Optical Instruments: The Microscope

Your unaided eye cannot resolve details smaller than about 0.1 mm, no matter how closely you look. The hard limit is the near point: anything closer than about 25 cm blurs out of focus. Optical instruments break past this limit by using lenses to bend light in just the right way, letting you see objects as though they were much larger and closer. The simplest such instrument is the magnifying glass, and building on the same idea with a second lens gives you the compound microscope.

What a Simple Microscope Really Does

A simple microscope (magnifying glass) is just a single converging lens with a short focal length. You hold the lens close to the object, keeping the object at the focal point or just inside it, and look through the lens from the other side.

Why does this help? Think about it this way: without a lens, the biggest a tiny object can appear is when you bring it to the near point ($D \cong 25$ cm). The angle it makes at your eye is as large as your eye can handle while still focusing. A magnifying lens lets you place the object much closer to your eye (closer than 25 cm), and the lens creates a virtual image that your eye can focus on comfortably. The object is still small, but it now fills a larger angle at your eye, so it looks bigger.

There are two ways to set up the viewing, and each gives a slightly different magnification.

Case 1: Image at the Near Point (Maximum Magnification)

Fig 9.23(a): A simple microscope producing the image at the near point

When the object sits slightly inside the focal length, the lens produces a virtual, erect, magnified image on the same side as the object. If you position the image exactly at the near point $D$, you get the maximum magnification.

Start with the general magnification formula for a lens:

$m = \frac{v}{u}$

Using the thin lens equation $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$, we can rewrite $\frac{1}{u}$ as $\frac{1}{v} - \frac{1}{f}$. Substituting:

$m = \frac{v}{u} = v \left(\frac{1}{v} - \frac{1}{f}\right) = 1 - \frac{v}{f}$

Now apply the sign convention. The image is virtual and on the same side as the object, so $v$ is negative. Its magnitude equals $D$ (the near point distance). So $v = -D$:

$m = 1 - \frac{(-D)}{f} = 1 + \frac{D}{f} \qquad \text{(9.39)}$

With $D = 25$ cm, a lens of focal length $f = 5$ cm gives:

$m = 1 + \frac{25}{5} = 1 + 5 = 6$

So the object appears six times larger than it would at the near point without a lens.

Notice that $m = h'/h$, the ratio of image size to object size. This also equals the ratio of the angle the image fills at your eye to the angle the object would fill if placed at $D$ for direct viewing. Be careful here: the reference angle is $h/D$ (what the object would subtend if placed at the near point), not $h/u$ (the angle the object actually subtends at the eye from its current position at distance $u$). The magnification tells you how much better the lens-assisted view is compared to the best your unaided eye could do.

The magnifying glass works by letting you bring the object closer than $D$, and the lens takes care of focusing.

Case 2: Image at Infinity (Comfortable Viewing)

Fig 9.23(b) and (c): Angular magnification, comparing the angle without a lens (b) to the angle with the lens when the object is at the focal point (c)

When you place the object exactly at the focal point of the lens, the image forms at infinity. Your eye views parallel rays, which means the eye muscles are fully relaxed. This is more comfortable for extended viewing, even though the magnification is slightly lower.

To find this magnification, we compare two angles.

Without the lens: the maximum angle the object can fill (while remaining in focus) is when it sits at the near point $D$:

$\tan \theta_o = \frac{h}{D} \approx \theta_o \qquad \text{(9.40)}$

(using the small-angle approximation, since the object is small compared to $D$).

With the lens: the object is now at the focal point, so $u = f$. The angle it fills at the lens is:

$\tan \theta_i = \frac{h}{f} \approx \theta_i \qquad \text{(9.41)}$

This follows from the geometry: when the object of height $h$ is at distance $f$ from the lens, the rays subtend an angle $h/f$ at the lens.

The angular magnification is the ratio of these two angles:

$m = \frac{\theta_i}{\theta_o} = \frac{h/f}{h/D} = \frac{D}{f} \qquad \text{(9.42)}$

This is exactly one less than the near-point magnification (Eq. 9.39). For $f = 5$ cm, the relaxed-eye magnification is $25/5 = 5$ versus the near-point value of 6. The difference is small, and the comfort gained makes the relaxed-eye setup the preferred choice.

For all the instruments that follow (compound microscope, telescope), we will assume the final image is at infinity, so that the viewer’s eye is relaxed.

Why a Single Lens is Not Enough

A single converging lens hits a practical wall at about $m \leq 9$. To get magnification of 9 with the image at the near point, you would need $f = 25/(9-1) \approx 3$ cm, which is already a very short focal length. Pushing further becomes impractical, because extremely short focal length lenses are difficult to make and the viewing becomes awkward.

The solution is to use two lenses, with the second one magnifying the already-magnified image produced by the first. This is the compound microscope.

The Compound Microscope: Two Lenses Working Together

Fig 9.24: Ray diagram for image formation by a compound microscope

A compound microscope uses two converging lenses mounted in a tube:

• The objective (the lens closest to the object): it has a very short focal length $f_o$. The object sits just beyond its focal point, so the objective forms a real, inverted, magnified intermediate image inside the tube.

• The eyepiece (the lens you look through): it has a short focal length $f_e$, but longer than the objective’s. The intermediate image lands at or just inside the eyepiece’s focal point. The eyepiece then works exactly like a simple microscope, magnifying this intermediate image further. The viewer sees a final virtual image that is greatly enlarged and inverted relative to the original object.

Magnification by the Objective

The objective creates the first level of magnification. From the ray diagram, using the geometry of the intermediate image:

$\tan \beta = \frac{h}{f_o} = \frac{h'}{L}$

Here $h$ is the object height, $h'$ is the intermediate image height, $f_o$ is the focal length of the objective, and $L$ is the tube length (the distance between the second focal point of the objective and the first focal point of the eyepiece).

Rearranging the two expressions for $\tan \beta$:

$\frac{h'}{h} = \frac{L}{f_o}$

So the linear magnification produced by the objective is:

$m_o = \frac{h'}{h} = \frac{L}{f_o} \qquad \text{(9.43)}$

The larger the tube length relative to the objective’s focal length, the more the objective magnifies.

Magnification by the Eyepiece

The eyepiece views the intermediate image and magnifies it further, just like a simple microscope.

If the final image is at the near point, using Eq. (9.39):

$m_e = 1 + \frac{D}{f_e} \qquad \text{[9.44(a)]}$

If the final image is at infinity (the preferred, relaxed-eye case), using Eq. (9.42):

$m_e = \frac{D}{f_e} \qquad \text{[9.44(b)]}$

Total Magnification of the Compound Microscope

The total magnification is the product of the objective’s magnification and the eyepiece’s magnification. For the image at infinity:

$m = m_o \times m_e = \frac{L}{f_o} \times \frac{D}{f_e} \qquad \text{(9.45)}$

This formula tells us clearly what you need for high magnification: both $f_o$ and $f_e$ should be as small as possible, and the tube length $L$ should be as large as possible.

In practice, it is hard to make focal lengths much shorter than about 1 cm, and making $L$ large requires long lenses and a long tube.

Worked Example: Computing Compound Microscope Magnification

Problem: A compound microscope has an objective with $f_o = 1.0$ cm, an eyepiece with $f_e = 2.0$ cm, and a tube length of $L = 20$ cm. Find the total magnification when the final image is at infinity ($D = 25$ cm).

Solution:

Step 1: Find the magnification due to the objective.

$m_o = \frac{L}{f_o} = \frac{20}{1.0} = 20$

The objective magnifies the object 20 times.

Step 2: Find the magnification due to the eyepiece (image at infinity).

$m_e = \frac{D}{f_e} = \frac{25}{2.0} = 12.5$

The eyepiece provides an additional magnification of 12.5.

Step 3: Multiply for the total magnification.

$m = m_o \times m_e = 20 \times 12.5 = 250$

The compound microscope produces a total magnification of 250, far beyond what any single lens could achieve.

Improving Image Quality in Real Microscopes

Beyond just the magnification number, several practical factors determine how clear and useful the image actually is. Proper illumination of the object is essential for visibility. In modern microscopes, the objective and eyepiece are not single lenses but multi-component lens systems. These compound lens assemblies are designed to correct for optical aberrations (defects in image formation that arise because real lenses do not focus all rays to a single perfect point). By using multiple lens elements together, each compensating for the flaws of the others, modern instruments deliver sharp, high-contrast images across the entire field of view.