Refraction through lenses

Key ideas

A converging (convex) lens refracts parallel rays so they meet at the focal point, a distance $f$ (the focal length) from the lens. Lenses form images whose position and size are found with two relationships.

The lens equation

$\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}$

• $f$ — focal length of the lens (m)
• $d_o$ — object distance (lens to object, m)
• $d_i$ — image distance (lens to image, m)

A positive $d_i$ means a real image (on the far side, can be projected onto a screen). A negative $d_i$ means a virtual image (same side as the object, seen by looking through the lens — e.g. a magnifying glass).

Magnification

$m = \frac{h_i}{h_o} = \frac{d_i}{d_o}$

• $h_i$ — image height, $h_o$ — object height
• $m > 1$ → image enlarged; $m < 1$ → image reduced
• A real image from a single converging lens is inverted; a virtual image is upright.

:::tip Take reciprocals last. Students often compute $\tfrac{1}{d_i}$ correctly then forget to flip it, quoting the reciprocal as the answer. After $\tfrac{1}{d_i} = 0.05$, the image distance is $d_i = 1/0.05 = 20$ cm — not $0.05$ cm. :::

Practice question

An object sits $20$ cm from a converging lens of focal length $5.0$ cm. Find the image distance.

Worked solution: $\tfrac{1}{d_i} = \tfrac{1}{5} - \tfrac{1}{20} = \tfrac{4-1}{20} = \tfrac{3}{20} = 0.15$, so $d_i = 1/0.15 = 6.7$ cm.