Thin lenses — lens equation, lens maker's equation: Question bank

The whole page leans on three facts you already built in the parent note. I number them here so the traps below can point back precisely:

$\frac{n _{2}}{v} - \frac{n _{1}}{u} = \frac{n _{2} - n _{1}}{R} (single-surface law)$

At surface 1 (air $\to$ glass, radius $R_{1}$ ) the object at $u$ makes an intermediate image at distance $v_{1}$ :

$\frac{n}{v _{1}} - \frac{1}{u} = \frac{n - 1}{R _{1}} (1)$

At surface 2 (glass $\to$ air, radius $R_{2}$ ) that same $v_{1}$ is the object, giving the final image at $v$ :

$\frac{1}{v} - \frac{n}{v _{1}} = \frac{1 - n}{R _{2}} (2)$

$\frac{1}{f} = (n - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}}) (lens maker’s)$

$\frac{1}{v} - \frac{1}{u} = \frac{1}{f} (Gaussian lens equation)$

Everything below just probes the edges of these facts. Four schematics anchor the trickiest ideas — read them before the reveals that point at them.

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

Recall Edge case with picture: object at infinity focuses at

$f$ Parallel rays ( $u \to - \infty$ , so $\frac{1}{u} \to 0$ ) collapse the lens equation to $\frac{1}{v} = \frac{1}{f}$ : they all meet at the focal point. This defines $f$ — see the figure.

True or false — justify

A "converging lens always converges light to a real image."

False — put the object inside the focus (

∣ u ∣ < f

, i.e.

- f < u < 0

), then

\frac{1}{v} = \frac{1}{f} + \frac{1}{u}

is (positive) + (larger-magnitude negative)

< 0

, so

v < 0

: a virtual, erect, magnified image (magnifying-glass mode, left branch of figure s04).

A "diverging lens can never make a real image of a real object."

True — for a real object

u < 0

and

f < 0

, so

\frac{1}{v} = \frac{1}{f} + \frac{1}{u}

is a sum of two negatives, giving

v < 0

always: the image is virtual, on the incoming side.

A "the lens maker's equation gives a different

f

depending on which face the light enters first."

False — flipping the lens sends

R_{1} \to - R_{2}

and

R_{2} \to - R_{1}

; substituting,

(n - 1) (\frac{1}{- R _{2}} - \frac{1}{- R _{1}}) = (n - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}})

, so

f

is unchanged. A thin lens has one focal length.

A "a symmetric biconvex lens has

R_{1} = R_{2}

False in sign — the magnitudes are equal but

R_{1} = + R

(centre on outgoing side) and

R_{2} = - R

(centre on incoming side, see s01); writing them equal drops the crucial minus.

A "if you submerge a glass converging lens in water, it stays converging with the same

f

Partly — it usually stays converging (glass still slower than water) but

f

grows because

(n_{glass} - 1)

is replaced by the smaller

(n_{glass} / n_{water} - 1)

; less relative slowing means weaker bending.

A "a flat glass plate has

f = 0

False — with

R_{1}, R_{2} \to \infty

both curvature terms vanish, so

\frac{1}{f} = 0

, meaning

f = \infty

(infinite focal length, zero power). It focuses nothing.

A "magnification

m = v / u

being negative means the image is smaller."

False — the sign of

m

encodes orientation (negative = inverted), while the magnitude

∣ m ∣

encodes size. A negative

m

can still be

∣ m ∣ > 1

(larger).

A "the intermediate image inside the lens derivation is a real image floating in the glass."

False —

v_{1}

is a bookkeeping construct; in the thin-lens limit the two surfaces coincide (figure s02), so

v_{1}

never physically materialises. It cancels precisely because it was never a real waypoint.

Spot the error

Someone writes the lens equation as

\frac{1}{v} + \frac{1}{u} = \frac{1}{f}

, copying the mirror form.

Wrong sign — Cartesian lenses use a minus:

\frac{1}{v} - \frac{1}{u} = \frac{1}{f}

. It only looks like the mirror "+" after you substitute the already-negative

u

of a real object.

A student computes

\frac{1}{f} = (n - 1) (\frac{1}{R _{1}} + \frac{1}{R _{2}})

for a biconvex lens because "both faces curve the same way."

Error — the formula has a minus,

(\frac{1}{R _{1}} - \frac{1}{R _{2}})

, and the sign convention already makes

R_{2} < 0

; the terms add only through that minus, not by writing a plus.

A student says "in the single-surface law I use

n_{1} / u

and

n_{2} / v

, but I'll just use

n = 1.5

on both sides."

Error — the two media are different (

n_{1}

= incoming medium,

n_{2}

= outgoing); that difference is the entire source of bending. Using one

n

everywhere gives

\frac{n _{2} - n _{1}}{R} = 0

(no refraction).

A student sets up surface 2 of the lens with object distance measured freshly from scratch, not equal to

v_{1}

Error — the thin-lens assumption is that both surfaces sit at one plane (s02), so surface 1's image (

v_{1}

) is surface 2's object at the same distance

v_{1}

; introducing a gap breaks the cancellation trick.

A student derives

i = α + ϕ

but writes

tan α = h / u

with

u

positive for a real object.

Error — here

α

is the incoming ray's small tilt and

h

the ray height at the surface; for a real object

u < 0

, so

α = h / (- u)

, and forgetting the sign of

u

flips the direction of that tilt.

A student claims paraxial rays and marginal (edge) rays focus at the same point, so aberrations don't exist.

Error — the derivation used

sin θ \approx θ

, valid only near the axis; edge rays violate it and focus closer, which is exactly spherical aberration (marginal rays cross the axis before the paraxial focus).

Why questions

Why does the

n / v_{1}

term cancel when we add equations (1) and (2)?

In (1) the intermediate image at

v_{1}

(in glass) appears as

+ n / v_{1}

; in (2) that same

v_{1}

(same point, same medium — figure s02) is the object and appears as

- n / v_{1}

, so adding (1)+(2) makes the two terms subtract to zero.

Why must light slow down in the glass for a lens to bend it at all?

Bending is driven by the factor

(n - 1)

; if

n = 1

the glass has the same speed as air, so

(n - 1) = 0

and

\frac{1}{f} = 0

— no speed change means no refraction (this is Snell's law with equal indices).

Why does a more curved surface (smaller

∣ R ∣

) bend light more strongly?

Because

1/ R

appears in the maker's equation: as

∣ R ∣

shrinks,

1/∣ R ∣

grows, so the same

(n - 1)

produces a larger

1/ f

— tighter curves give the rays a steeper normal to bend around.

Why do we derive the single-surface law first instead of attacking the whole lens at once?

A lens is two refracting surfaces; deriving the bending at one curved face once lets us reuse it twice, which is exactly the building-block strategy of the parent note.

Why is the focal length the special case

u \to - \infty

in the derivation?

Parallel incoming rays correspond to an object infinitely far away, so

\frac{1}{u} \to 0

and the lens equation collapses to

\frac{1}{v} = \frac{1}{f}

— by definition

f

is where parallel light meets (figure s03).

Why does the ray through the optical centre stay undeviated, giving

m = v / u

At the centre the two lens faces are locally parallel (like a thin flat plate), so the ray exits parallel to its entry with negligible offset; similar triangles (heights

h

and

h^{'}

over distances

u

and

v

) then force

h^{'} / h = v / u

Edge cases

Object placed exactly at the focal point of a converging lens (

u = - f

): where's the image?

\frac{1}{v} = \frac{1}{f} + \frac{1}{- f} = 0

, so there is no finite image — the emerging rays are exactly parallel and the image forms at infinity (the reverse of figure s03).

Object at the optical centre (

u \to 0^{-}

): what happens?

\frac{1}{v} = \frac{1}{f} + \frac{1}{u} \to - \infty

(dominated by

1/ u

), so

v \to 0

: image coincides with the object at the lens,

m \to 1

— degenerate, no real separation forms.

Virtual object (

u > 0

): converging rays are heading toward a point behind the lens before they hit it — can this give a real image?

Yes — with

u > 0

and converging

f > 0

\frac{1}{v} = \frac{1}{f} + \frac{1}{u}

is a sum of two positives, so

v > 0

: a real image on the outgoing side, formed even nearer than

f

. This is the right branch of figure s04 and the regime relevant to lens combinations, where one lens hands converging rays to the next.

A plano-convex lens has one flat face (

R_{2} \to \infty

): does it still converge?

Yes — the flat face contributes

1/ R_{2} \to 0

, so

\frac{1}{f} = (n - 1) / R_{1}

; one curved face alone still gives finite positive

f

. Curvature on either face suffices.

A concavo-convex (meniscus) lens where

∣ R_{1} ∣ < ∣ R_{2} ∣

: converging or diverging?

Using the

R > 0

-if-centre-on-outgoing-side rule (s01), both faces here have the same-signed

R

, so it comes down to the sign of

(\frac{1}{R _{1}} - \frac{1}{R _{2}})

: the more sharply curved face (smaller

∣ R ∣

, larger

1/∣ R ∣

) wins. Sharper convex

\to

converging; sharper concave

\to

diverging — the net curvature decides, not the shape name.

n \to 1

(glass index approaches air), what does the lens do?

(n - 1) \to 0

\frac{1}{f} \to 0

f \to \infty

: the lens becomes optically invisible, transmitting light straight through — the limit where glass and air are indistinguishable.

For a real object with a converging lens, when exactly does the image flip from real to virtual?

At the crossover

u = - f

(object at the focus): for

∣ u ∣ > f

the image is real and inverted; for

∣ u ∣ < f

it becomes virtual and erect. The focal point is the boundary between the two regimes (the vertical asymptote in figure s04).

Connections

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

Question bank — Thin lenses — lens equation, lens maker's equation

True or false — justify

Spot the error

Why questions

Edge cases

Connections

Concept Map