Optical instruments — microscope, telescope, human eye comparison

Question

Compare the compound microscope and astronomical telescope in terms of: (a) purpose, (b) focal length of objective vs eyepiece, (c) magnifying power formula, and (d) image characteristics. Why can't we use a telescope as a microscope?

Solution — Step by Step

Understand the purpose of each instrument

Compound microscope: magnifies tiny nearby objects (cells, bacteria). Object is very close to the objective.
Astronomical telescope: magnifies distant large objects (planets, stars). Object is essentially at infinity.
Human eye: has a built-in lens system with focal length that adjusts (accommodation). Normal near point = 25 cm, far point = infinity.

Compare focal lengths and magnifying power

Property	Compound Microscope	Astronomical Telescope
Objective focal length	Short ( $f_o$ small)	Long ( $f_o$ large)
Eyepiece focal length	Short ( $f_e$ small)	Short ( $f_e$ small)
Magnifying power	$m = \frac{L}{f_o} \times \frac{D}{f_e}$	$m = \frac{f_o}{f_e}$
Tube length	$L \approx v_o + f_e$	$L \approx f_o + f_e$
Final image	Virtual, inverted	Virtual, inverted

Here $L$ is tube length, $D = 25$ cm (near point), $v_o$ is image distance from objective.

Why a telescope cannot work as a microscope

A telescope's objective has a large focal length — it forms a real image of distant objects near its focus. If you place a nearby tiny object in front of it, the image forms far behind the lens (if at all), and the magnification is poor. The optics are designed for a completely different scenario. Similarly, a microscope's short-focal-length objective cannot focus parallel rays from distant objects.

Why This Works

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Both the microscope and telescope use two convex lenses — but the design philosophy is opposite. A microscope needs a short-focal-length objective to produce high magnification of a nearby object. A telescope needs a long-focal-length objective to collect light from a distant object and produce a bright image.

🎯 Exam Insider

NEET loves asking the magnifying power formula for both instruments. The most common trap: the microscope formula changes depending on whether the final image is at the near point ( $D$ ) or at infinity. At the near point: $m = -\frac{v_o}{u_o}\left(1 + \frac{D}{f_e}\right)$ . At infinity: $m = -\frac{L}{f_o} \cdot \frac{D}{f_e}$ . Always check which case the question specifies.

Alternative Method

💡 Expert Tip

A quick way to remember: microscope magnification depends on tube length, telescope magnification depends only on focal length ratio. If a question gives tube length, it is a microscope problem. If it gives only focal lengths of objective and eyepiece, it is likely a telescope problem.

For the telescope, to make magnification large, make $f_o$ large and $f_e$ small. For the microscope, make both $f_o$ and $f_e$ small and tube length $L$ large.

Common Mistake

⚠️ Common Mistake

Forgetting the sign convention in microscope magnification. The compound microscope produces an inverted image, so the magnification is negative. Many students write $m = \frac{L}{f_o} \times \frac{D}{f_e}$ without the negative sign. In CBSE boards, the negative sign is expected. In JEE/NEET MCQs, the options usually give the magnitude — but read carefully whether they ask for magnification or magnifying power (magnitude).

Optical Instruments — Quick Reference

Simple magnifier: $m = 1 + \frac{D}{f}$ (image at near point) or $m = \frac{D}{f}$ (image at infinity)

Compound microscope: $m = -\frac{v_o}{u_o}\left(1 + \frac{D}{f_e}\right)$ or $m = -\frac{L}{f_o} \cdot \frac{D}{f_e}$

Astronomical telescope: $m = -\frac{f_o}{f_e}$ (normal adjustment)

Tube length of telescope: $L = f_o + f_e$

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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