How big something looks depends on the angle it covers at your eye, not on its actual size. To see a small thing in more detail, you bring it closer, but your eye cannot focus closer than its near point, about 25 cm (see the eye). A magnifying glass lets you bring it closer anyway, and a microscope, a two-lens system, does it twice.
Key idea
A simple magnifier of focal length $f$ (in cm) gives angular magnification $$M = \frac{25\ \text{cm}}{f}$$ with the object at its focal point (the eye relaxed), or $1 + \frac{25\ \text{cm}}{f}$ with the image at the near point. A compound microscope's total magnification is the product $$M = m_{\text{obj}}\,M_{\text{eye}} \approx -\frac{L}{f_{\text{obj}}}\cdot\frac{25\ \text{cm}}{f_{\text{eye}}},$$ where $L$ is the distance from the objective's focal point to the image it forms (the tube length, often 16 cm).
Why it works
Without a lens, the best you can do is hold an object of height $h$ at 25 cm, where it covers the angle $\theta_0 = h/25\ \text{cm}$. With the object at a magnifier's focal point, the rays leave parallel, as if from a far-off object covering the angle $\theta = h/f$. The ratio is $\theta/\theta_0 = 25\ \text{cm}/f$. In a microscope, the objective first makes a real image $|m_{\text{obj}}|$ times as large as the object, and the eyepiece is a magnifier for that image. The image is $|m_{\text{obj}}|$ times larger, and the eyepiece multiplies its angle by $M_{\text{eye}}$: the factors multiply. When the object is just outside the objective's focal point, $d_i \approx L + f_{\text{obj}}$ and $|m_{\text{obj}}| = \frac{d_i - f_{\text{obj}}}{f_{\text{obj}}} = \frac{L}{f_{\text{obj}}}$.
Worked examples
Worked example: A magnifying glass
A magnifier has $f = 5.0$ cm. Find its angular magnification with a relaxed eye, and with the image at the near point.
- Relaxed: $M = \frac{25}{5.0} = 5.0$.
- Image at the near point: $M = 1 + \frac{25}{5.0} = 6.0$.
Answer: $5.0\times$ relaxed, $6.0\times$ at the near point
Worked example: Objective times eyepiece
A microscope has a $40\times$ objective and a $10\times$ eyepiece. What is its total magnification?
- The eyepiece magnifies the image the objective has already enlarged: $40 \times 10 = 400$.
- The image is inverted, so with signs $M = -400$.
Answer: $400\times$ (inverted)
Worked example: From the lens equation
An objective with $f = 0.50$ cm holds an object 0.52 cm away. The eyepiece has $f = 2.5$ cm. Find the total magnification.
- Objective: $\frac{1}{d_i} = \frac{1}{0.50} - \frac{1}{0.52} = 0.0769$, so $d_i = 13.0$ cm.
- $m_{\text{obj}} = -\frac{13.0}{0.52} = -25$.
- Eyepiece: $M_{\text{eye}} = \frac{25}{2.5} = 10$.
- $M = (-25)(10) = -250$.
Answer: $M = -250$
Common mistake
Adding the two magnifications ($40 + 10 = 50$). The eyepiece magnifies an image that is already 40 times enlarged, so the factors multiply. Check with a picture: if the objective's image is 40 times the object's size and the eyepiece makes anything look 10 times bigger, the result is 400 times.
Common mistake
Expecting a magnifier to work with the object outside its focal length. Then the lens makes a real, inverted image on the far side, not the enlarged upright view you want; keep the object at or inside $f$.