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The Timpson Lyceum · University Physics II (2026) – Electricity & Magnetism

Chapter 34 — Geometric Optics

Problems with worked solutions · 15 points · 3 extra credit

Chapter 34 — Geometric Optics

Core: 15 points · Extra credit: 3 points. Show your reasoning; partial credit is given.

Use the sign conventions of this chapter: for mirrors and lenses, object distance ss is positive when the object is on the incoming (light-side) of the surface; image distance s′s' is positive when the image is on the outgoing side (real image); the radius RR is positive when its center of curvature is on the outgoing side. Lateral magnification is m=−s′sm=-\dfrac{s'}{s}; m>0m>0 means upright, m<0m<0 means inverted.

  1. 1. (1 pt •) A person stands 1.5 m1.5\,\text{m} in front of a flat bathroom mirror.

    (a) How far behind the mirror is the image, and is it real or virtual?

    (b) What is the lateral magnification, and is the image upright or inverted?

    Solution

    1.

    (a) A plane mirror can be treated as a spherical mirror with R→∞R\to\infty, so f→∞f\to\infty and 1s′=−1s\dfrac{1}{s'}=-\dfrac{1}{s}, giving s′=−s=−1.5 ms'=-s=-1.5\,\text{m}. The negative sign means the image is 1.5 m1.5\,\text{m} behind the mirror and is virtual (no light actually converges there).

    (b) m=−s′s=−−1.51.5=+1m=-\dfrac{s'}{s}=-\dfrac{-1.5}{1.5}=+1. The image is the same size, upright, and (as a plane mirror) reversed front-to-back.

  2. 2. (2 pts ••) A concave makeup mirror has radius of curvature R=24 cmR=24\,\text{cm}. A small object 4.0 cm4.0\,\text{cm} tall is placed on the axis.

    (a) Find the focal length.

    (b) With the object 36 cm36\,\text{cm} from the mirror, find the image distance, the magnification, and the image height. State whether the image is real or virtual, upright or inverted.

    (c) Now move the object to 8.0 cm8.0\,\text{cm} from the mirror. Recompute the image distance and magnification, and describe the image (this is how the mirror is actually used).

    Solution

    2.

    (a) f=R2=242=12 cmf=\dfrac{R}{2}=\dfrac{24}{2}=12\,\text{cm} (concave, so RR and ff are positive).

    (b) 1s′=1f−1s=112−136=3−136=236\dfrac{1}{s'}=\dfrac{1}{f}-\dfrac{1}{s}=\dfrac{1}{12}-\dfrac{1}{36}=\dfrac{3-1}{36}=\dfrac{2}{36}, so s′=18 cms'=18\,\text{cm}.

    m=−s′s=−1836=−0.50m=-\dfrac{s'}{s}=-\dfrac{18}{36}=-0.50, and h′=mh=(−0.50)(4.0)=−2.0 cmh'=m h=(-0.50)(4.0)=-2.0\,\text{cm}.

    Since s′>0s'>0 the image is real; m<0m<0 so it is inverted and half-size.

    (c) 1s′=112−18=2−324=−124\dfrac{1}{s'}=\dfrac{1}{12}-\dfrac{1}{8}=\dfrac{2-3}{24}=-\dfrac{1}{24}, so s′=−24 cms'=-24\,\text{cm}.

    m=−−248=+3.0m=-\dfrac{-24}{8}=+3.0. The image is virtual (behind the mirror), upright, and magnified 3×3\times — exactly what a makeup mirror does when the face is inside the focal point.

  3. 3. (2 pts ••) A convex security mirror has radius of curvature of magnitude 40 cm40\,\text{cm}. A 6.0 cm6.0\,\text{cm}-tall object stands 30 cm30\,\text{cm} in front of it.

    (a) Find the focal length (mind the sign).

    (b) Find the image distance, the magnification, and the image height, and describe the image.

    Solution

    3.

    (a) A convex mirror has its center of curvature behind the mirror, so R=−40 cmR=-40\,\text{cm} and f=R2=−20 cmf=\dfrac{R}{2}=-20\,\text{cm}.

    (b) 1s′=1f−1s=1−20−130=−3−260=−560\dfrac{1}{s'}=\dfrac{1}{f}-\dfrac{1}{s}=\dfrac{1}{-20}-\dfrac{1}{30}=\dfrac{-3-2}{60}=-\dfrac{5}{60}, so s′=−12 cms'=-12\,\text{cm}.

    m=−s′s=−−1230=+0.40m=-\dfrac{s'}{s}=-\dfrac{-12}{30}=+0.40, and h′=(0.40)(6.0)=2.4 cmh'=(0.40)(6.0)=2.4\,\text{cm}.

    The image is virtual (behind the mirror), upright, and reduced — a convex mirror always gives an upright, diminished virtual image, which is why it widens the field of view.

  4. 4. (2 pts ••) A long glass rod (n=1.50n=1.50) ends in a convex spherical surface of radius R=10 cmR=10\,\text{cm} facing the air (n=1.00n=1.00). A small object 3.0 mm3.0\,\text{mm} tall sits in the air on the axis, 30 cm30\,\text{cm} from the vertex of the surface.

    (a) Use the refraction relation for a single spherical surface to find where the image forms inside the glass.

    (b) Find the lateral magnification (use m=−n1s′n2sm=-\dfrac{n_1 s'}{n_2 s}) and the image height, and state whether the image is real or virtual, upright or inverted.

    Solution

    4.

    (a) Single spherical surface: n1s+n2s′=n2−n1R\dfrac{n_1}{s}+\dfrac{n_2}{s'}=\dfrac{n_2-n_1}{R}, with n1=1.00n_1=1.00, n2=1.50n_2=1.50, R=+10 cmR=+10\,\text{cm} (center of curvature is on the outgoing/glass side), s=30 cms=30\,\text{cm}.

    n2s′=n2−n1R−n1s=0.5010−1.0030=0.0500−0.0333=0.01667 cm−1\dfrac{n_2}{s'}=\dfrac{n_2-n_1}{R}-\dfrac{n_1}{s}=\dfrac{0.50}{10}-\dfrac{1.00}{30}=0.0500-0.0333=0.01667\,\text{cm}^{-1}.

    s′=1.500.01667=90 cms'=\dfrac{1.50}{0.01667}=90\,\text{cm}. The image forms 90 cm90\,\text{cm} inside the glass.

    (b) m=−n1s′n2s=−(1.00)(90)(1.50)(30)=−9045=−2.0m=-\dfrac{n_1 s'}{n_2 s}=-\dfrac{(1.00)(90)}{(1.50)(30)}=-\dfrac{90}{45}=-2.0, so h′=(−2.0)(3.0 mm)=−6.0 mmh'=(-2.0)(3.0\,\text{mm})=-6.0\,\text{mm}.

    Since s′>0s'>0 the image is real (light really converges inside the glass); m<0m<0 so it is inverted and twice as tall.

  5. 5. (2 pts ••) A thin converging lens has focal length f=15 cmf=15\,\text{cm}. A 2.0 cm2.0\,\text{cm}-tall object is placed on its axis.

    (a) With the object 45 cm45\,\text{cm} from the lens, find the image distance, the magnification, and the image height, and describe the image.

    (b) Now place the same object 10 cm10\,\text{cm} from the lens. Find the image distance and magnification, and describe the image (this is the lens used as a magnifier).

    Solution

    5.

    (a) 1s′=1f−1s=115−145=3−145=245\dfrac{1}{s'}=\dfrac{1}{f}-\dfrac{1}{s}=\dfrac{1}{15}-\dfrac{1}{45}=\dfrac{3-1}{45}=\dfrac{2}{45}, so s′=22.5 cms'=22.5\,\text{cm}.

    m=−s′s=−22.545=−0.50m=-\dfrac{s'}{s}=-\dfrac{22.5}{45}=-0.50, and h′=(−0.50)(2.0)=−1.0 cmh'=(-0.50)(2.0)=-1.0\,\text{cm}.

    s′>0s'>0: real; m<0m<0: inverted, half-size (a projector-type configuration).

    (b) 1s′=115−110=2−330=−130\dfrac{1}{s'}=\dfrac{1}{15}-\dfrac{1}{10}=\dfrac{2-3}{30}=-\dfrac{1}{30}, so s′=−30 cms'=-30\,\text{cm}.

    m=−−3010=+3.0m=-\dfrac{-30}{10}=+3.0. The image is virtual (same side as the object), upright, and magnified 3×3\times — the lens acting as a simple magnifier (object inside the focal point).

  6. 6. (2 pts ••) A thin diverging lens has focal length f=−20 cmf=-20\,\text{cm}. A 5.0 cm5.0\,\text{cm}-tall object is placed 30 cm30\,\text{cm} from it.

    (a) Find the image distance and magnification.

    (b) Give the image height and describe the image (real/virtual, upright/inverted, enlarged/reduced).

    Solution

    6.

    (a) 1s′=1f−1s=1−20−130=−3−260=−560\dfrac{1}{s'}=\dfrac{1}{f}-\dfrac{1}{s}=\dfrac{1}{-20}-\dfrac{1}{30}=\dfrac{-3-2}{60}=-\dfrac{5}{60}, so s′=−12 cms'=-12\,\text{cm}.

    m=−s′s=−−1230=+0.40m=-\dfrac{s'}{s}=-\dfrac{-12}{30}=+0.40.

    (b) h′=(0.40)(5.0)=2.0 cmh'=(0.40)(5.0)=2.0\,\text{cm}. The image is virtual, upright, and reduced — a diverging lens always produces an upright, diminished virtual image for a real object.

  7. 7. (2 pts ••) A thin lens is ground from glass of index n=1.60n=1.60.

    (a) The lens is biconvex, with front radius of magnitude 20 cm20\,\text{cm} and back radius of magnitude 30 cm30\,\text{cm}. Using the sign convention, assign signs to R1R_1 and R2R_2 and use the lensmaker's equation to find its focal length. Is it converging or diverging?

    (b) A second lens is planoconvex, with one flat face and one curved face of radius of magnitude 12 cm12\,\text{cm}, made of the same glass. Find its focal length.

    Solution

    7.

    (a) Sign convention: for the front (first) surface the center of curvature lies on the outgoing side, so R1=+20 cmR_1=+20\,\text{cm}; for the back surface the center lies on the incoming side, so R2=−30 cmR_2=-30\,\text{cm}.

    1f=(n−1)(1R1−1R2)=(1.60−1)(120−1−30)=0.60(120+130)\dfrac{1}{f}=(n-1)\left(\dfrac{1}{R_1}-\dfrac{1}{R_2}\right)=(1.60-1)\left(\dfrac{1}{20}-\dfrac{1}{-30}\right)=0.60\left(\dfrac{1}{20}+\dfrac{1}{30}\right).

    120+130=3+260=560=112\dfrac{1}{20}+\dfrac{1}{30}=\dfrac{3+2}{60}=\dfrac{5}{60}=\dfrac{1}{12}, so 1f=0.6012=0.050 cm−1\dfrac{1}{f}=\dfrac{0.60}{12}=0.050\,\text{cm}^{-1} and f=+20 cmf=+20\,\text{cm}. Positive ff: converging.

    (b) Planoconvex with the flat face (R→∞R\to\infty, so 1/R=01/R=0) and a curved face of magnitude 12 cm12\,\text{cm}. Take the curved surface as R1=+12 cmR_1=+12\,\text{cm} and the flat as R2=∞R_2=\infty:

    1f=(1.60−1)(112−0)=0.6012=0.050 cm−1\dfrac{1}{f}=(1.60-1)\left(\dfrac{1}{12}-0\right)=\dfrac{0.60}{12}=0.050\,\text{cm}^{-1}, so f=+20 cmf=+20\,\text{cm}. (The result is the same whichever face light hits first — the two radii just swap roles.)

  8. 8. (2 pts ••) Two thin converging lenses are mounted on a common axis, 50 cm50\,\text{cm} apart. The first has f1=15 cmf_1=15\,\text{cm}, the second f2=12 cmf_2=12\,\text{cm}. A 1.0 cm1.0\,\text{cm}-tall object is placed 30 cm30\,\text{cm} in front of the first lens. Treating the image formed by the first lens as the object for the second:

    (a) Find the location and magnification of the intermediate image formed by lens 1.

    (b) Find the location of the final image (measured from lens 2).

    (c) Find the overall lateral magnification and the final image height, and state whether the final image is real or virtual, upright or inverted.

    Solution

    8.

    (a) Lens 1: 1s1′=1f1−1s1=115−130=2−130=130\dfrac{1}{s_1'}=\dfrac{1}{f_1}-\dfrac{1}{s_1}=\dfrac{1}{15}-\dfrac{1}{30}=\dfrac{2-1}{30}=\dfrac{1}{30}, so s1′=30 cms_1'=30\,\text{cm} (real image, 30 cm30\,\text{cm} to the right of lens 1).

    m1=−s1′s1=−3030=−1.0m_1=-\dfrac{s_1'}{s_1}=-\dfrac{30}{30}=-1.0.

    (b) The intermediate image is 30 cm30\,\text{cm} right of lens 1; lens 2 is 50 cm50\,\text{cm} right of lens 1, so this image is 50−30=20 cm50-30=20\,\text{cm} to the left of lens 2 — a real object for lens 2 with s2=+20 cms_2=+20\,\text{cm}.

    1s2′=1f2−1s2=112−120=5−360=260=130\dfrac{1}{s_2'}=\dfrac{1}{f_2}-\dfrac{1}{s_2}=\dfrac{1}{12}-\dfrac{1}{20}=\dfrac{5-3}{60}=\dfrac{2}{60}=\dfrac{1}{30}, so s2′=30 cms_2'=30\,\text{cm}. The final image is 30 cm30\,\text{cm} to the right of lens 2.

    (c) m2=−s2′s2=−3020=−1.5m_2=-\dfrac{s_2'}{s_2}=-\dfrac{30}{20}=-1.5, so the overall magnification is m=m1m2=(−1.0)(−1.5)=+1.5m=m_1 m_2=(-1.0)(-1.5)=+1.5.

    h′=mh=(+1.5)(1.0)=1.5 cmh'=m h=(+1.5)(1.0)=1.5\,\text{cm}. Since s2′>0s_2'>0 the final image is real; m>0m>0 so it is upright relative to the original object, magnified 1.5×1.5\times.

  9. Extra Credit (3 pts •••)

    Derive the lensmaker's equation from the refraction relation for a single spherical surface, treating a thin lens of index nn (in air) as two refracting surfaces of radii R1R_1 and R2R_2 separated by negligible thickness.

    (a) Apply the single-surface relation to the first surface for an object at distance s1s_1, obtaining the intermediate image distance s1′s_1'.

    (b) Argue why, for a thin lens, the object distance for the second surface is s2=−s1′s_2=-s_1', then apply the single-surface relation to the second surface (now going from glass back to air) to obtain the final image distance s2′s_2'.

    (c) Add the two results to eliminate s1′s_1' and show that

    1s1+1s2′=(n−1)(1R1−1R2),\frac{1}{s_1}+\frac{1}{s_2'}=(n-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right),

    so that the focal length (the image distance for s1→∞s_1\to\infty) satisfies 1f=(n−1)(1R1−1R2)\dfrac{1}{f}=(n-1)\left(\dfrac{1}{R_1}-\dfrac{1}{R_2}\right).

    Solution

    Extra Credit

    Model the thin lens as two spherical refracting surfaces of radii R1R_1 and R2R_2, of glass index nn surrounded by air (nair=1n_{\text{air}}=1), separated by a thickness we take to zero.

    (a) First surface (air → glass). The single-surface refraction relation is

    nas+nbs′=nb−naR,\frac{n_a}{s}+\frac{n_b}{s'}=\frac{n_b-n_a}{R},

    with na=1n_a=1, nb=nn_b=n, radius R1R_1, object distance s1s_1:

    1s1+ns1′=n−1R1.(1)\frac{1}{s_1}+\frac{n}{s_1'}=\frac{n-1}{R_1}. \qquad (1)

    Here s1′s_1' is where the light would focus inside the glass if the second surface were absent.

    (b) Second surface (glass → air). For a thin lens the two vertices coincide, so the image produced by surface 1 acts as the object for surface 2 at the same location. If s1′s_1' is measured to the right of the (common) vertex, then as an object for the second surface it lies on the outgoing side, i.e. it is a virtual object with s2=−s1′s_2=-s_1'. Now light travels from glass into air, so na=nn_a=n, nb=1n_b=1, radius R2R_2:

    ns2+1s2′=1−nR2.\frac{n}{s_2}+\frac{1}{s_2'}=\frac{1-n}{R_2}.

    Substituting s2=−s1′s_2=-s_1':

    −ns1′+1s2′=1−nR2=−(n−1)1R2.(2)-\frac{n}{s_1'}+\frac{1}{s_2'}=\frac{1-n}{R_2}=-(n-1)\frac{1}{R_2}. \qquad (2)

    (c) Add (1) and (2). The ± n/s1′\pm\, n/s_1' terms cancel:

    1s1+1s2′=n−1R1−(n−1)1R2=(n−1)(1R1−1R2).\frac{1}{s_1}+\frac{1}{s_2'}=\frac{n-1}{R_1}-(n-1)\frac{1}{R_2}=(n-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right).

    Writing s≡s1s\equiv s_1 for the original object and s′≡s2′s'\equiv s_2' for the final image, this is exactly the thin-lens equation 1s+1s′=1f\dfrac{1}{s}+\dfrac{1}{s'}=\dfrac{1}{f} with

    1f=(n−1)(1R1−1R2),\frac{1}{f}=(n-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right),

    since by definition ff is the image distance s′s' when the object is infinitely far away (1/s→01/s\to 0). This is the lensmaker's equation. ■\blacksquare