Physics / Unit 8: Light and Optics
8.5 Lenses
Notes
A lens uses refraction to form images.
- A converging (convex) lens is thicker in the middle and brings parallel rays to a focal point; $f$ is positive.
- A diverging (concave) lens is thinner in the middle and spreads rays out; $f$ is negative. It always makes a smaller, upright, virtual image.
- Thin lens equation: $\displaystyle \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}$ and $\displaystyle m = -\frac{d_i}{d_o}$. For lenses, a positive $d_i$ is a real image on the far side.
- Object farther than $f$ from a converging lens: real, inverted image (cameras, projectors, your eye). Object closer than $f$: virtual, upright, enlarged image (magnifying glass).
- Example: $f = 10 \text{ cm}$, $d_o = 5 \text{ cm}$: $\displaystyle \frac{1}{d_i} = \frac{1}{10} - \frac{1}{5} = -\frac{1}{10}$, so $d_i = -10 \text{ cm}$ and $m = +2$.
Practice
An object is 30 cm from a converging lens with $f = 10 \text{ cm}$. What is the image distance?
$\displaystyle \frac{1}{d_i} = \frac{1}{10} - \frac{1}{30} = \frac{2}{30}$, so $d_i = 15 \text{ cm}$.
An object is $30 \text{ cm}$ from a lens and its image forms $15 \text{ cm}$ away ($d_o = 30 \text{ cm}$, $d_i = 15 \text{ cm}$). What is the magnification?
$\displaystyle m = -\frac{15}{30} = -0.5$: real, inverted and half the size.
A magnifying glass works by placing the object:
Inside the focal length, a converging lens forms an enlarged, upright, virtual image.
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