Optical Instruments

Lesson

Part 1 — making tiny things look big: the magnifying glass and the compound microscope.

In this lesson you will

Define magnifying power as β/α and derive it for a simple microscope with the image at D and at infinity
Draw the ray diagram of a compound microscope and derive its magnifying power with the final image at D
Use the expression for a compound microscope with the final image at infinity

Concept

Angular size and magnifying power

How big an object looks depends on the angle it subtends at the eye. The eye cannot focus closer than the near point, the least distance of distinct vision, $D = 25$ cm. The magnifying power of an instrument is $M = \dfrac{\beta}{\alpha} = \dfrac{\text{angle subtended at the eye by the final image}}{\text{angle subtended by the object without the instrument}}$

For a microscope, α is the angle when the object is held at D, the closest the unaided eye can view it clearly.
For small angles, an angle is (height)/(distance).

Table defining the two angles in the magnifying power beta over alpha: for a microscope, alpha is the angle of the object held at 25 centimetres; for a telescope, the angle of the object where it is; beta is always the angle of the final image at the eye
Instrument	α (without the instrument)	β (with the instrument)
Microscope	Object held at D = 25 cm from the eye	Final image at the eye
Telescope	Object at its actual (very large) distance	Final image at the eye

Diagram

Simple microscope: image at D

Hold a convex lens close to the eye with the object inside F. The image is virtual, erect and enlarged; put it at D: $v = -D$ . With the eye at the lens, $\beta = \dfrac{h}{|u|}$ and $\alpha = \dfrac{h}{D}$ , so $M = \dfrac{D}{|u|}$ . From $\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$ with $v = -D$ : $\dfrac{1}{|u|} = \dfrac{1}{f} + \dfrac{1}{D}$ , hence $M = 1 + \dfrac{D}{f}$

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