Power of a lens, combination of lenses: Exercises

Before line one, two reminders that every problem needs:

Before you touch a problem, make sure you can read the two combination rules, not just recite them. The next box builds both from a single picture so nothing on this page is a black box.

Intuition Where the two combination rules come from (read once, then reuse everywhere)

The contact rule ( $1/ F = 1/ f_{1} + 1/ f_{2}$ ). Send an object through lens 1; it makes an image at $v_{1}$ . Because the lenses touch, that image is instantly the object for lens 2. Writing the Thin Lens Equation for each lens and adding them makes the internal $v_{1}$ cancel:

\;\Rightarrow\; \frac{1}{v}-\frac{1}{u}=\frac{1}{f_1}+\frac{1}{f_2}.$$ Comparing with $\tfrac{1}{v}-\tfrac{1}{u}=\tfrac{1}{F}$ gives $\tfrac{1}{F}=\tfrac{1}{f_1}+\tfrac{1}{f_2}$, i.e. $P=P_1+P_2$. **WHY it works:** two bends with no gap between them stack into one bend that is their *sum*. **The separation rule ($P=P_1+P_2-dP_1P_2$).** Now pull the lenses $d$ apart. A ray bent by lens 1 travels the gap $d$ and reaches lens 2 at a **different height** than it left lens 1. Lens 2's bend depends on that height, so the two bends no longer add cleanly; carrying the height-change through the algebra produces exactly the correction $-d\,P_1P_2$. **Sanity check:** put $d=0$ (touching) and the term vanishes, recovering $P=P_1+P_2$. ✓ Look at the figure below to *see* the height change. ![[deepdives/dd-physics-2.5.07-d4-s01.png]]

We will also use the Thin Lens Equation to find image positions, so fix its sign convention now.

Level 1 — Recognition

(Can you read the definition off correctly and convert units?)

Exercise 1.1

A convex lens has focal length $f = 50 cm$ . State its power in dioptres.

Recall Solution 1.1

WHAT: apply $P = 1/ f$ . WHY the first move is unit conversion: the dioptre is defined for metres, so we must convert before dividing. $f = 50 cm = 0.50 m$ $P = \frac{1}{0.50} = + 2 D$ Positive because the lens is convex (converging). Answer: $P = + 2 D$ .

Exercise 1.2

A lens is labelled $- 4 D$ . Is it convex or concave, and what is its focal length in cm?

Recall Solution 1.2

WHAT the sign tells us: $P < 0 \Rightarrow f < 0 \Rightarrow$ concave (diverging) lens. WHY invert: $P = 1/ f \Rightarrow f = 1/ P$ . $f = \frac{1}{- 4} = - 0.25 m = - 25 cm$ Answer: concave lens, $f = - 25 cm$ .

Exercise 1.3

Convert a power of $+ 2.5 D$ into a focal length in metres and in centimetres.

Recall Solution 1.3

WHY invert: power and focal length are reciprocals ( $P = 1/ f$ ), so to recover $f$ from a given $P$ we simply flip the relation, $f = 1/ P$ . That is the only algebra that undoes the reciprocal. $f = \frac{1}{P} = \frac{1}{2.5} = 0.40 m = 40 cm$ Positive power $\Rightarrow$ converging lens. Answer: $f = 0.40 m = 40 cm$ .

Level 2 — Application

(Plug into the contact and separation formulas.)

Exercise 2.1

A convex lens $P_{1} = + 6 D$ is placed in contact with a concave lens $f_{2} = - 20 cm$ . Find the net power and the equivalent focal length.

Recall Solution 2.1

Step 1 — get $P_{2}$ . $f_{2} = - 0.20 m$ , so $P_{2} = \frac{1}{- 0.20} = - 5 D .$ Step 2 — add (they touch, so powers add): $P = P_{1} + P_{2} = 6 + (- 5) = + 1 D .$ WHY addition is legal: in contact there is no gap, so the ray leaves lens 1 and enters lens 2 at essentially the same height — two bends stack into one sum (see the derivation box at the top). Step 3 — equivalent focal length: $F = \frac{1}{P} = \frac{1}{1} = + 1 m = + 100 cm .$ Positive $\Rightarrow$ the combination still converges, but only weakly. Answer: $P = + 1 D$ , $F = + 1 m$ .

Exercise 2.2

Two convex lenses $f_{1} = 10 cm$ and $f_{2} = 40 cm$ are placed in contact. Find $F$ .

Recall Solution 2.2

WHY the reciprocals add: the two lenses touch, so (as derived in the top box) their powers add — and power is $1/ f$ . Writing $P = P_{1} + P_{2}$ with $P = 1/ F$ , $P_{1} = 1/ f_{1}$ , $P_{2} = 1/ f_{2}$ is literally the statement $\frac{1}{F} = \frac{1}{f _{1}} + \frac{1}{f _{2}}$ . So we may work with reciprocals directly. In metres: $\frac{1}{F} = \frac{1}{0.10} + \frac{1}{0.40} = 10 + 2.5 = 12.5 D .$ $F = \frac{1}{12.5} = 0.08 m = 8 cm .$ Notice: the combined $F$ ( $8 cm$ ) is smaller than either individual $f$ . Two converging pushes make a stronger, shorter-focus lens. Answer: $F = 8 cm$ ( $P = 12.5 D$ ).

Exercise 2.3

Two convex lenses $f_{1} = 20 cm$ , $f_{2} = 25 cm$ are separated by $d = 5 cm$ . Find the equivalent power and focal length.

Recall Solution 2.3

WHY the separation formula: $d \neq = 0$ , so the ray lands at a different height on lens 2 — the $- d P_{1} P_{2}$ correction is needed (derived in the top box; see figure s01). In metres: $P_{1} = 1/0.20 = 5 D$ , $P_{2} = 1/0.25 = 4 D$ , $d = 0.05 m$ . $P = P_{1} + P_{2} - d P_{1} P_{2} = 5 + 4 - (0.05) (5) (4) = 9 - 1 = 8 D .$ $F = \frac{1}{8} = 0.125 m = 12.5 cm .$ Answer: $P = 8 D$ , $F = 12.5 cm$ .

Level 3 — Analysis

(Reason backwards, or interpret a result.)

Exercise 3.1

An opticians' cabinet has a $+ 3 D$ lens. What single lens must you place in contact with it to make the pair behave like a plane glass plate (zero net power)?

Recall Solution 3.1

WHAT "plane glass" means: it neither converges nor diverges, so $P = 0$ . WHY contact addition: in contact, $P = P_{1} + P_{2} = 0$ . $0 = 3 + P_{2} \Rightarrow P_{2} = - 3 D .$ $f_{2} = \frac{1}{- 3} = - 0.3333 m \approx - 33.3 cm .$ You need a ==concave lens of $- 3 D$ == ( $f \approx - 33.3 cm$ ). Its diverging push exactly cancels the converging push. Answer: $P_{2} = - 3 D$ , $f_{2} = - \frac{1}{3} m \approx - 33.3 cm$ .

Exercise 3.2

A combination of a convex ( $+ 5 D$ ) and a concave ( $- 8 D$ ) lens in contact is used. Is the combination converging or diverging? Find $F$ .

Recall Solution 3.2

$P = 5 + (- 8) = - 3 D .$ $P < 0 \Rightarrow$ the concave lens wins; the pair is diverging. $F = \frac{1}{- 3} = - 0.3333 m \approx - 33.3 cm .$ Interpretation: even though one lens converges, the stronger diverging power dominates the sum. Answer: diverging, $F = - \frac{1}{3} m \approx - 33.3 cm$ .

Exercise 3.3

For two lenses separated by distance $d$ , at what separation does the combination become afocal ( $P = 0$ , i.e. $F \to \infty$ )? Take $f_{1} = 15 cm$ , $f_{2} = 30 cm$ (both convex).

Recall Solution 3.3

WHAT afocal means: parallel rays in $\to$ parallel rays out; the equivalent power is zero. This is the geometry of a telescope (see Microscope and Telescope). Set $P = 0$ in the separation formula: $0 = P_{1} + P_{2} - d P_{1} P_{2} \Rightarrow d = \frac{P _{1} + P _{2}}{P _{1} P _{2}} = \frac{1}{P _{2}} + \frac{1}{P _{1}} = f_{1} + f_{2} .$ WHAT it looks like — read the figure below. A horizontal ray enters lens 1 (top of the figure) and is bent down to lens 1's focus, marked in red. Because the separation is chosen as $d = f_{1} + f_{2}$ , that same red point is also lens 2's front focus, so the ray leaving that focus and striking lens 2 emerges horizontal again. The two focal points have merged into one — that single shared focus is the whole reason the system is afocal. Trace the red dot: it is the hinge that makes "parallel in" become "parallel out." $d = f_{1} + f_{2} = 0.15 + 0.30 = 0.45 m = 45 cm .$ Answer: $d = f_{1} + f_{2} = 45 cm$ .

Figure — Power of a lens, combination of lenses

Level 4 — Synthesis

(Combine power with magnification, or with the lensmaker's relation.)

Exercise 4.1

A convex lens ( $f_{1} = 12 cm$ ) and a concave lens ( $f_{2} = - 18 cm$ ) are in contact. An object is placed $24 cm$ in front of the pair. Find the image position and the magnification.

Recall Solution 4.1

Step 1 — equivalent lens. In contact, use focal lengths (metres): $\frac{1}{F} = \frac{1}{0.12} + \frac{1}{- 0.18} = 8.333 - 5.556 = 2.778 D .$ $F = \frac{1}{2.778} = 0.36 m = 36 cm .$ WHY treat as one lens: in contact, the pair is a single equivalent lens of focal length $F$ , so we can use the ordinary Thin Lens Equation once. Step 2 — thin lens equation. Using the sign convention fixed above (real object in front $\Rightarrow u < 0$ ), take $u = - 24 cm$ , $F = + 36 cm$ : $\frac{1}{v} - \frac{1}{u} = \frac{1}{F} \Rightarrow \frac{1}{v} = \frac{1}{36} + \frac{1}{- 24} = \frac{2 - 3}{72} = - \frac{1}{72} .$ $v = - 72 cm .$ Negative $v$ $\Rightarrow$ image is on the same side as the object: a virtual, erect image. Step 3 — magnification. $m = v / u = (- 72) / (- 24) = + 3$ . Answer: $v = - 72 cm$ (virtual, same side), $m = + 3$ (erect, magnified $3 \times$ ).

Exercise 4.2

A biconvex lens is made of glass with refractive index $n = 1.5$ and both radii of curvature equal to $20 cm$ in magnitude. Using the sign convention $R_{1} = + 20 cm$ , $R_{2} = - 20 cm$ , find its power. Then find the power if it is immersed so that only $n - 1$ effectively halves.

Recall Solution 4.2

Part A — lensmaker's power (from Lensmaker's Equation): $P = (n - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}}), R in metres .$ $P = (1.5 - 1) (\frac{1}{0.20} - \frac{1}{- 0.20}) = 0.5 (5 + 5) = 0.5 \times 10 = + 5 D .$ So $f = 1/5 = 0.20 m = 20 cm$ . Part B — halve the $(n - 1)$ factor. Power is directly proportional to $(n - 1)$ ; halving it halves the power: $P^{'} = \frac{1}{2} \times 5 = + 2.5 D (f^{'} = 40 cm) .$ WHY: the surface curvatures didn't change; only the bending strength per surface did, which lives entirely in the $(n - 1)$ factor. Answer: $P = + 5 D$ ; immersed $P^{'} = + 2.5 D$ .

Level 5 — Mastery

(Multi-step design and limiting-case reasoning.)

Exercise 5.1

A student wants a converging system of power exactly $+ 4 D$ but only has two convex lenses of $f_{1} = 25 cm$ and $f_{2} = 50 cm$ . They may separate them by a distance $d$ . Find every value of $d$ (in cm) that gives $P = + 4 D$ , and check each is physically sensible.

Recall Solution 5.1

Set up. $P_{1} = 1/0.25 = 4 D$ , $P_{2} = 1/0.50 = 2 D$ . $P = P_{1} + P_{2} - d P_{1} P_{2} = 4 + 2 - d (4) (2) = 6 - 8 d .$ Solve $6 - 8 d = 4$ : $8 d = 2 \Rightarrow d = 0.25 m = 25 cm .$ Check physicality: $d = 25 cm$ is positive and finite — the lenses sit $25 cm$ apart, perfectly buildable. WHY only one answer: $P$ is linear in $d$ here, so the equation $6 - 8 d = 4$ has exactly one root. (At $d = 0$ we'd get $6 D$ ; as $d$ grows the power falls; it passes through $4 D$ once at $d = 25 cm$ , and reaches $0$ at $d = f_{1} + f_{2} = 75 cm$ .) Answer: $d = 25 cm$ (unique).

Exercise 5.2

Three thin lenses in contact have powers $+ 10 D$ , $- 5 D$ , and $+ 2 D$ . (a) Find the net power and focal length. (b) A myopic eye needs a corrective power of $- 3 D$ ; which single lens from this set (or which pair in contact) provides it?

Recall Solution 5.2

(a) In contact, all powers add: $P = 10 + (- 5) + 2 = + 7 D .$ $F = \frac{1}{7} = 0.1429 m \approx 14.3 cm (converging) .$ (b) We need $- 3 D$ . No single lens gives it. Try pairs in contact: $10 + (- 5) = + 5, 10 + 2 = + 12, (- 5) + 2 = - 3 ✓$ The $- 5 D$ and $+ 2 D$ lenses in contact give exactly $- 3 D$ — the diverging power a short-sighted eye needs (see Defects of Vision). Answer: (a) $+ 7 D$ , $F \approx 14.3 cm$ ; (b) the $- 5 D$ and $+ 2 D$ lenses together.

Exercise 5.3 (limiting case)

In the separation formula, what happens to the equivalent power $P$ as $d \to \infty$ for two convex lenses? Interpret physically.

Recall Solution 5.3

$P = P_{1} + P_{2} - d P_{1} P_{2} .$ As $d \to \infty$ , the $- d P_{1} P_{2}$ term dominates and (for two positive powers) $P \to - \infty$ . $F = \frac{1}{P} \to 0^{-} .$ Physical reading: when the second lens is extremely far away, a ray bent by lens 1 has moved enormously off-axis by the time it reaches lens 2, so lens 2 deflects it drastically — the system becomes a strong (formally diverging) system. The formula never "breaks"; it just tells you the two-lens model stops being a useful single equivalent lens once $d$ is huge. This is why the clean $P = P_{1} + P_{2}$ rule is reserved for contact ( $d = 0$ ), and why real instruments keep $d$ modest and comparable to the focal lengths (see Microscope and Telescope). Answer: $P \to - \infty$ , $F \to 0^{-}$ ; the combination becomes formally strongly diverging.

Connections

Power of a lens, combination of lenses — the parent note these exercises test.
Thin Lens Equation — used in §4.1 to locate the image and to derive the contact rule.
Lensmaker's Equation — powers §4.2 from radii and $n$ .
Magnification of Lenses — the "multiply" rule in §4.
Microscope and Telescope — afocal systems (§3.3, §5.3).
Defects of Vision — corrective powers in dioptres (§5.2).
Resistors in Parallel — the same reciprocal-addition mathematics.

Exercises — Power of a lens, combination of lenses