Optics problem solving guide

1. The thin lens equation

The single most important relation:

$\frac{1}{f}=\frac{1}{d_o}+\frac{1}{d_i}$

Rearrange to solve for whichever unknown you need. Almost every numerical question starts here.

Sign conventions

Quantity	Positive	Negative
$d_o$	Object on incoming side (real)	Object on outgoing side (virtual)
$d_i$	Image on outgoing side (real)	Image on incoming side (virtual)
$f$	Converging lens	Diverging lens
$m$	Upright image	Inverted image

Get these wrong and everything falls apart. Draw the diagram first and assign signs before touching the equation.

Magnification

$m=-\frac{d_i}{d_o}$

Total magnification for a multi-lens system is the product of individual magnifications.

Example 1.1 — Real image from a converging lens

Q: An object is 40.0 cm from a converging lens of focal length 15.0 cm. Find the image distance and magnification.

A:

$\frac{1}{d_i}=\frac{1}{f}-\frac{1}{d_o}=\frac{1}{15.0}-\frac{1}{40.0}=\frac{40.0-15.0}{15.0\times 40.0}=\frac{25.0}{600};d_i=\frac{600}{25}$

$d_i=24.0\text{ cm}$

Positive $d_i$ → real image on the far side of the lens.

$m=-\frac{d_i}{d_o}=-\frac{24.0}{40.0}=-0.600$

Inverted, demagnified.

Sanity check: object beyond $2f$ (30 cm), so image should be between $f$ and $2f$ (15–30 cm). 24.0 cm ✓

Example 1.2 — Virtual image from a diverging lens

Q: An object is 25.0 cm from a diverging lens with $f=-20.0$ cm. Where is the image?

A:

$\frac{1}{d_i}=\frac{1}{-20.0}-\frac{1}{25.0}=-0.0500-0.0400=-0.0900$

$d_i=-11.1\text{ cm}$

Negative → virtual image, same side as the object. This is always the case for diverging lenses with real objects.

$m=-\frac{-11.1}{25.0}=+0.444$

Upright, reduced — characteristic of a diverging lens.

Example 1.3 — Finding focal length from image data

Q: A sharp image forms 88.0 mm behind a converging lens when the object is 508 mm away. What is the focal length?

A:

$\frac{1}{f}=\frac{1}{d_o}+\frac{1}{d_i}=\frac{1}{508}+\frac{1}{88.0}=0.001969+0.01136=0.01333$

$f=75.0\text{ mm}$

Sanity check: $d_i<d_o$ and both positive, so a real image closer to the lens than the object — consistent with $d_o\gg 2f$.

---

2. Multi-lens systems

Method

1. Solve the thin lens equation for lens 1 to find $d_{i1}$.
2. Compute the object distance for lens 2: $d_{o2}=\text{separation}-d_{i1}$.
3. If $d_{i1}$ lands beyond lens 2, the converging rays haven’t met yet — virtual object. The sign of $d_{o2}$ goes negative.
4. Solve the thin lens equation for lens 2.
5. Multiply magnifications: $m_{\text{total}}=m_1\times m_2$.

Virtual objects

This is where most mistakes happen. If lens 1 produces an image past lens 2, lens 2 intercepts converging rays before they focus. From lens 2’s frame, the “object” is behind it — hence $d_{o2}<0$.

Quick check: if $d_{i1}>\text{lens separation}$, you have a virtual object for lens 2.

Object at infinity

Parallel rays from infinity focus at $f$. So $d_{i1}=f_1$, then proceed to lens 2 as above. This is the standard starting point for telescope/compound lens problems.

Example 2.1 — Virtual object (image past lens 2)

Q: Two converging lenses ($f_1=f_2=25.0$ cm) are 16.5 cm apart. Object is 35.0 cm in front of lens 1. Find the final image position and total magnification.

A:

Lens 1:

$\frac{1}{d_{i1}}=\frac{1}{25.0}-\frac{1}{35.0}=\frac{35.0-25.0}{25.0\times 35.0}=\frac{10.0}{875}⟹d_{i1}=87.5\text{ cm}$

This is past lens 2 (87.5 > 16.5), so virtual object for lens 2:

$d_{o2}=-(87.5-16.5)=-71.0\text{ cm}$

Lens 2:

$\frac{1}{d_{i2}}=\frac{1}{25.0}-\frac{1}{-71.0}=0.0400+0.01408=0.05408⟹d_{i2}=18.5\text{ cm}$

Magnification:

$m_1=-\frac{87.5}{35.0}=-2.50,m_2=-\frac{18.5}{-71.0}=-0.260$

$m_{\text{total}}=(-2.50)(-0.260)=+0.651$

Final image: real, 18.5 cm behind lens 2, upright, demagnified.

Example 2.2 — Object at infinity through a diverging-converging pair

Q: A converging lens ($f_1=20.0$ cm) is followed by a diverging lens ($f_2=-33.5$ cm), separated by 14.0 cm. Object at infinity. Where is the final image?

A:

Object at infinity → lens 1 focuses at $d_{i1}=f_1=20.0$ cm.

This image falls $20.0-14.0=6.0$ cm behind lens 2 → virtual object: $d_{o2}=-6.0$ cm.

$\frac{1}{d_{i2}}=\frac{1}{-33.5}-\frac{1}{-6.0}=-0.02985+0.1667=0.1368$

$d_{i2}=7.3\text{ cm behind lens 2}$

Real image. The diverging lens doesn’t diverge the beam here because the virtual object produces a converging contribution that dominates.

Example 2.3 — Real object for lens 2 (image between the lenses)

Q: A converging lens ($f_1=12.0$ cm) and a diverging lens ($f_2=-18.0$ cm) are 30.0 cm apart. Object is 24.0 cm in front of lens 1. Find the final image.

A:

Lens 1:

$\frac{1}{d_{i1}}=\frac{1}{12.0}-\frac{1}{24.0}=\frac{1}{24.0}⟹d_{i1}=24.0\text{ cm}$

$d_{i1}<30.0$, so the image forms between the lenses → real object for lens 2:

$d_{o2}=30.0-24.0=+6.0\text{ cm}$

$\frac{1}{d_{i2}}=\frac{1}{-18.0}-\frac{1}{6.0}=-0.0556-0.1667=-0.2222$

$d_{i2}=-4.5\text{ cm}$

Negative → virtual image, 4.5 cm in front of lens 2. The diverging lens pushed the image back between the lenses.

Magnification:

$m_1=-\frac{24.0}{24.0}=-1.00,m_2=-\frac{-4.5}{6.0}=+0.75$

$m_{\text{total}}=(-1.00)(+0.75)=-0.75$

Inverted, slightly demagnified.

---

3. The lensmaker’s equation

$\frac{1}{f}=(n_{\text{rel}}-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$

where $n_{\text{rel}}=n_{\text{lens}}/n_{\text{medium}}$.

Key consequences

• Immersion matters: changing the surrounding medium changes $n_{\text{rel}}$ and therefore $f$. If $n_{\text{medium}}=n_{\text{lens}}$, the lens has zero optical power.
• Mirrors are immune: $f=R/2$ depends only on geometry. The medium doesn’t matter.
• Inverted lenses: if $n_{\text{lens}}<n_{\text{medium}}$ (e.g. air pocket in water), $(n_{\text{rel}}-1)<0$ and every shape does the opposite — biconcave converges, biconvex diverges.

Planoconvex lens

One flat surface ($R=\infty$) simplifies the equation:

$\frac{1}{f}=\frac{n-1}{R}$

For sign convention: if the curved surface’s centre of curvature is on the transmission side, $R_2$ is negative.

Example 3.1 — Chromatic aberration in a planoconvex lens

Q: A planoconvex lens has $R=15.3$ cm (flat on one side). For red light $n=1.5106$ and yellow light $n=1.5226$. Object at 66.0 cm. Find the image locations for each colour.

A:

Flat surface → $R_1=\infty$, curved surface → $R_2=-15.3$ cm.

$\frac{1}{f}=(n-1)\left(0-\frac{1}{-15.3}\right)=\frac{n-1}{15.3}$

Red: $1/f_r=0.5106/15.3=0.03337⟹f_r=29.97$ cm

Yellow: $1/f_y=0.5226/15.3=0.03416⟹f_y=29.27$ cm

Image positions using thin lens equation with $d_o=66.0$ cm:

Red: $1/d_i=0.03337-0.01515=0.01822⟹d_i=54.9$ cm

Yellow: $1/d_i=0.03416-0.01515=0.01901⟹d_i=52.6$ cm

Yellow image forms 2.3 cm closer — this separation is chromatic aberration.

Example 3.2 — Lens immersed in water

Q: A biconvex glass lens ($n=1.52$, $R_1=+20.0$ cm, $R_2=-20.0$ cm) is used in air, then immersed in water ($n_w=1.33$). Find $f$ in each case.

A:

In air ($n_{\text{rel}}=1.52$):

$\frac{1}{f}=(1.52-1)\left(\frac{1}{20.0}-\frac{1}{-20.0}\right)=0.52\times \frac{2}{20.0}=0.0520$

$f_{\text{air}}=19.2\text{ cm}$

In water ($n_{\text{rel}}=1.52/1.33=1.143$):

$\frac{1}{f}=(1.143-1)\left(\frac{2}{20.0}\right)=0.143\times 0.100=0.01430$

$f_{\text{water}}=69.9\text{ cm}$

Focal length increases by a factor of ~3.6. The lens is much weaker underwater because the refractive index contrast is smaller.

Example 3.3 — Air lens in water

Q: An air-filled biconvex plastic shell ($R_1=+10.0$ cm, $R_2=-10.0$ cm) is submerged in water ($n_w=1.33$). Is it converging or diverging? Find $f$.

A:

$n_{\text{lens}}=1.00$ (air), $n_{\text{medium}}=1.33$, so $n_{\text{rel}}=1.00/1.33=0.752$.

$\frac{1}{f}=(0.752-1)\left(\frac{1}{10.0}-\frac{1}{-10.0}\right)=(-0.248)(0.200)=-0.0496$

$f=-20.2\text{ cm}$

Negative → diverging. A biconvex air lens in water diverges light (inverted behaviour). To converge, you’d need a biconcave shape.

---

4. Aberrations

Spherical aberration

Marginal rays focus at a different point than paraxial rays. Minimised by:

• Distributing bending across both surfaces (curved side faces the parallel rays for a planoconvex lens).
• Reducing aperture (squinting — blocks marginal rays, increases depth of field, sharpens image at the cost of brightness).

Chromatic aberration

Dispersion means $n$ varies with $\lambda$: $n_{\text{violet}}>n_{\text{red}}$, so $|f_{\text{violet}}|<|f_{\text{red}}|$ for both converging and diverging lenses.

Achromatic doublet: a converging element (crown glass, low dispersion) cemented to a diverging element (flint glass, high dispersion). The dispersions cancel while net power remains positive. Combined focal length for thin lenses in contact:

$\frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2}$

Example 4.1 — Achromatic doublet

Q: An achromatic doublet uses a diverging lens ($f_1=-28.0$ cm) cemented to a converging lens ($f_2=+25.0$ cm). Is the combination converging or diverging? Find the net focal length.

A:

$\frac{1}{f}=\frac{1}{-28.0}+\frac{1}{25.0}=-0.03571+0.04000=0.00429$

$f=233\text{ cm}$

Positive → converging, but very weakly. The diverging element nearly cancels the converging element’s power — that’s the point. The two elements are made of different glass types (different dispersion curves), so chromatic aberration cancels while net positive power remains.

Example 4.2 — Focal length shift due to dispersion

Q: A symmetric biconvex lens has $f=18.0$ cm for yellow light ($n_y=1.523$). Find $f$ for violet light ($n_v=1.545$).

A:

For a symmetric biconvex lens: $1/f=(n-1)(2/R)$. The ratio of focal lengths:

$\frac{f_y}{f_v}=\frac{n_v-1}{n_y-1}=\frac{0.545}{0.523}=1.042$

$f_v=\frac{f_y}{1.042}=\frac{18.0}{1.042}=17.3\text{ cm}$

Violet focuses 0.7 cm closer — the longitudinal chromatic aberration.

Example 4.3 — Planoconvex orientation for minimum spherical aberration

Q: A planoconvex lens is used to focus a collimated laser beam. Should the curved or flat surface face the beam? Why?

A:

The curved surface should face the incoming parallel beam. With this orientation, the first surface partially refracts the rays inward, and the second (flat) surface completes the bending. The total angular deviation is shared roughly equally between the two surfaces.

If the flat side faces the beam, rays pass through the first surface undeviated (normal incidence on a flat surface), and all bending occurs at the curved exit surface. Snell’s law is nonlinear — concentrating all refraction at one surface produces larger higher-order (spherical) aberration terms than splitting the same total deviation across two surfaces.

---

5. Diffraction and resolution

Single slit

Minima at: $\sin \theta =\frac{m\lambda }{a}$, where $m=\pm 1,\pm 2,\dots$

• Wider slit ($a\uparrow$) → narrower pattern.
• Longer wavelength ($\lambda \uparrow$) → wider pattern.
• If $a<\lambda$, no minima exist — light spreads in all directions.

Rayleigh criterion (circular aperture)

${\theta }_{\min }=1.22\frac{\lambda }{D}$

Applies to lenses and spherical mirrors (both have finite apertures). Does not apply to plane mirrors (no convergence through an aperture).

Resolution of a microscope

$d_{\min }=\frac{0.61\lambda }{\text{NA}}$

Shorter $\lambda$ → better resolution. Violet light gives the best optical resolution. Visible light ($\sim$ 400–700 nm) cannot resolve atoms ($\sim 10^{-10}$ m) — you need electron microscopy or scanning probe methods.

Immersion

Placing a high-$n$ fluid between the final element and the sample:

• Increases NA ($=n\sin \theta$), improving resolution.
• Reduces effective wavelength (${\lambda }_{\text{eff}}={\lambda }_0/n$), narrowing diffraction patterns.

This is the principle behind immersion lithography (water, $n=1.44$, enabling sub-40 nm features at 193 nm).

Example 5.1 — Single slit width from diffraction pattern

Q: Violet light ($\lambda =415$ nm) produces a central diffraction peak 8.20 cm wide on a screen 2.85 m away. Find the slit width.

A:

Half-width of the central max = $4.10$ cm $=0.0410$ m. This corresponds to the first minimum.

$\sin \theta \approx \tan \theta =\frac{y}{L}=\frac{0.0410}{2.85}=0.01439$

$a=\frac{\lambda }{\sin \theta }=\frac{415\times 10^{-9}}{0.01439}=2.88\times 10^{-5}\text{ m}=28.8\text{ µm}$

Example 5.2 — Rayleigh criterion for a telescope

Q: A telescope has a 25.0 cm diameter objective. What is the minimum angular separation it can resolve at $\lambda =550$ nm? Could it resolve two stars separated by 0.5 arcseconds?

A:

${\theta }_{\min }=1.22\frac{\lambda }{D}=1.22\times \frac{550\times 10^{-9}}{0.250}=2.68\times 10^{-6}\text{ rad}$

Convert to arcseconds: $2.68\times 10^{-6}\times \frac{180}{\pi }\times 3600=0.55^{\prime \prime }$

The 0.5” separation is slightly below the 0.55” limit, so the stars are not quite resolvable at this wavelength with this aperture.

Example 5.3 — Effect of immersion on single-slit diffraction

Q: A single slit ($a=40.0$ µm) is illuminated with 600 nm light. Find the angular half-width of the central maximum in air and in water ($n=1.33$).

A:

First minimum at $\sin \theta =\lambda /a$.

In air:

$\sin \theta =\frac{600\times 10^{-9}}{40.0\times 10^{-6}}=0.01500⟹\theta =0.859°$

In water (${\lambda }_{\text{eff}}=600/1.33=451$ nm):

$\sin \theta =\frac{451\times 10^{-9}}{40.0\times 10^{-6}}=0.01128⟹\theta =0.646°$

The pattern narrows by a factor of $n$. This is exactly the principle behind immersion lithography — shorter effective wavelength means tighter diffraction features and better resolution.

---

6. Compound instruments

Microscope magnification

$M=-\frac{L}{f_o}\cdot \frac{N}{f_e}$

where $L$ is the barrel length minus both focal lengths ($L=\ell -f_o-f_e$) and $N=25$ cm (near point). The negative sign indicates an inverted final image.

Example 6.1 — Microscope magnification

Q: A microscope has $f_o=0.65$ cm, $f_e=1.50$ cm, barrel length 17.5 cm. Final image at infinity. Find the magnification.

A:

$L=17.5-0.65-1.50=15.35\text{ cm}$

$M=-\frac{15.35}{0.65}\times \frac{25}{1.50}=-23.6\times 16.7=-394$

Magnification $\approx 390\times$, inverted.

Example 6.2 — Designing for a target magnification

Q: You need $|M|=200\times$ using an eyepiece with $f_e=2.50$ cm and barrel length 16.0 cm. What objective focal length is required?

A:

$|M|=\frac{L}{f_o}\cdot \frac{N}{f_e}$

Approximate $L\approx 16.0-2.50=13.5$ cm (since $f_o\ll L$):

$200=\frac{13.5}{f_o}\times \frac{25}{2.50}=\frac{13.5}{f_o}\times 10.0$

$f_o=\frac{135}{200}=0.675\text{ cm}$

Iterate: $L=16.0-0.675-2.50=12.825$ cm → $M=(12.825/0.675)(10.0)=190$. Slightly under, so reduce $f_o$ to ~0.64 cm for exactly 200×.

Example 6.3 — Useful vs empty magnification

Q: The microscope from 6.1 uses white light with NA $=1.25$ (oil immersion). What is the smallest resolvable feature? Is $390\times$ useful or empty magnification?

A:

Best resolution at $\lambda =400$ nm (violet):

$d_{\min }=\frac{0.61\times 400\text{ nm}}{1.25}=195\text{ nm}\approx 0.20\text{ µm}$

At the near point (25 cm), the eye resolves ~70 µm. At $390\times$, a 0.20 µm feature appears as $0.20\times 390=78$ µm — just above the eye’s limit. The magnification is useful: matched to the diffraction limit.

At $1000\times$ with the same NA, features would appear bigger but no new detail would emerge — empty magnification.

---

7. Problem-solving checklist

1. Draw the diagram. Label every surface, distance, and index.
2. Assign signs using the convention table above before substituting.
3. Identify the equation. Thin lens eq for image problems, lensmaker’s eq for focal length / material problems, diffraction relations for resolution problems.
4. Watch for virtual objects in multi-lens setups — the most common source of sign errors.
5. Sanity check the result. Object beyond $2f$ → image between $f$ and $2f$, etc. Positive $m$ → upright, $|m|<1$ → demagnified.
6. Units. Keep everything in the same unit system throughout. Convert at the end.