Simple and compound microscopes

Reading the print on a medicine strip

The print on a medicine strip is tiny. You hold it close, and it becomes a little larger. It becomes still larger when you hold a magnifying glass over it. What did the glass do? It did not make the print bigger. It let you bring the print closer than $25\ \text{cm}$ and still see it sharply.

What is magnifying power?

The size of the picture on the retina depends on the angle that the object makes at the eye. Something close looks big because it makes a larger angle. So optical instruments are compared by the angle, not by the height of the image.

The magnifying power (or angular magnification) is

Here $\beta$ is the angle that the image makes at the eye. The angle $\alpha$ is the one that the object makes when it is seen without the instrument, at the nearest comfortable place. That place is the near point, $D = 25\ \text{cm}$. For a small object of height $h$, $\alpha = \dfrac{h}{D}$.

The simple microscope

A convex lens of short focal length is a simple microscope, or magnifying glass. The object is placed between the lens and its focus. Then the lens forms a virtual, erect and enlarged image.

In the picture, the object is just inside the focus. The image is at the near point, $25\ \text{cm}$ from the lens, on the same side as the object.

The image at the near point. The angle that the image makes is the same as the angle of the object at the lens, $\beta = \dfrac{h}{|u|}$. So

From the lens formula with $v = -D$, $\dfrac{1}{|u|} = \dfrac{1}{D} + \dfrac{1}{f}$. Put this in:

The image at infinity. This is the relaxed setting. The eye does not strain. The object is at the focus, $|u| = f$, so $\beta = \dfrac{h}{f}$ and

The image at the near point gives one more than the relaxed setting, but the eye works harder. A shorter focal length gives a larger magnifying power. A single lens is used up to about $10$ times, because stronger lenses are very small and show distorted images.

Example. A lens has $f = 5.0\ \text{cm}$. At the near point, $m = 1 + \dfrac{25\ \text{cm}}{5.0\ \text{cm}} = 6.0$. At infinity, $m = \dfrac{25\ \text{cm}}{5.0\ \text{cm}} = 5.0$. For the near-point image, the object distance is $|u| = \dfrac{25 \times 5.0}{25 + 5.0}\ \text{cm} = 4.17\ \text{cm}$. This is the picture above: the image is $6$ times taller than the object.

Start with the defaults. Read the two magnifying powers. Then move the focal length to $2.5\ \text{cm}$. How much larger do the two values become? Is the difference between them still $1$?

Practise with a magnifying glass

The compound microscope

A magnifying glass cannot show a cell or a bacterium. We need much more. The compound microscope uses two convex lenses in turn.

• The objective has a very short focal length, about a millimetre or a centimetre. The object is just beyond its focus, and the objective forms a real, inverted and much larger image. We call it $I_1$.
• The eyepiece has a longer focal length, a few centimetres. The first image $I_1$ lies just inside its focus. The eyepiece acts as a simple microscope on $I_1$, and forms a virtual final image.

In the picture, the objective has $f_o = 2.0\ \text{cm}$. The object is $2.5\ \text{cm}$ from it, and the first image is $10\ \text{cm}$ on the other side. The final rays leave the eyepiece parallel, so the final image is at infinity.

The magnifying power. The eyepiece magnifies the image that the objective made. So the two magnifications multiply:

For the objective, $m_o = \dfrac{v_o}{u_o}$. It is negative because the first image is upside down. The final image is also upside down compared with the object.

• The final image at the near point. The eyepiece works as a magnifier with $m_e = 1 + \dfrac{D}{f_e}$. So $M = \dfrac{v_o}{u_o}\left(1 + \dfrac{D}{f_e}\right)$.
• The final image at infinity. Now $m_e = \dfrac{D}{f_e}$. Let $L$ be the distance from the second focus of the objective to the first focus of the eyepiece. The first image is then $L$ beyond the focus of the objective. By the lens formula, $m_o = -\dfrac{L}{f_o}$, exactly. So

What NCERT calls the tube length. NCERT calls this distance $L$ the tube length, and it is not the distance between the two lenses. The two lenses are $f_o + L + f_e$ apart. Many older microscopes had a fixed distance of $160\ \text{mm}$ between the lenses, which is also called the tube length. When a question gives a tube length, read it to see which meaning is used.

Example. The microscope in the picture has $f_o = 2.0\ \text{cm}$, $f_e = 5.0\ \text{cm}$ and $L = 8.0\ \text{cm}$. The object is $2.5\ \text{cm}$ from the objective.

1. The objective. $m_o = -\dfrac{L}{f_o} = -\dfrac{8.0\ \text{cm}}{2.0\ \text{cm}} = -4.0$. Check with the lens formula: $v_o = 10\ \text{cm}$ and $\dfrac{v_o}{u_o} = \dfrac{10\ \text{cm}}{-2.5\ \text{cm}} = -4.0$.
2. The eyepiece. $m_e = \dfrac{D}{f_e} = \dfrac{25\ \text{cm}}{5.0\ \text{cm}} = 5.0$.
3. The microscope. $M = (-4.0)(5.0) = -20$. The final image is $20$ times larger, and upside down.
4. The lenses. They are $f_o + L + f_e = 2.0\ \text{cm} + 8.0\ \text{cm} + 5.0\ \text{cm} = 15\ \text{cm}$ apart.

Example, a design. We want a microscope with $|M| = 150$ from an objective of $f_o = 1.0\ \text{cm}$ and $L = 15\ \text{cm}$. We want $f_e$, so $f_e = \dfrac{L}{f_o}\cdot\dfrac{D}{M} = 15 \times \dfrac{25\ \text{cm}}{150} = 2.5\ \text{cm}$.

Short focal lengths for both lenses give a large magnifying power. A large $L$ also helps.

Start with the defaults. Read the three magnifying powers. Now halve $f_o$. Does $M$ double? Then halve $f_e$ as well.

Practise with a microscope

Magnification is not enough: resolution

Suppose we keep adding magnification. Will we see every detail? No. Light is a wave, and it spreads out when it passes an opening, which is called diffraction. Two points that are too close give two overlapping blobs, and no lens can separate them again.

The smallest separation that a microscope objective can show as two points is

Here $\lambda$ is the wavelength of the light. The refractive index of the medium between the object and the objective is $n$. The half-angle of the cone of light that enters the objective is $\beta$. The product $n\sin\beta$ is the numerical aperture. It is also written as $0.61\,\lambda$ divided by the numerical aperture, which is the same thing.

NCERT writes the limit in the first form. Many other books write the second. A smaller $d_{\min}$ means a better microscope. To improve it we can:

• use light of a shorter wavelength, such as blue or ultraviolet light;
• fill the space between the object and the objective with oil, which has $n \approx 1.5$. This is an oil immersion objective.

Example. For green light of $550\ \text{nm}$ and a dry objective with a numerical aperture of $0.65$, $d_{\min} = \dfrac{0.61 \times 550\ \text{nm}}{0.65} = 516\ \text{nm}$. An oil objective of numerical aperture $1.4$ gives $\dfrac{0.61 \times 550\ \text{nm}}{1.4} = 240\ \text{nm}$.

Magnification and resolution are different. A-Level and IB students learn this clearly. Magnification is the size of the image divided by the size of the object. Resolution is how close two points can be and still look separate. The eye itself can just separate two points that are about $0.10\ \text{mm}$ apart at $25\ \text{cm}$. So for the objective to be useful, the microscope must magnify the smallest separation, $d_{\min}$, to at least $0.10\ \text{mm}$. For $d_{\min} = 240\ \text{nm}$ that needs a magnification of $\dfrac{0.10\ \text{mm}}{240\ \text{nm}} \approx 420$. More magnification than the resolution can use only makes a bigger blur. It is called empty magnification.

Practise with the limit of resolution

Try it in Python

The program prints the magnifying power of three magnifying glasses and of four compound microscopes.

Common mistakes

• Adding the two magnifications. For a compound microscope they multiply.
• Using the formula $1 + \dfrac{D}{f}$ for the relaxed eye. For the image at infinity the magnifying power is $\dfrac{D}{f}$.
• Mixing up the two meanings of tube length. NCERT uses the distance $L$ between the focal points. The distance between the lenses is $f_o + L + f_e$.
• Putting the object at the focus of the objective. The object must be a little beyond $f_o$. Then the first image is real and far away.
• Thinking that more magnification always shows more detail. The limit of resolution decides how much detail exists.
• Forgetting the sign. The final image of a compound microscope is inverted, so $M$ is negative. The size is the magnitude.

Remember these

• Magnifying power $m = \dfrac{\beta}{\alpha}$, an angular ratio, with $D = 25\ \text{cm}$.
• Simple microscope: $m = 1 + \dfrac{D}{f}$ (image at the near point) and $m = \dfrac{D}{f}$ (image at infinity).
• Compound microscope: $M = m_om_e$. For the image at infinity, $M = -\dfrac{L}{f_o}\cdot\dfrac{D}{f_e}$.
• The objective has a short focal length, and the first image is real, inverted and magnified. The eyepiece magnifies it.
• Limit of resolution: $d_{\min} = \dfrac{1.22\,\lambda}{2n\sin\beta}$. Shorter wavelength and oil immersion make it smaller.
• Magnification gives size. Resolution gives detail.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. A watchmaker uses a lens of focal length $3.0\ \text{cm}$ with the image at his near point. What is the magnifying power, and how far is the watch from the lens?

2. Why does a compound microscope use an objective of very short focal length and not a long one?

3. A microscope has an eyepiece of $4.0\ \text{cm}$ and an objective of $0.80\ \text{cm}$, with $L = 16\ \text{cm}$. The final image is at infinity. What is $|M|$, and how far apart are the two lenses?

4. Two students argue. One says: "Increase the magnification and I will see the virus in my school microscope." Why is she wrong?

5. What is the smallest separation that an oil immersion objective with $n\sin\beta = 1.25$ can resolve with light of wavelength $480\ \text{nm}$?

References

• NCERT Physics Part II (Class 12), Chapter 9, the section on the microscope.
• OpenStax University Physics Volume 3, Chapter 2, the section on the microscope.
• OpenStax College Physics 2e, Chapter 26, the section on microscopes.