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

Chapter 33 — The Nature & Propagation of Light

Problems with worked solutions · 13 points · 3 extra credit

Chapter 33 — The Nature & Propagation of Light

Core: 13 points · Extra credit: 3 points. Show your reasoning; partial credit is given.

1. 1. (1 pt •) A beam of light travels through a transparent liquid whose index of refraction is $n=1.36$.

   (a) Find the speed of the light in the liquid.

   (b) If the light has vacuum wavelength $589\,\text{nm}$, find its wavelength in the liquid. Does the frequency change?

   Solution

   1.

   (a) The speed in a medium is $v=\dfrac{c}{n}=\dfrac{3.00\times10^{8}}{1.36}\approx 2.2\times10^{8}\,\text{m/s}$.

   (b) The wavelength shortens by the factor $n$: $\lambda=\dfrac{\lambda_0}{n}=\dfrac{589\,\text{nm}}{1.36}\approx 433\,\text{nm}$. The frequency does not change — it is set by the source; only $v$ and $\lambda$ change so that $v=f\lambda$ still holds.

2. 2. (2 pts ••) A ray of light in air strikes the flat surface of a block of crown glass ($n=1.52$) at an angle of incidence of $40^\circ$ measured from the normal.

   (a) Find the angle of refraction inside the glass.

   (b) The ray then travels through the block and hits the far face, which is parallel to the first. Find the angle at which it emerges back into the air, and state its direction relative to the original incoming ray.

   Solution

   2.

   (a) Snell's law: $n_1\sin\theta_1=n_2\sin\theta_2$, so $\sin\theta_2=\frac{(1.00)\sin 40^\circ}{1.52}=\frac{0.643}{1.52}=0.423.$ $\theta_2=\sin^{-1}(0.423)\approx 25^\circ$.

   (b) The two faces are parallel. Applying Snell's law at the second (glass→air) surface with incidence angle $25^\circ$ reverses part (a): $\sin\theta=1.52\sin 25^\circ=1.00\cdot\sin\theta_\text{out}$, giving $\theta_\text{out}=40^\circ$. The emerging ray is parallel to the original incoming ray (same direction), merely displaced sideways by the slab.

3. 3. (2 pts ••) Light inside a block of glass ($n_1=1.52$) reaches a boundary.

   (a) Find the critical angle for total internal reflection if the other side of the boundary is air.

   (b) Find the critical angle if the other side is instead water ($n_2=1.33$). Explain physically why this critical angle is larger.

   Solution

   3.

   (a) At the critical angle $\sin\theta_c=\dfrac{n_2}{n_1}=\dfrac{1.00}{1.52}=0.658$, so $\theta_c=\sin^{-1}(0.658)\approx 41^\circ$.

   (b) $\sin\theta_c=\dfrac{1.33}{1.52}=0.875$, so $\theta_c\approx 61^\circ$. It is larger because the indices are closer together: with less of a mismatch the ray bends less at the boundary, so it must strike at a steeper (larger) angle before refraction reaches $90^\circ$ and total internal reflection sets in.

4. 4. (2 pts ••) An optical fiber has a glass core of index $n_\text{core}=1.48$ surrounded by cladding of index $n_\text{clad}=1.46$.

   (a) Find the critical angle at the core–cladding boundary.

   (b) A ray travels down the fiber and strikes the core–cladding wall at $82^\circ$ from the normal. Is it totally internally reflected? Justify with a comparison to part (a).

   Solution

   4.

   (a) $\sin\theta_c=\dfrac{n_\text{clad}}{n_\text{core}}=\dfrac{1.46}{1.48}=0.9865$, so $\theta_c=\sin^{-1}(0.9865)\approx 81^\circ$ (more precisely $80.6^\circ$).

   (b) The ray strikes at $82^\circ$, which is greater than the critical angle $80.6^\circ$, so the ray is totally internally reflected and stays trapped in the core. (A ray at, say, $79^\circ$ would partly leak into the cladding.)

5. 5. (2 pts ••) A narrow beam of white light in air strikes the surface of a piece of glass at $60^\circ$ from the normal. The glass has index $n=1.514$ for red light and $n=1.528$ for violet light.

   (a) Find the angle of refraction for each color inside the glass.

   (b) Find the angular spread between the red and violet rays inside the glass. Which color bends more, and why?

   Solution

   5.

   (a) Snell's law at $\theta_1=60^\circ$ ($\sin 60^\circ=0.866$):

   Red: $\sin\theta_2=\dfrac{0.866}{1.514}=0.5720\Rightarrow\theta_2\approx 34.9^\circ$.

   Violet: $\sin\theta_2=\dfrac{0.866}{1.528}=0.5668\Rightarrow\theta_2\approx 34.5^\circ$.

   (b) The angular spread is $34.89^\circ-34.53^\circ\approx 0.37^\circ$. Violet bends more (smaller refraction angle) because the glass has a higher index for violet light — this wavelength dependence of $n$ is dispersion.

6. 6. (2 pts ••) Unpolarized light of intensity $I_0$ passes through two ideal polarizing filters. The transmission axis of the second is rotated $30^\circ$ relative to the first.

   (a) Find the intensity transmitted through the first filter alone, in terms of $I_0$.

   (b) Find the intensity transmitted through both filters.

   (c) At what angle between the two axes would the final intensity be exactly $10\%$ of $I_0$?

   Solution

   6.

   (a) An ideal polarizer transmits exactly half of unpolarized light, regardless of its axis orientation: $I_1=\tfrac12 I_0$.

   (b) Malus's law at the second filter: $I_2=I_1\cos^2 30^\circ=\tfrac12 I_0\,(0.866)^2=\tfrac12 I_0(0.75)\approx 0.38\,I_0$.

   (c) Require $\tfrac12\cos^2\theta=0.10$, so $\cos^2\theta=0.20$, $\cos\theta=0.447$, and $\theta=\cos^{-1}(0.447)\approx 63^\circ$.

7. 7. (2 pts ••) A beam of light in air reflects off the flat top surface of a pane of glass ($n=1.52$).

   (a) Find Brewster's angle — the angle of incidence at which the reflected light is completely polarized.

   (b) When light is incident at Brewster's angle, find the angle of refraction of the transmitted ray, and verify that the reflected and refracted rays are perpendicular to each other.

   Solution

   7.

   (a) Brewster's angle: $\tan\theta_p=\dfrac{n_2}{n_1}=\dfrac{1.52}{1.00}=1.52$, so $\theta_p=\tan^{-1}(1.52)\approx 57^\circ$.

   (b) Refraction at $\theta_p$: $\sin\theta_2=\dfrac{\sin 56.66^\circ}{1.52}=\dfrac{0.8358}{1.52}=0.5499$, so $\theta_2\approx 33.3^\circ$. Then $\theta_p+\theta_2=56.7^\circ+33.3^\circ=90.0^\circ$, so the reflected and refracted rays are perpendicular. (This is the defining geometry of Brewster's angle: the reflected ray is $90^\circ$ from the refracted ray, and the reflected light is fully polarized parallel to the surface.)

8. Extra Credit (3 pts •••)

   Three ideal polarizing filters are stacked. Unpolarized light of intensity $I_0$ enters the first, whose transmission axis defines the reference direction $0^\circ$. The third filter's axis is fixed at $90^\circ$ (crossed with the first). The middle filter's axis is at an adjustable angle $\theta$ (with $0^\circ<\theta<90^\circ$).

   (a) Using Malus's law at each stage, derive an expression for the final transmitted intensity $I(\theta)$ in terms of $I_0$ and $\theta$.

   (b) Note that with the middle filter removed, the crossed first-and-third pair transmits nothing. Explain in one or two sentences how inserting the middle filter lets light through.

   (c) Use calculus to find the angle $\theta$ that maximizes $I(\theta)$, and compute the maximum transmitted intensity as a fraction of $I_0$.

   Solution

   Extra Credit

   (a) Stage by stage:

   • After filter 1 (unpolarized in): $I_1=\tfrac12 I_0$, polarized along $0^\circ$.

   • After filter 2 (axis at $\theta$), Malus's law: $I_2=I_1\cos^2\theta=\tfrac12 I_0\cos^2\theta$, now polarized along $\theta$.

   • After filter 3 (axis at $90^\circ$), the angle between the incoming polarization ($\theta$) and this axis is $90^\circ-\theta$: $I_3=I_2\cos^2(90^\circ-\theta)=\tfrac12 I_0\cos^2\theta\,\sin^2\theta.$

   Using $\cos\theta\sin\theta=\tfrac12\sin 2\theta$, $I(\theta)=\tfrac12 I_0\left(\tfrac12\sin 2\theta\right)^2=\frac{I_0}{8}\sin^2 2\theta.$

   (b) With filters 1 and 3 crossed, the light reaching filter 3 is polarized along $0^\circ$, exactly perpendicular to the $90^\circ$ axis, so $\cos^2 90^\circ=0$ transmits nothing. The middle filter re-projects the polarization onto its own intermediate axis $\theta$; the emerging light is polarized along $\theta$, which now has a nonzero component along the $90^\circ$ axis. Each polarizer only passes the projection of the field, so an intermediate step gives the light a "foothold" that lets some through.

   (c) Maximize $I(\theta)=\dfrac{I_0}{8}\sin^2 2\theta$. Differentiate with respect to $\theta$: $\frac{dI}{d\theta}=\frac{I_0}{8}\cdot 2\sin 2\theta\cdot\cos 2\theta\cdot 2=\frac{I_0}{2}\sin 2\theta\cos 2\theta=\frac{I_0}{4}\sin 4\theta.$ Setting $\dfrac{dI}{d\theta}=0$ gives $\sin 4\theta=0$, so $4\theta=180^\circ$, i.e. $\theta=45^\circ$ (the interior solution; $\theta=0,90^\circ$ are minima with zero transmission). The second derivative is negative there, confirming a maximum. The peak transmitted intensity is $I_\text{max}=\frac{I_0}{8}\sin^2(90^\circ)=\frac{I_0}{8}=0.125\,I_0.$ So the best the crossed pair can do, with an optimally-oriented middle filter, is transmit one-eighth of the original unpolarized intensity.