Geometric and Physical Optics — AP Physics 2 Phys 2 Study Guide

For: AP Physics 2 candidates sitting AP Physics 2.

Covers: Reflection, Snell’s law of refraction, total internal reflection, mirror and lens image formation, diffraction, interference, and single-slit and double-slit pattern analysis as specified in the AP Physics 2 CED.

You should already know: AP Physics 1 or equivalent.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Physics 2 style for educational use. They are not reproductions of past College Board papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official College Board mark schemes for grading conventions.

1. What Is Geometric and Physical Optics?

Geometric optics models light as straight-line rays to describe how light interacts with surfaces and transparent media, explaining phenomena like reflection, refraction, and image formation by mirrors and lenses. Physical optics treats light as a wave to explain effects that the ray model cannot, including diffraction, interference, and polarization. Combined, these two subfields make up 15-20% of the AP Physics 2 exam content, with equal weight given to multiple-choice and free-response questions on the topic.

2. Reflection and Snell's law of refraction

Reflection

When light hits a smooth, opaque surface, it undergoes specular reflection, following the law of reflection: $${\theta }_i={\theta }_r$$ Both the incident angle ${\theta }_i$ and reflected angle ${\theta }_r$ are measured relative to the normal, an imaginary line perpendicular to the surface at the point of incidence. Diffuse reflection from rough surfaces follows the same rule, but scattered rays reflect at different angles due to uneven surface topography.

Refraction

When light passes between two transparent media with different optical densities, it changes speed and bends, a phenomenon called refraction. The refractive index $n$ of a medium quantifies how much it slows light: $$n=\frac{c}{v}$$ where $c=3\times 10^8\text{m/s}$ is the speed of light in a vacuum, and $v$ is the speed of light in the medium. The index of refraction is always $\ge 1$, with air approximated as $n=1.00$ for AP Physics 2 calculations.

Snell’s Law describes the relationship between incident and refracted angles: $$n_1\sin {\theta }_1=n_2\sin {\theta }_2$$ where $n_1$ and ${\theta }_1$ are the refractive index and incident angle in the first medium, and $n_2$ and ${\theta }_2$ are the values in the second medium. If $n_2>n_1$, light bends toward the normal; if $n_2<n_1$, light bends away from the normal.

Worked Example

A light ray travels from air ($n_1=1.00$) into a block of quartz ($n_2=1.54$) at an incident angle of $35^{\circ }$ from the normal. Calculate the refracted angle.

1. Rearrange Snell’s Law to solve for $\sin {\theta }_2$: $\sin {\theta }_2=\frac{n_1\sin {\theta }_1}{n_2}$
2. Substitute values: $\sin {\theta }_2=\frac{1.00\times \sin (35^{\circ })}{1.54}=\frac{0.5736}{1.54}\approx 0.372$
3. Solve for ${\theta }_2$: ${\theta }_2=\arcsin (0.372)\approx 21.8^{\circ }$

Exam tip: Examiners often draw surfaces at oblique angles to trick you into measuring angles from the surface instead of the normal. Always draw the normal line explicitly before calculating angles.

3. Total internal reflection

Total internal reflection (TIR) occurs when light travels from a higher-refractive-index medium to a lower-refractive-index medium, and the incident angle exceeds a threshold critical angle ${\theta }_c$. At angles above ${\theta }_c$, no refracted ray forms, and 100% of the light reflects back into the original medium.

The critical angle is the incident angle where the refracted angle equals $90^{\circ }$. Substitute ${\theta }_2=90^{\circ }$ ($\sin 90^{\circ }=1$) into Snell’s Law to get the critical angle formula: $$\sin {\theta }_c=\frac{n_2}{n_1}$$ where $n_1$ is the refractive index of the medium the light is traveling from, and $n_2$ is the index of the medium it would enter if refraction occurred.

TIR is the basis of fiber optic communication, where light signals bounce along glass or plastic fibers with almost no energy loss, and prismatic binoculars, which use TIR to fold the light path and reduce device size.

Worked Example

Calculate the critical angle for light traveling from diamond ($n=2.42$) to air ($n=1.00$).

1. Substitute into the critical angle formula: $\sin {\theta }_c=\frac{1.00}{2.42}\approx 0.413$
2. Solve for ${\theta }_c$: ${\theta }_c=\arcsin (0.413)\approx 24.4^{\circ }$ This low critical angle explains why diamonds sparkle: most light entering the diamond undergoes repeated TIR before exiting, creating high brightness and dispersion.

4. Mirrors and lenses — image formation

All mirrors and lenses follow the same core equations, with sign conventions that determine the type, orientation, and size of the image formed. The AP Physics 2 standard sign convention is used here:

• Object distance $d_o$ is always positive for real objects (the default for all AP exam questions)
• Focal length $f$ is positive for converging optical elements (concave mirrors, convex lenses) and negative for diverging elements (convex mirrors, concave lenses)
• Image distance $d_i$ is positive for real images (formed in front of mirrors, behind lenses) and negative for virtual images (formed behind mirrors, in front of lenses)
• Magnification $m$: positive values mean upright images, negative values mean inverted images; $|m|>1$ means magnified, $|m|<1$ means reduced.

The mirror/lens equation relates focal length, object distance, and image distance: $$\frac{1}{f}=\frac{1}{d_o}+\frac{1}{d_i}$$ The magnification formula is: $$m=-\frac{d_i}{d_o}=\frac{h_i}{h_o}$$ where $h_i$ is image height and $h_o$ is object height.

Worked Example

A diverging lens has a focal length of $-15$ cm. A 4 cm tall object is placed 30 cm to the left of the lens. Find the image position, height, orientation, and type.

1. Substitute into the lens equation: $\frac{1}{-15}=\frac{1}{30}+\frac{1}{d_i}$
2. Rearrange to solve for $d_i$: $\frac{1}{d_i}=-\frac{1}{15}-\frac{1}{30}=-\frac{3}{30}=-\frac{1}{10}$, so $d_i=-10$ cm
3. Calculate magnification: $m=-\frac{(-10)}{30}=+0.33$
4. Image height: $h_i=m\times h_o=0.33\times 4=1.32$ cm Result: The image is 10 cm to the left of the lens, 1.32 cm tall, upright, and virtual.

5. Diffraction and interference

Diffraction is the bending of light around obstacles or through narrow openings, a phenomenon that can only be explained by the wave model of light. Huygens’ Principle describes this: every point on a wavefront acts as a source of secondary spherical wavelets that spread out at the speed of light in the medium.

Interference occurs when two or more coherent light waves (same frequency, constant phase difference) overlap, following the principle of superposition:

• Constructive interference: Wave crests align with crests, troughs align with troughs, and the combined amplitude is the sum of individual amplitudes, producing bright fringes. This occurs when the path difference $\Delta r$ (difference in distance traveled by the two waves) is an integer multiple of wavelength: $\Delta r=m\lambda$, where $m=0,\pm 1,\pm \mathrm{2...}$ (the order of the fringe).
• Destructive interference: Wave crests align with troughs, and the combined amplitude is the difference of individual amplitudes, producing dark fringes. This occurs when the path difference is a half-integer multiple of wavelength: $\Delta r=(m+\frac{1}{2})\lambda$.

Coherence is required for a stable interference pattern: without matching frequency and constant phase difference, fringes will shift rapidly and not be visible.

Worked Example

Two coherent light waves with wavelength 600 nm have a path difference of 1500 nm. State the type of interference that occurs.

1. Divide path difference by wavelength: $\frac{\Delta r}{\lambda }=\frac{1500}{600}=2.5=2+\frac{1}{2}$
2. The path difference is a half-integer multiple of wavelength, so destructive interference occurs.

6. Single-slit and double-slit patterns

Double-slit interference (Young’s experiment)

When monochromatic light passes through two narrow, closely spaced slits, the diffracted light from each slit overlaps and produces an interference pattern of equally spaced bright and dark fringes on a screen placed behind the slits. The condition for bright fringes is: $$d\sin \theta =m\lambda$$ where $d$ is the distance between the two slits, $\theta$ is the angle from the central maximum to the fringe, and $m$ is the order of the fringe. For small angles ($\theta <10^{\circ }$, the standard on AP exams), $\sin \theta \approx \tan \theta =\frac{y}{L}$, where $y$ is the distance from the central maximum to the fringe, and $L$ is the distance from the slits to the screen. Rearranged, the position of bright fringes is: $$y=\frac{m\lambda L}{d}$$ The spacing between adjacent bright fringes is constant: $\Delta y=\frac{\lambda L}{d}$.

Single-slit diffraction

When light passes through a single narrow slit, it produces a diffraction pattern with a wide, bright central maximum, and narrower, dimmer secondary maxima on either side. The condition for dark fringes (minima) is: $$a\sin \theta =m\lambda$$ where $a$ is the width of the single slit, and $m=\pm 1,\pm \mathrm{2...}$. Using the small angle approximation, the position of dark fringes is $y=\frac{m\lambda L}{a}$. The central maximum spans the distance between the $m=+1$ and $m=-1$ dark fringes, so its width is $2y_1=\frac{2\lambda L}{a}$.

Worked Example

A double-slit setup has a slit separation of 0.2 mm, and the screen is 2 m away from the slits. The distance between the central maximum and the second order bright fringe is 1.1 cm. Calculate the wavelength of the light used.

1. Rearrange the double-slit fringe position formula to solve for $\lambda$: $\lambda =\frac{yd}{mL}$
2. Substitute values (convert all units to meters): $y=0.011$ m, $d=0.2\times 10^{-3}$ m, $m=2$, $L=2$ m
3. Calculate: $\lambda =\frac{0.011\times 0.2\times 10^{-3}}{2\times 2}=5.5\times 10^{-7}$ m = 550 nm.

7. Common Pitfalls (and how to avoid them)

• Wrong move: Measuring incident/refracted angles from the medium surface instead of the normal. Why students do it: Diagrams often draw surfaces at angles, making the surface easier to reference than the normal. Correct move: Explicitly draw the normal (perpendicular line) at the point of incidence before calculating any angles.
• Wrong move: Applying total internal reflection when light travels from a lower to higher refractive index medium. Why students do it: They memorize the critical angle formula without remembering its context. Correct move: First check the direction of light travel: TIR only occurs when $n_1>n_2$, no critical angle exists otherwise.
• Wrong move: Mixing up focal length sign conventions for mirrors and lenses. Why students do it: Convex lenses have positive f, but convex mirrors have negative f, leading to easy confusion. Correct move: Memorize the rule: "Converging elements (concave mirror, convex lens) = positive f; diverging elements (convex mirror, concave lens) = negative f".
• Wrong move: Using the double-slit bright fringe formula for single-slit bright fringes. Why students do it: They mix up $a$ (single slit width) and $d$ (double slit separation), and forget the single-slit formula solves for dark fringes, not bright. Correct move: Write down variable definitions first: $a$ = single slit width, $d$ = double slit separation; the standard single-slit formula only applies to dark minima.
• Wrong move: Assuming negative magnification means a virtual image. Why students do it: Students associate negative values with virtual properties. Correct move: Image distance sign tells you if an image is real or virtual; magnification sign only tells you if it is upright or inverted, the two are independent.

8. Practice Questions (AP Physics 2 Style)

Question 1

A light ray travels from sapphire ($n=1.77$) into an unknown transparent liquid. The incident angle is $37^{\circ }$ from the normal, and the refracted angle is $49^{\circ }$. (a) Calculate the refractive index of the liquid. (b) What is the critical angle for the sapphire-liquid interface, if any?

Solution

(a) Use Snell’s Law: $n_1\sin {\theta }_1=n_2\sin {\theta }_2$ Rearrange to solve for $n_2$: $n_2=\frac{n_1\sin {\theta }_1}{\sin {\theta }_2}=\frac{1.77\times \sin (37^{\circ })}{\sin (49^{\circ })}=\frac{1.77\times 0.6018}{0.7547}\approx 1.41$ (b) TIR is possible because light travels from higher n (sapphire) to lower n (liquid). Use the critical angle formula: $\sin {\theta }_c=\frac{n_2}{n_1}=\frac{1.41}{1.77}\approx 0.7966$, so ${\theta }_c=\arcsin (0.7966)\approx 52.8^{\circ }$

Question 2

A concave mirror has a focal length of 25 cm. A 5 cm tall object is placed 40 cm in front of the mirror. (a) Find the position of the image formed. (b) Calculate the image height and orientation. (c) State if the image is real or virtual, and where it is located relative to the mirror.

Solution

(a) Use the mirror equation: $\frac{1}{f}=\frac{1}{d_o}+\frac{1}{d_i}$ $f=+25$ cm (concave mirror is converging), $d_o=40$ cm $\frac{1}{d_i}=\frac{1}{25}-\frac{1}{40}=\frac{8-5}{200}=\frac{3}{200}$, so $d_i\approx 66.7$ cm (b) Magnification: $m=-\frac{d_i}{d_o}=-\frac{66.7}{40}\approx -1.67$ Image height: $h_i=m\times h_o=-1.67\times 5=-8.35$ cm The image is 8.35 cm tall, and the negative magnification means it is inverted. (c) Positive $d_i$ means the image is real, located 66.7 cm in front of the mirror.

Question 3

A single slit of width 0.04 mm is illuminated with 620 nm red light, and the diffraction pattern is projected onto a screen 1.8 m away. (a) Calculate the width of the central bright maximum. (b) If the single slit is replaced with a double slit with separation 0.2 mm, calculate the spacing between adjacent bright fringes on the same screen.

Solution

(a) Central maximum width is twice the distance to the first dark fringe ($m=1$): $y_1=\frac{m\lambda L}{a}=\frac{1\times 620\times 10^{-9}\times 1.8}{0.04\times 10^{-3}}=0.0279$ m = 2.79 cm Central width = $2y_1=5.58$ cm (b) Double-slit fringe spacing: $\Delta y=\frac{\lambda L}{d}=\frac{620\times 10^{-9}\times 1.8}{0.2\times 10^{-3}}=0.00558$ m = 0.558 cm

9. Quick Reference Cheatsheet

10. What's Next

Mastery of geometric and physical optics is critical for success in later AP Physics 2 units, particularly modern physics, where you will extend the wave model of light to explore wave-particle duality and photon behavior. Optics is also frequently combined with electromagnetism and energy topics in free-response questions, so the sign conventions and formula fluency you build here will directly improve your performance on cross-unit exam problems.

If you struggle with any of the concepts, sign conventions, or practice problems in this guide, reach out to Ollie, our AI tutor, for personalized explanations, extra practice questions, or step-by-step walkthroughs tailored to your learning gaps. You can also find more AP Physics 2 study resources and full-length practice tests on the homepage to fully prepare for your exam.