Microscopes

Microscopes are designed to resolve and magnify nearby small objects. Two basic types: the simple magnifier (a single converging lens) and the compound microscope (two lenses).

Concept

Least distance of distinct vision $D = 25$ cm (standard value for a normal eye).

Simple magnifier (magnifying glass)

A converging lens of short focal length $f$. The object is placed just inside the focal point so that a virtual, erect, magnified image forms at the near point or at infinity.

• Image at near point ($D$): $M = 1 + \frac{D}{f}$
• Image at infinity (relaxed eye): $M = \frac{D}{f}$

Compound microscope

Two converging lenses: objective (short focal length $f_o$) close to the specimen, and eyepiece ($f_e$) close to the eye. The objective forms a real, inverted, enlarged intermediate image; the eyepiece then acts as a simple magnifier on this image.

Magnification:

$M = m_o \times m_e$

Approximately (for image at infinity by eyepiece, and tube length $L \approx v_o \approx$ length between lenses minus focal lengths):

$M \approx -\frac{L}{f_o}\cdot\frac{D}{f_e}$

A short $f_o$ and short $f_e$ together give a large magnification.

Derivation

For the simple magnifier at near point:

The angle subtended by the object at the eye, without the lens, viewed at $D$, is $\theta_0 = h/D$. With the lens producing a virtual image at $D$ of magnification $m = 1 - v/f$... using $v = -D$ (image at near point on the same side), $1/(-D) - 1/u = 1/f$ gives $1/u = -1/D - 1/f$. The angular size of the image (or equivalently of the object close to the lens) is $\theta = h/|u|$. Then

$M = \frac{\theta}{\theta_0} = \frac{D}{|u|} = D\left(\frac{1}{D} + \frac{1}{f}\right) = 1 + \frac{D}{f}$

For the eye relaxed (image at infinity), put $v = -\infty$, so $|u| = f$, giving $M = D/f$.

Worked Example

A compound microscope has $f_o = 0.5$ cm, $f_e = 5$ cm, tube length $L = 15$ cm, $D = 25$ cm. Magnification?

$M \approx -\frac{L}{f_o}\cdot\frac{D}{f_e} = -\frac{15}{0.5}\cdot\frac{25}{5} = -30 \times 5 = -150$

Negative sign indicates the image is inverted.

Common Confusions

• The two formulas ($1 + D/f$ vs $D/f$) differ depending on where the final image forms; problems usually state "for normal (relaxed) eye" (use $D/f$) or "for least distance of distinct vision" (use $1 + D/f$).
• For the compound microscope, magnification is the product, not the sum.
• $f_o$ must be small for high resolution and magnification; $f_e$ also small but a little larger to give a comfortable working distance.

Key Takeaways

• Simple magnifier: $M = 1 + D/f$ (near point) or $D/f$ (infinity).
• Compound microscope: $M \approx (L/f_o)(D/f_e)$.
• Short focal lengths give strong magnification; the objective does the heavy lifting on resolution.