Compound microscope and telescope — magnification formulas and adjustments

Question

Compare the magnification formulas of a compound microscope and an astronomical telescope for (a) normal adjustment and (b) final image at the least distance of distinct vision (D = 25 cm).

(CBSE Class 12 / JEE Main / NEET pattern)

Solution — Step by Step

flowchart TD
    A["Optical Instruments"] --> B["Compound Microscope\n(for tiny objects)"]
    A --> C["Astronomical Telescope\n(for distant objects)"]
    B --> D["Objective: short f, large aperture\nEyepiece: short f"]
    C --> E["Objective: large f, large aperture\nEyepiece: short f"]
    D --> F["M = m₀ × mₑ\n= (L/f₀)(D/fₑ) or\n(L/f₀)(1 + D/fₑ)"]
    E --> G["M = -f₀/fₑ or\n-(f₀/fₑ)(1 + fₑ/D)"]

The objective forms a real, magnified image (linear magnification $m_o$ ), and the eyepiece acts as a simple magnifier (angular magnification $m_e$ ).

Normal adjustment (final image at infinity, relaxed eye):

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

where $L$ = tube length (distance between $f_o$ and $f_e$ focal points), $D = 25$ cm.

Final image at D (least distance of distinct vision):

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

This gives slightly higher magnification because the eyepiece is working harder.

The objective forms a real image at its focal point, and the eyepiece magnifies that image.

Normal adjustment (final image at infinity):

M = -\frac{f_o}{f_e}

Tube length $= f_o + f_e$ .

Final image at D:

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

Tube length $= f_o + \frac{f_e D}{f_e + D}$ .

Feature	Compound Microscope	Astronomical Telescope
Object distance	Very close (near $f_o$ )	Very far (at infinity)
$f_o$	Very small (few mm)	Very large (metres)
$f_e$	Small	Small
Normal adjustment $M$	$\frac{L}{f_o} \cdot \frac{D}{f_e}$	$\frac{f_o}{f_e}$
Image orientation	Inverted	Inverted
Resolution limited by	Diffraction of light	Atmospheric turbulence

Why This Works

Both instruments work in two stages: the objective creates an intermediate image, and the eyepiece magnifies it. The key difference is that a microscope needs to magnify a nearby tiny object (so $f_o$ must be small), while a telescope needs to collect light from a faint distant object (so the objective must have a large aperture and long focal length).

For the telescope, magnification is simply the ratio of focal lengths because the object is at infinity — the angular size of the intermediate image is fixed by $f_o$ , and the eyepiece magnification is $f_o/f_e$ .

Alternative Method — Resolving Power

When questions ask about “quality” rather than magnification:

Resolving power of microscope: $\frac{2n\sin\theta}{\lambda}$ (depends on numerical aperture)

Resolving power of telescope: $\frac{D}{1.22\lambda}$ (depends on objective diameter)

For NEET, the most tested formula is the telescope in normal adjustment: $M = f_o/f_e$ , tube length $= f_o + f_e$ . Also remember: to increase magnification, use a large $f_o$ and small $f_e$ . For microscope: both $f_o$ and $f_e$ should be small, and $L$ should be large.

Common Mistake

Students mix up “normal adjustment” and “image at D.” Normal adjustment means the final image is at infinity (eye is relaxed) — this gives slightly lower magnification but is more comfortable for prolonged viewing. Image at D means the final image is at 25 cm — higher magnification but the eye strains. JEE/NEET problems always specify which case — read carefully, because the formulas differ.

Question

Solution — Step by Step

Why This Works

Alternative Method — Resolving Power

Common Mistake

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