In the biology laboratory, a student lifts a thin skin from an onion, stains it and slides it under the microscope. Through the eyepiece the skin becomes a wall of bricks, each brick a cell with a dark dot at its centre. When she pushes the slide to the left, the cells move to the right. The eyepiece is marked "10×" and the objective "10×".
Two lenses, two jobs
A compound microscope has two convex lenses at the ends of a tube. The lens near the object is the objective. It has a very short focal length, often a few millimetres. The lens at the eye is the eyepiece. It has a moderate focal length, a few centimetres, and works as a simple microscope.
The object is placed just outside the focus of the objective. The objective then forms a real, inverted and much enlarged image, \(I_1\), inside the tube. The eyepiece is set so that \(I_1\) falls just inside its focus, or on it. The eyepiece then treats \(I_1\) as an object for a magnifying glass. It forms a virtual final image, enlarged again, at \(D\) or at infinity.
Why is the intermediate image real while the final image is virtual? The object lies outside the focus of the objective, so the objective's image is real and inverted. That image lies inside the focus of the eyepiece, so the eyepiece's image is virtual. The eyepiece keeps the orientation of \(I_1\). So the final image is inverted with respect to the object.
In a compound microscope, where do the object and the intermediate image \(I_1\) lie?
Magnifying power
The two lenses multiply their effects. The objective makes \(I_1\) larger than the object by its linear magnification \(m_o = v_o/u_o\). The eyepiece then enlarges the angle of \(I_1\), as a magnifying glass does, by its magnifying power \(M_e\). The derivation below shows that the total is the product.
\[ \alpha = \dfrac{h}{D} \]
Why. the object, of height \(h\), seen by the unaided eye at the least distance of distinct vision.
\[ h_1 = m_o h, \qquad m_o = \dfrac{v_o}{u_o} \]
Why. the objective forms \(I_1\) of height \(h_1\); \(m_o\) is its linear magnification, negative because \(I_1\) is inverted.
\[ \beta = M_e \dfrac{h_1}{D} \]
Why. the eyepiece is a simple microscope with \(I_1\) as its object. \(h_1/D\) is the angle \(I_1\) would make at \(D\), and the eyepiece enlarges it \(M_e\) times.
\[ M = \dfrac{\beta}{\alpha} = \dfrac{M_e m_o h/D}{h/D} = m_o M_e \]
Read it. divide step 3 by step 1 and put in \(h_1\) from step 2. The height and \(D\) cancel.
Put in the numbers of Fig 4.1. The objective has \(f_o = 1\,\text{cm}\) and \(u_o = -1.1\,\text{cm}\). The lens formula gives \(\dfrac{1}{v_o} = \dfrac{1}{1} - \dfrac{1}{1.1} = \dfrac{0.1}{1.1}\), so \(v_o = +11\,\text{cm}\). Then \(m_o = \dfrac{11}{-1.1} = -10\). With the final image at \(D\), the eyepiece gives \(M_e = 1 + 25/5 = 6\). So \(M = -10 \times 6 = -60\).
The minus sign records that the final image is inverted, like the onion cells that move the wrong way. Its size, \(60\), says the cells look sixty times larger than at \(25\,\text{cm}\). With the final image at infinity, \(M_e = 25/5 = 5\) and \(M = -50\). A microscope's engraved numbers are sizes: an objective marked "10×" and an eyepiece marked "10×" give about \(100\) times.
A microscope has an objective of focal length \(2\,\text{cm}\) and an eyepiece of focal length \(6.25\,\text{cm}\). An object is placed \(2.5\,\text{cm}\) from the objective, and the final image is at \(25\,\text{cm}\). Find the magnifying power and the separation of the lenses.
Objective: \(f_o = 2\,\text{cm}\), \(u_o = -2.5\,\text{cm}\). Eyepiece: \(f_e = 6.25\,\text{cm}\), \(v_e = -25\,\text{cm}\).
\(\dfrac{1}{v_o} = \dfrac{1}{2} - \dfrac{1}{2.5} = \dfrac{1}{10}\), so \(v_o = 10\,\text{cm}\) and \(m_o = \dfrac{10}{-2.5} = -4\).
\(M_e = 1 + \dfrac{25}{6.25} = 5\), so \(M = -4 \times 5 = -20\). For the eyepiece, \(\dfrac{1}{u_e} = -\dfrac{1}{25} - \dfrac{1}{6.25} = -\dfrac{5}{25}\), so \(u_e = -5\,\text{cm}\).
The lenses are \(v_o + |u_e| = 10 + 5 = 15\,\text{cm}\) apart, and the microscope magnifies twenty times, with an inverted image.
The objective of a microscope forms an inverted image \(20\) times the size of the object. The eyepiece acts as a simple microscope of magnifying power \(6\). Find the magnifying power of the microscope, with its sign.
A microscope has \(f_o = 1\,\text{cm}\) and \(f_e = 4\,\text{cm}\), with the lenses \(15\,\text{cm}\) apart and the final image at infinity. Find the magnifying power, with its sign.
Tube length and normal adjustment
The distance between the objective and the eyepiece is the tube length of the microscope in this book. It is the objective's image distance plus the eyepiece's object distance, \(L = v_o + |u_e|\). For the final image at \(D\), the eyepiece of \(5\,\text{cm}\) has \(\dfrac{1}{u_e} = -\dfrac{1}{25} - \dfrac{1}{5}\), so \(|u_e| = 4.17\,\text{cm}\). The tube length is \(L = 11 + 4.17 = 15.2\,\text{cm}\).
For long viewing, the final image is put at infinity so that the eye stays relaxed. This setting is called normal adjustment. Then \(I_1\) sits at the focus of the eyepiece, \(|u_e| = f_e\), and \(L = v_o + f_e = 11 + 5 = 16\,\text{cm}\). Switch the chips of Fig 4.1: the eyepiece moves out by \(0.83\,\text{cm}\), and \(M\) falls from \(-60\) to \(-50\).
The NCERT writes the objective's magnification as \(m_o = L/f_o\). Its \(L\) is the gap between the second focus of the objective and the first focus of the eyepiece. In normal adjustment that gap is \(v_o - f_o = 11 - 1 = 10\,\text{cm}\), and \(10/1\) gives \(|m_o| = 10\), exactly. So \(M \approx -\dfrac{L}{f_o}\cdot\dfrac{D}{f_e}\). This book, like the workbook, calls the full lens separation the tube length.
Now the choice of focal lengths makes sense. The objective's magnification is \(v_o/u_o\), and it is large only when \(|u_o|\) is small and close to \(f_o\). A very short \(f_o\) therefore gives a large \(m_o\) with \(I_1\) only a few centimetres away, and the tube stays short. The eyepiece has a moderate \(f_e\). It then gives a comfortable, fairly large view of \(I_1\) with small aberrations.
In the microscope of Fig 4.1 (\(f_o = 1\,\text{cm}\), \(u_o = -1.1\,\text{cm}\), \(f_e = 5\,\text{cm}\)), what is the tube length in normal adjustment, in cm?
Why is the objective of a compound microscope given a very short focal length?
The English scientist Robert Hooke built a compound microscope with three lenses. He lit his specimens with an oil lamp shining through a glass ball of water. In 1665 he published Micrographia, a book of large, careful drawings of a flea, a fly's eye and a thin slice of cork. The cork was made of tiny empty boxes, and Hooke called them cells, after the small rooms of monks. The onion cells of the school laboratory carry his word.
A hospital laboratory looks for malaria parasites in a drop of blood. It uses a \(100\times\) objective and a \(10\times\) eyepiece, about a thousand times in all. The objective's focal length is under \(2\,\text{mm}\), and its front lens nearly touches the slide. A drop of cedar oil, of nearly the same refractive index as glass, fills the gap. Light then goes from slide to lens without bending away at an air gap, and the image stays bright and sharp.
- Adding the two magnifications instead of multiplying them.The eyepiece enlarges the angle of an image that is already enlarged: \(M = m_o M_e\).
- Dropping the sign of \(m_o\).The objective's image is inverted, so \(m_o = v_o/u_o < 0\), and the microscope's \(M\) is negative.
- Taking the tube length as \(v_o\) alone, or as \(f_o + f_e\).The separation is \(v_o + |u_e|\): \(15.2\,\text{cm}\) for the image at \(D\), \(16\,\text{cm}\) in normal adjustment, in the example.