Notes
Reflection
- Law of reflection: angle of incidence = angle of reflection, measured from the normal.
- Plane mirrors make upright, virtual images the same size, as far behind the mirror as the object is in front.
Refraction
- Light slows down in materials: $\displaystyle n = \frac{c}{v}$, where $n$ is the index of refraction.
- Snell's law: $n_1\sin\theta_1 = n_2\sin\theta_2$. Going into a higher-$n$ material, light bends toward the normal.
- Total internal reflection happens going from higher $n$ to lower $n$ when the angle exceeds the critical angle: $\displaystyle \sin\theta_c = \frac{n_2}{n_1}$. Used in fiber optics.
- Dispersion: different colors bend by different amounts (prisms, rainbows).
Mirrors and lenses
- Converging (concave mirror, convex lens): $f > 0$. Diverging (convex mirror, concave lens): $f < 0$.
- A positive $s_i$ means a real image (light actually meets there; inverted). A negative $s_i$ means a virtual image (upright).
- For a mirror, $\displaystyle f = \frac{R}{2}$.
| Object position (converging lens) | Image |
|---|---|
| Beyond $2f$ | Real, inverted, smaller |
| At $2f$ | Real, inverted, same size |
| Between $f$ and $2f$ | Real, inverted, larger |
| Inside $f$ | Virtual, upright, larger (magnifier) |
- Diverging lenses and convex mirrors always make virtual, upright, smaller images.
Practice
Angle of incidence equals angle of reflection.
$\displaystyle n = \frac{3.0 \times 10^8}{2.0 \times 10^8}$.
Glass has a higher index of refraction.
It requires going from higher $n$ to lower $n$ beyond the critical angle.
$\displaystyle \frac{1}{s_i} = \frac{1}{10} - \frac{1}{30} = \frac{2}{30}$.
$\displaystyle M = -\frac{15}{30} = -0.5$.
That gives a virtual, enlarged image.
That's why they're used as wide-angle security mirrors.
Free response
Write your answer first, then open the scoring guide.
Free-response question
A 2.0 cm tall object is placed 15 cm in front of a converging lens with focal length 10 cm.
(a) Calculate the image distance.
Show the answer and scoring guide
$\displaystyle \frac{1}{s_i} = \frac{1}{10} - \frac{1}{15} = \frac{1}{30}$, so $s_i = 30$ cm (on the other side of the lens).
Check off the points you earned:
(b) Calculate the magnification and image height, and describe the image.
Show the answer and scoring guide
$\displaystyle M = -\frac{30}{15} = -2.0$. Image height $= -4.0$ cm. The image is real, inverted, and twice as tall.
Check off the points you earned:
(c) The object moves to 5.0 cm from the lens. Describe how the image changes.
Show the answer and scoring guide
Now the object is inside $f$: $\displaystyle \frac{1}{s_i} = \frac{1}{10} - \frac{1}{5} = -\frac{1}{10}$, so $s_i = -10$ cm. The image becomes virtual, upright, and enlarged ($M = 2.0$), on the same side as the object.
Check off the points you earned: