A student examines a slide of pond algae with a compound microscope…

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A student examines a slide of pond algae with a compound microscope. The figure shows its two lenses, both of which can be treated as thin. The objective has a focal length of $1.20\ \text{cm}$ , and the slide is placed $1.50\ \text{cm}$ in front of it. The eyepiece has a focal length of $5.0\ \text{cm}$ . The student moves the eyepiece until the final image is a virtual image at the least distance of distinct vision, $25\ \text{cm}$ from the eye. Treat the eye as touching the eyepiece. The magnifying power of a microscope is the angle that the final image makes at the eye, divided by the angle that the object would make if placed at the same distance, $25\ \text{cm}$ .

A schematic of a compound microscope along a horizontal axis, not to scale. A small arrow on the left is the object on the slide, 1.50 cm in front of the objective lens, which is marked with a focal length of 1.20 cm. A tube joins the objective to the eyepiece lens on the right, marked with a focal length of 5.0 cm. An eye touches the eyepiece. A note says the final image is virtual and lies 25 cm from the eye. No rays are drawn.

At what distance from the objective does the objective form its image?
What is the magnitude of the linear magnification of the objective?
At what distance from the eyepiece does this first image lie?
What is the magnitude of the total magnifying power?
What is the distance between the two lenses?

The worked solution

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Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Plan. Find the image of the objective first. It is the object for the eyepiece. Then multiply the two magnifications.

(a) Objective. Here $u = -1.50\ \text{cm}$ and $f = 1.20\ \text{cm}$ . So $\dfrac{1}{v} = \dfrac{1}{1.20} - \dfrac{1}{1.50} = 0.1667\ \text{cm}^{-1}$ , which gives $v = 6.00\ \text{cm}$ .

(b) Objective magnification. $m_o = \dfrac{v}{u} = \dfrac{6.00}{-1.50} = -4.0$ . The magnitude is $4.0$ , and the image is inverted.

(c) Eyepiece. Here $v = -25\ \text{cm}$ and $f = 5.0\ \text{cm}$ . So $\dfrac{1}{u} = \dfrac{1}{v} - \dfrac{1}{f} = -\dfrac{1}{25} - \dfrac{1}{5.0} = -0.2400\ \text{cm}^{-1}$ , which gives $u = -4.17\ \text{cm}$ . The first image lies $4.17\ \text{cm}$ from the eyepiece.

(d) Total magnifying power. The eyepiece gives $m_e = 1 + \dfrac{D}{f_e} = 1 + \dfrac{25}{5.0} = 6.0$ . So $M = m_o \times m_e = 4.0 \times 6.0 = 24.0$ in magnitude.

(e) Distance between the lenses. $6.00\ \text{cm} + 4.17\ \text{cm} = 10.17\ \text{cm}$ .

Answer: (a) $6.00\ \text{cm}$ . (b) $4.0$ . (c) $4.17\ \text{cm}$ . (d) $24.0$ . (e) $10.17\ \text{cm}$ .

Check. Take a $2.0\ \text{mm}$ object. The objective makes an image $8.0\ \text{mm}$ tall, and the eyepiece makes it $48.0\ \text{mm}$ tall at $25\ \text{cm}$ . Seen from the same $25\ \text{cm}$ , that is $\dfrac{48.0}{2.0} = 24.0$ times the angle of the object.

Watch out. If the final image were at infinity, the eyepiece would give $\dfrac{D}{f_e} = 5.0$ and the total would be $20.0$ , not $24.0$ . The usual short formula $\dfrac{L}{f_o}\dfrac{D}{f_e}$ also needs a different $L$ , the gap between the two focal points, so it is not used here.

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A student examines a slide of pond algae with a compound microscope…

The worked solution

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