Notes
Electromagnetic waves
- Light is a transverse EM wave; in a vacuum all EM waves move at $c = 3.0 \times 10^8 \text{ m/s}$, and $c = f\lambda$.
- Spectrum by increasing frequency and energy: radio, microwave, infrared, visible (red to violet), ultraviolet, X-ray, gamma.
Reflection
- Angle of incidence $=$ angle of reflection, measured from the normal.
- Plane mirror: virtual, upright, same-size image, as far behind the mirror as the object is in front.
- Curved mirrors: $\displaystyle f = \frac{R}{2}$, $\displaystyle \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}$, $\displaystyle m = -\frac{d_i}{d_o}$. Concave can form real images; convex always forms small virtual images.
Refraction and lenses
- $\displaystyle n = \frac{c}{v}$ and Snell's law $n_1\sin\theta_1 = n_2\sin\theta_2$. Into higher $n$: bends toward the normal.
- Total internal reflection beyond the critical angle, $\displaystyle \sin\theta_c = \frac{n_2}{n_1}$.
- Lenses use the same equation. Converging lenses form real images (object beyond $f$) or magnified virtual images (object inside $f$); diverging lenses form small virtual images.
Color
- Prisms disperse white light; violet bends most. Objects show the colors they reflect.
Topics in this unit
Each topic has its own short notes and practice.
Practice
Wavelength increases as frequency decreases, from gamma toward radio.
$\displaystyle f = \frac{3.0 \times 10^8}{6.5 \times 10^{-7}} \approx 4.6 \times 10^{14} \text{ Hz}$.
Light bends as it leaves the water, so the fish appears shallower.
$\sin\theta_2 = 1.33 \times 0.5 = 0.665$, so $\theta_2 \approx 42^\circ$; it bends away from the normal.
For very large $d_o$, $\displaystyle \frac{1}{d_o} \approx 0$, so $d_i \approx f$.
Light hitting the inner wall beyond the critical angle is completely reflected.
Free response
Write your answer first, then open the scoring guide.
Free-response question
A student places a 2.0 cm tall candle 15 cm in front of a converging lens with a focal length of 10 cm.
(a) Find the image distance.
Show the answer and scoring guide
$\displaystyle \frac{1}{d_i} = \frac{1}{10} - \frac{1}{15} = \frac{3 - 2}{30} = \frac{1}{30}$, so $d_i = 30 \text{ cm}$ on the far side of the lens.
Check off the points you earned:
(b) Find the magnification and the height of the image.
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$\displaystyle m = -\frac{d_i}{d_o} = -\frac{30}{15} = -2$. Image height $= -2 \times 2.0 = -4.0 \text{ cm}$ (4.0 cm tall, inverted).
Check off the points you earned:
(c) Describe the image (real or virtual, upright or inverted, larger or smaller). Then explain what kind of image forms if the candle is moved to 5 cm from the lens.
Show the answer and scoring guide
The image is real, inverted and larger. At 5 cm the candle is inside the focal length: $\displaystyle \frac{1}{d_i} = \frac{1}{10} - \frac{1}{5} = -\frac{1}{10}$, so $d_i = -10 \text{ cm}$. The image is virtual, upright and enlarged ($m = +2$), like a magnifying glass.
Check off the points you earned: