) A plano-convex lens $R_{1} = \infty$ (flat front), $R_{2} = - 20$ cm, $n = 1.5$ . Find $f$ .

Forecast (a): Flat surfaces have no curvature to bend rays toward a common point ⇒ $1/ f = 0$ , $f = \infty$ , no focusing. (b): Only one face works, so expect a longer $f$ than a biconvex of the same $R$ .

Step 1 — Part (a): take the $R \to \infty$ limit. Why this step? $1/ R \to 0$ as $R \to \infty$ , so each curvature term dies. $\frac{1}{f} = (0.5) (\frac{1}{\infty} - \frac{1}{\infty}) = (0.5) (0 - 0) = 0 \Rightarrow f = \infty$ A flat plate is optically inert (it shifts rays sideways but does not focus).

Step 2 — Part (b): one flat, one curved. Why this step? Show that a single dead surface just drops out. $\frac{1}{f} = (0.5) (\frac{1}{\infty} - \frac{1}{- 20}) = (0.5) (0 + \frac{1}{20}) = \frac{1}{40} \Rightarrow f = + 40 cm$

Verify: (a) $1/ f = 0$ ✔ — matches the parent's "flat plate does nothing." (b) $f = + 40$ cm ✔, converging, and longer than the biconvex $\pm 20$ (which gave $20$ cm) because only half the curvature is doing work ✔.

Recall

Recall Quick self-test across the matrix

Object beyond $2 f$ of converging lens — image nature? ::: Real, inverted, diminished (Ex 2) Object at $2 f$ — magnification? ::: $m = - 1$ , real inverted same-size (Ex 3) Object at the focal point — where is the image? ::: At infinity, $1/ v = 0$ (Ex 6a) Converging lens, object inside $f$ — image nature? ::: Virtual, erect, enlarged — magnifier (Ex 7) Diverging lens, any object — image nature? ::: Always virtual, erect, diminished (Ex 5) Flat plate ( $R \to \infty$ ) focal length? ::: $f = \infty$ , $1/ f = 0$ , no focusing (Ex 8a)

Connections

Thin lenses — lens equation, lens maker's equation (parent)
Refraction at a single spherical surface
Snell's law & paraxial approximation
Magnification and image formation
Power of a lens (dioptres)
Lens combinations & equivalent focal length
Spherical mirrors — mirror equation
Lens aberrations (chromatic, spherical)