AP Physics 2 Geometric and Physical Optics — Worked Answer Explanations

Unit 5 · 16% of the AP exam · 12 questions explained

1. Question 1 · Easy

   Snell's law describes refraction at the interface between two media. If a ray travels from medium 1 (index $n_1$) into medium 2 (index $n_2>n_1$), what happens to the angle of refraction ${\theta }_2$ compared to the angle of incidence ${\theta }_1$?

   • A
     ${\theta }_2>{\theta }_1$ (ray bends away from the normal)

     Why not A: Bending away from the normal occurs when going into a less dense medium ($n_2<n_1$); entering a denser medium bends toward the normal.
   • B
     ${\theta }_2<{\theta }_1$ (ray bends toward the normal)Correct
   • C
     ${\theta }_2={\theta }_1$ (no bending occurs)

     Why not C: No bending only occurs at normal incidence (${\theta }_1=0$) or when $n_1=n_2$.
   • D
     ${\theta }_2=90°$ (total internal reflection occurs)

     Why not D: Total internal reflection requires $n_1>n_2$ (going from dense to less dense medium) and incidence beyond the critical angle — opposite conditions.
   Explanation

   Snell's law: $n_1\sin {\theta }_1=n_2\sin {\theta }_2$

   Solving for ${\theta }_2$: $\sin {\theta }_2=\frac{n_1}{n_2}\sin {\theta }_1$

   If $n_2>n_1$, then $\frac{n_1}{n_2}<1$, so $\sin {\theta }_2<\sin {\theta }_1$, meaning ${\theta }_2<{\theta }_1$. The ray bends toward the normal when entering a denser (higher $n$) medium. Light slows down in the denser medium, causing it to bend inward — like a car's wheels slowing on one side before the other.

   Key takeaway

   Snell's law: $n_1\sin\theta_1 = n_2\sin\theta_2$; entering denser medium ($n$ increases) bends ray toward normal.

2. Question 2 · Easy

   Total internal reflection can occur when light travels from:

   • A
     Air into glass, at any angle of incidence

     Why not A: Going from air (low $n$) to glass (high $n$) causes refraction toward the normal — no total internal reflection is possible.
   • B
     Glass into air, at angles exceeding the critical angleCorrect
   • C
     Glass into air, at angles smaller than the critical angle

     Why not C: Below the critical angle, light refracts out (and partially reflects); TIR only occurs above the critical angle.
   • D
     Air into water, when the beam is nearly parallel to the surface

     Why not D: Air has a lower index than water; TIR requires traveling from the higher-index medium to the lower-index medium.
   Explanation

   Total internal reflection (TIR) requires:

   1. Light traveling from a higher index medium to a lower index medium ($n_1>n_2$)
   2. Angle of incidence greater than the critical angle ${\theta }_c$

   The critical angle is defined by $\sin {\theta }_c=n_2/n_1$ (when ${\theta }_2=90°$). For glass ($n\approx 1.5$) to air ($n=1$): $\sin {\theta }_c=1/1.5\approx 0.667$, so ${\theta }_c\approx 42°$.

   TIR is the principle behind fiber optics, prism binoculars, and diamond sparkle.

   Key takeaway

   TIR occurs only when going from high-$n$ to low-$n$ medium at angles greater than the critical angle $\theta_c = \sin^{-1}(n_2/n_1)$.

3. Question 3 · Easy

   An object is placed $30\mathrm{cm}$ in front of a concave mirror with focal length $f=10\mathrm{cm}$. Using the mirror equation $\frac{1}{f}=\frac{1}{d_o}+\frac{1}{d_i}$, what is the image distance $d_i$?

   • A
     $d_i=-15\mathrm{cm}$

     Why not A: A negative image distance would indicate a virtual image behind the mirror; the object is outside the focal length, so the image is real and in front.
   • B
     $d_i=15\mathrm{cm}$Correct
   • C
     $d_i=20\mathrm{cm}$

     Why not C: This would result from subtracting instead of using $1/d_i=1/f-1/d_o$.
   • D
     $d_i=40\mathrm{cm}$

     Why not D: This may result from adding $d_o$ and $f$ or from an arithmetic error in the fractions.
   Explanation

   Using the mirror equation:
   $$\frac{1}{d_i}=\frac{1}{f}-\frac{1}{d_o}=\frac{1}{10}-\frac{1}{30}=\frac{3}{30}-\frac{1}{30}=\frac{2}{30}=\frac{1}{15}$$
   $$d_i=15\mathrm{cm}$$

   Positive $d_i$ confirms a real image in front of the mirror. The magnification is $m=-d_i/d_o=-15/30=-0.5$ — the image is inverted and half the size of the object. Object outside the focal length of a concave mirror always gives a real, inverted image.

   Key takeaway

   Mirror equation: $1/f = 1/d_o + 1/d_i$; positive $d_i$ means real image in front of mirror for a concave mirror.

4. Question 4 · Easy

   A converging (convex) lens with focal length $f=20\mathrm{cm}$ has an object placed $10\mathrm{cm}$ from it (inside the focal length). What type of image does the lens produce?

   • A
     Real, inverted, and magnified

     Why not A: Real inverted images from a converging lens occur when the object is outside the focal length, not inside.
   • B
     Virtual, upright, and magnifiedCorrect
   • C
     Real, upright, and diminished

     Why not C: Converging lenses cannot form real upright images of real objects; real images are always inverted.
   • D
     No image is formed when the object is inside the focal length.

     Why not D: An image is still formed — it is a virtual image on the same side as the object.
   Explanation

   Using the thin lens equation: $\frac{1}{f}=\frac{1}{d_o}+\frac{1}{d_i}$

   $$\frac{1}{d_i}=\frac{1}{f}-\frac{1}{d_o}=\frac{1}{20}-\frac{1}{10}=\frac{1}{20}-\frac{2}{20}=-\frac{1}{20}$$
   $$d_i=-20\mathrm{cm}$$

   Negative $d_i$ indicates a virtual image on the same side as the object. Magnification: $m=-d_i/d_o=-(-20)/10=+2$ — the image is upright and magnified. This is exactly how a magnifying glass works: place the object inside the focal length to get an enlarged virtual image.

   Key takeaway

   Object inside focal length of converging lens: $d_i$ is negative (virtual image), upright and magnified — a magnifying glass.

5. Question 5 · Medium

   In a double-slit interference experiment, the slits are separated by $d=0.5\mathrm{mm}$ and a screen is $L=2.0\mathrm{m}$ away. For light of wavelength $\lambda =500\mathrm{nm}$, what is the distance between adjacent bright fringes?

   • A
     $\Delta y=1.0\mathrm{mm}$

     Why not A: This is off by a factor of 2; likely used $d=0.25\mathrm{mm}$ or $L=1.0\mathrm{m}$ instead of the given values.
   • B
     $\Delta y=2.0\mathrm{mm}$Correct
   • C
     $\Delta y=0.5\mathrm{mm}$

     Why not C: This confuses $d$ and $L$; the fringe spacing $\Delta y=\lambda L/d$ has $L$ in the numerator.
   • D
     $\Delta y=4.0\mathrm{mm}$

     Why not D: This doubles the correct answer, possibly from using $2\lambda L/d$ instead of $\lambda L/d$.
   Explanation

   Fringe spacing for double-slit interference (small-angle approximation):
   $$\Delta y=\frac{\lambda L}{d}$$

   Substituting:
   $$\Delta y=\frac{(500\times 10^{-9}\mathrm{m})(2.0\mathrm{m})}{0.5\times 10^{-3}\mathrm{m}}=\frac{10^{-6}}{0.5\times 10^{-3}}=\frac{10^{-6}}{5\times 10^{-4}}=2\times 10^{-3}\mathrm{m}=2.0\mathrm{mm}$$

   Constructive interference (bright fringes) occurs where path difference $\Delta =m\lambda$ for integers $m$; the fringe spacing depends on the wavelength, screen distance, and slit separation.

   Key takeaway

   Double-slit fringe spacing: $\Delta y = \lambda L/d$; larger $\lambda$ or $L$ spreads fringes; larger $d$ compresses them.

6. Question 6 · Medium

   Light travels from glass ($n_g=1.5$) into water ($n_w=1.33$) at an angle of incidence of $60°$. What is the angle of refraction in water? ($\sin 60°\approx 0.866$)

   • A
     ${\theta }_2\approx 40.6°$ (ray bends toward the normal)

     Why not A: Light bending toward the normal occurs when entering a denser medium; here the ray goes from glass ($n=1.5$) to water ($n=1.33$), a less dense medium, so it bends away from the normal.
   • B
     Total internal reflection occurs; no transmitted ray.

     Why not B: The critical angle for glass-to-water is ${\sin }^{-1}(1.33/1.5)\approx 62.5°$. Since $60°<62.5°$, TIR does not occur and the ray is transmitted.
   • C
     ${\theta }_2\approx 77.8°$ (ray bends away from the normal)Correct
   • D
     ${\theta }_2=60°$ (no bending at the glass-water interface)

     Why not D: No bending only occurs when $n_1=n_2$; since glass and water have different indices, the ray changes direction at the interface.
   Explanation

   Step 1: Check for TIR. Critical angle for glass-to-water:
   $$\sin {\theta }_c=\frac{n_w}{n_g}=\frac{1.33}{1.5}\approx 0.887\Rightarrow {\theta }_c\approx 62.5°$$

   Since $60°<62.5°$, TIR does not occur.

   Step 2: Apply Snell's law: $n_g\sin {\theta }_1=n_w\sin {\theta }_2$
   $$\sin {\theta }_2=\frac{n_g\sin 60°}{n_w}=\frac{1.5\times 0.866}{1.33}=\frac{1.299}{1.33}\approx 0.977$$
   $${\theta }_2={\sin }^{-1}(0.977)\approx 77.8°$$

   The ray bends away from the normal (since $n_w<n_g$), consistent with entering a less optically dense medium.

   Key takeaway

   Always check TIR first ($\theta_i > \theta_c?$); if not, apply Snell's law — going to lower $n$ bends the ray away from the normal.

7. Question 7 · Medium

   A diffraction grating with $500lines/mm$ is illuminated with light of wavelength $\lambda =600\mathrm{nm}$. At what angle does the first-order ($m=1$) maximum appear?

   • A
     $\theta \approx 17.5°$Correct
   • B
     $\theta \approx 30°$

     Why not B: This would apply if $d\sin \theta =\lambda$ gave $\sin \theta =0.5$, which requires $d=1200\mathrm{nm}$, not the given $2000\mathrm{nm}$.
   • C
     $\theta \approx 8.6°$

     Why not C: This may result from using the number of lines (500) directly instead of the slit spacing $d=1/500\mathrm{mm}$.
   • D
     $\theta \approx 35.1°$

     Why not D: This is the second-order maximum; using $m=2$ instead of $m=1$.
   Explanation

   Grating spacing: $d=\frac{1}{500lines/mm}=0.002\mathrm{mm}=2000\mathrm{nm}$

   Grating equation for maxima: $d\sin \theta =m\lambda$

   For $m=1$:
   $$\sin \theta =\frac{m\lambda }{d}=\frac{(1)(600\mathrm{nm})}{2000\mathrm{nm}}=0.30$$
   $$\theta ={\sin }^{-1}(0.30)\approx 17.5°$$

   For the second order ($m=2$): $\sin \theta =0.60$, $\theta \approx 36.9°$. The grating equation has the same form as double-slit constructive interference but for many equally-spaced slits.

   Key takeaway

   Diffraction grating: $d\sin\theta = m\lambda$; convert lines/mm to spacing $d$ first, then solve for $\theta$.

8. Question 8 · Medium

   A convex (diverging) mirror always produces what type of image for a real object?

   • A
     Real, inverted, diminished

     Why not A: Real images are formed in front of a mirror; a convex mirror's focal point is behind the mirror (virtual), so it cannot form real images.
   • B
     Real, upright, magnified

     Why not B: Convex mirrors never form real images of real objects; all images are virtual.
   • C
     Virtual, inverted, diminished

     Why not C: Convex mirror images are virtual and diminished, but they are always upright (same orientation as object), not inverted.
   • D
     Virtual, upright, diminishedCorrect
   Explanation

   A convex (diverging) mirror has a negative focal length ($f<0$). Using the mirror equation:
   $$\frac{1}{d_i}=\frac{1}{f}-\frac{1}{d_o}$$

   Since $f<0$ and $d_o>0$, we get $1/d_i<0$, so $d_i<0$ always — the image is virtual (behind the mirror). The magnification $m=-d_i/d_o>0$ (since $d_i<0$) — image is upright. And $|d_i|<d_o$, so $|m|<1$ — image is diminished.

   Convex mirrors are used as side-view mirrors (wide field of view) and security mirrors — always virtual, upright, small.

   Key takeaway

   Convex mirror: always virtual, upright, diminished — regardless of object position. Focal length is negative.

9. Question 9 · Hard

   A thin film of soap ($n=1.35$) in air has a thickness $t=350\mathrm{nm}$. For what wavelength of visible light (in air) does the film appear brightest (constructive interference in reflection)?

   • A
     $\lambda =945\mathrm{nm}$ (infrared — not visible)

     Why not A: This uses $2t$ alone without accounting for the phase shift at the air-soap interface, and gives an infrared wavelength.
   • B
     $\lambda =630\mathrm{nm}$ (orange-red)Correct
   • C
     $\lambda =472.5\mathrm{nm}$ (blue-violet)

     Why not C: This uses $2nt/m$ with $m=2$, giving the second-order condition, which is a destructive condition in the reflection geometry.
   • D
     $\lambda =315\mathrm{nm}$ (ultraviolet — not visible)

     Why not D: This uses a higher-order condition $m=3$ without the half-wave shift correction.
   Explanation

   For a soap film in air: light reflects from the air-soap interface (phase shift of $\pi$, since soap has higher $n$ than air) and from the soap-air interface (no phase shift, going to lower $n$).

   Net phase shifts: one reflection gets a $\lambda /2$ shift → constructive reflection requires:
   $$2nt=\left(m-\frac{1}{2}\right)\lambda \text{or equivalently}2nt=(2m-1)\frac{\lambda }{2}$$

   For the lowest-order constructive interference ($m=1$):
   $$\lambda =\frac{2\times 1.35\times 350\mathrm{nm}}{1/2\cdot 1}=\frac{2nt}{1/2}$$

   Wait — using the standard form $2nt=(m-\frac{1}{2})\lambda$:
   $$\lambda =\frac{2nt}{m-1/2}=\frac{2(1.35)(350)}{0.5}=\frac{945}{0.5}=1890\mathrm{nm}(m=1,\text{not visible})$$
   For $m=2$: $\lambda =945/1.5=630\mathrm{nm}$ — orange-red, visible! ✓

   The soap film appears orange-red in reflection.

   Key takeaway

   Thin film: one phase shift → constructive reflection: $2nt = (m-\frac{1}{2})\lambda$; find $m$ giving a visible wavelength.

10. Question 10 · Hard

    An object is placed $15\mathrm{cm}$ to the left of a converging lens with $f_1=10\mathrm{cm}$. A second converging lens with $f_2=20\mathrm{cm}$ is placed $40\mathrm{cm}$ to the right of the first lens. Where is the final image formed relative to the second lens?

    • A
      $20\mathrm{cm}$ to the right of lens 2 (real image)

      Why not A: This applies $f_2=20\mathrm{cm}$ as the image distance directly, ignoring the lens equation entirely.
    • B
      $20\mathrm{cm}$ to the left of lens 2 (virtual image)Correct
    • C
      $13.3\mathrm{cm}$ to the right of lens 2 (real image)

      Why not C: This uses the original object distance ($15\mathrm{cm}$) as $d_{o2}$ for lens 2, forgetting to compute the intermediate image position first.
    • D
      No image is formed — the object is at $f_2$.

      Why not D: The intermediate image falls $10\mathrm{cm}$ from lens 2, not at $f_2=20\mathrm{cm}$; an image is formed, though it is virtual.
    Explanation

    Step 1: Image from lens 1 (object at $d_{o1}=15\mathrm{cm}$, $f_1=10\mathrm{cm}$):
    $$\frac{1}{d_{i1}}=\frac{1}{10}-\frac{1}{15}=\frac{3-2}{30}=\frac{1}{30}\Rightarrow d_{i1}=30\mathrm{cm}$$
    Real image is $30\mathrm{cm}$ to the right of lens 1.

    Step 2: Object distance for lens 2. Lens 2 is $40\mathrm{cm}$ to the right of lens 1, so the intermediate image is $40-30=10\mathrm{cm}$ to the left of lens 2:
    $$d_{o2}=+10\mathrm{cm}\text{(real object)}$$

    Step 3: Image from lens 2 ($d_{o2}=10\mathrm{cm}$, $f_2=20\mathrm{cm}$):
    $$\frac{1}{d_{i2}}=\frac{1}{20}-\frac{1}{10}=\frac{1-2}{20}=-\frac{1}{20}\Rightarrow d_{i2}=-20\mathrm{cm}$$

    Negative: virtual image $20\mathrm{cm}$ to the left of lens 2.

    Key takeaway

    Two-lens systems: use lens 1 image as lens 2 object; subtract intermediate image position from lens separation to find $d_{o2}$.

11. Question 11 · Hard

    In single-slit diffraction, the first minimum occurs at angle $\theta$ given by $a\sin \theta =\lambda$, where $a$ is the slit width. If the slit width is doubled while keeping the wavelength constant, the central diffraction maximum becomes:

    • A
      Twice as wide and half as bright

      Why not A: Doubling the slit width halves the angle to the first minimum, making the central maximum narrower, not wider.
    • B
      Half as wide and more brightCorrect
    • C
      The same width but twice as bright

      Why not C: The width does change when $a$ changes; width is not independent of slit size.
    • D
      Half as wide and half as bright

      Why not D: Narrowing the central maximum concentrates the same (actually more) light, increasing brightness — intensity scales with $a^2$ in the limit, so narrower but much brighter.
    Explanation

    First minimum at: $\sin {\theta }_1=\lambda /a$

    If $a\to 2a$: $\sin {\theta }_1^{\prime }=\lambda /(2a)$ — the angle is halved, making the central maximum narrower (half as wide).

    The intensity at the center scales as $a^2$ (since more slit area contributes more amplitude). With double the slit width, the amplitude doubles and intensity quadruples at the center — the central maximum is much brighter.

    General principle: wider slit → narrower diffraction pattern (less diffraction). This is the wave optics analog of the geometric optics limit: for very wide slits, light goes essentially straight through with minimal spreading.

    Key takeaway

    Wider slit → narrower (half-width) diffraction pattern but brighter central maximum; $\theta_1 = \sin^{-1}(\lambda/a)$.

12. Question 12 · Hard

    A fish is $1.0\mathrm{m}$ below the surface of water ($n_w=1.33$). A person looks straight down at the fish from air ($n_{\text{air}}=1.0$). Due to refraction, the fish appears to be at what depth below the surface?

    • A
      $d_{\text{apparent}}\approx 0.75\mathrm{m}$Correct
    • B
      $d_{\text{apparent}}\approx 1.33\mathrm{m}$

      Why not B: This multiplies the actual depth by $n_w$; apparent depth is actual depth divided by $n$, not multiplied.
    • C
      $d_{\text{apparent}}=1.0\mathrm{m}$ (no distortion)

      Why not C: Refraction at the water surface always makes objects appear shallower than they actually are when viewed from air.
    • D
      $d_{\text{apparent}}\approx 0.50\mathrm{m}$

      Why not D: This would correspond to $n_w=2.0$; for $n_w=1.33$, the apparent depth is $1.0/1.33\approx 0.75\mathrm{m}$.
    Explanation

    For near-normal viewing, the apparent depth formula applies:
    $$d_{\text{apparent}}=\frac{d_{\text{actual}}}{n_w}\times n_{\text{air}}=\frac{d_{\text{actual}}}{n_w}$$

    (when viewing from air, $n_{\text{air}}=1$)

    $$d_{\text{apparent}}=\frac{1.0\mathrm{m}}{1.33}\approx 0.75\mathrm{m}$$

    Derivation sketch: Light rays from the fish diverge upward. At the surface, Snell's law bends them away from the normal (going air is less dense). To the observer, rays appear to come from a point 0.75 m below — the object appears shallower. This is why spearfishers must aim below where they see the fish.

    Key takeaway

    Apparent depth formula: $d_{\text{apparent}} = d_{\text{actual}}/n$; objects in water always appear shallower when viewed from air.