Compound microscope and telescope — magnification formulas and adjustments

Question

What are the magnification formulas for a compound microscope and an astronomical telescope, and how do the normal adjustment and infinity adjustment differ?

Solution — Step by Step

A compound microscope has two convex lenses:

Objective (short focal length $f_o$ , near the object): forms a real, magnified, inverted image
Eyepiece (focal length $f_e$ , near the eye): acts as a magnifying glass on the image from the objective

The image formed by the objective falls just inside the focal point of the eyepiece.

At normal adjustment (final image at near point $D = 25$ cm):

M = m_o \times m_e = \frac{v_o}{u_o} \times \left(1 + \frac{D}{f_e}\right)

When the object is placed very close to $f_o$ and the tube length $L$ is the distance between the lenses:

M \approx \frac{L}{f_o} \times \frac{D}{f_e}

At infinity adjustment (final image at infinity — relaxed eye):

M = \frac{L}{f_o} \times \frac{D}{f_e}

where $L$ = distance between the second focal point of objective and the first focal point of eyepiece (often called the tube length or the separation minus the focal lengths).

Two convex lenses again, but now:

Objective: large focal length $f_o$ (collects light from distant objects)
Eyepiece: short focal length $f_e$

At normal adjustment (final image at infinity — most common case):

M = -\frac{f_o}{f_e}

The negative sign indicates an inverted image. The length of the telescope = $f_o + f_e$ .

At near point adjustment (final image at $D$ ):

M = -\frac{f_o}{f_e}\left(1 + \frac{f_e}{D}\right)

Length = $f_o + u_e$ where $u_e$ is the object distance for the eyepiece (found using lens formula with $v_e = -D$ ).

Property	Compound Microscope	Astronomical Telescope
Purpose	Small nearby objects	Distant large objects
$f_o$	Very small	Very large
$f_e$	Small	Small
$M$ at normal setting	$\frac{L}{f_o} \times \frac{D}{f_e}$	$\frac{f_o}{f_e}$
Image	Inverted, magnified	Inverted, angular magnification
Length	$f_o + f_e + L$	$f_o + f_e$

flowchart TD
    A["Optical Instrument Problem"] --> B{"Which instrument?"}
    B -->|"Microscope"| C{"Adjustment type?"}
    B -->|"Telescope"| D{"Adjustment type?"}
    C -->|"Normal: image at D"| E["M = L/fo times 1 + D/fe"]
    C -->|"Infinity: relaxed eye"| F["M = L/fo times D/fe"]
    D -->|"Normal: image at infinity"| G["M = fo/fe, length = fo + fe"]
    D -->|"Near point: image at D"| H["M = fo/fe times 1 + fe/D"]

Why This Works

Both instruments use two-stage magnification. The objective creates an intermediate image, and the eyepiece magnifies that image further. The total magnification is the product of the two individual magnifications.

For the telescope, the “magnification” is angular magnification (ratio of angles subtended at the eye), because we cannot bring distant stars closer — we can only make them appear larger in angle.

Alternative Method

For quick problem solving, remember that the “normal adjustment” for a telescope means the final image is at infinity (relaxed viewing). But for a microscope, “normal adjustment” means the final image is at the near point ( $D = 25$ cm). This naming convention confuses students, but there is logic: the “normal” way to view through each instrument is what gives the most comfortable or practical viewing experience.

Common Mistake

Students swap the magnification formulas — they use $f_o/f_e$ for the microscope and $L \cdot D/(f_o \cdot f_e)$ for the telescope. Remember: the telescope formula is simpler because the object is at infinity, so the intermediate image forms at $f_o$ automatically. The microscope needs $L$ (tube length) because the object is at a finite distance and the intermediate image position depends on the geometry. CBSE 2024 boards and NEET 2023 both had questions on this.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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