Optical instruments — microscope, telescope, human eye comparison

Question

Compare the compound microscope and astronomical telescope in terms of their construction, magnifying power formulas, and final image position. Why does a telescope need a large objective while a microscope needs a small objective?

(CBSE 12 & NEET — frequently asked comparison)

Solution — Step by Step

Construction comparison

Feature	Compound Microscope	Astronomical Telescope
Objective	Short focal length, small aperture	Long focal length, large aperture
Eyepiece	Short focal length	Short focal length
Object distance	Just beyond $f_o$	At infinity (very far)
Image formation	Real, magnified intermediate image	Real, diminished intermediate image
Purpose	Magnify nearby tiny objects	See distant objects more clearly

Magnifying power formulas

Microscope (image at near point $D = 25$ cm): $M = -\frac{L}{f_o}\left(1 + \frac{D}{f_e}\right)$

where $L$ = tube length (distance between lenses minus focal lengths).

Telescope (image at infinity — normal adjustment): $M = -\frac{f_o}{f_e}$

For telescope at near point: $M = -\dfrac{f_o}{f_e}\left(1 + \frac{f_e}{D}\right)$

Why the objectives differ

Microscope needs short $f_o$ to produce high magnification ( $M \propto 1/f_o$ )
Telescope needs large $f_o$ for high magnification ( $M \propto f_o$ ) and large aperture to collect more light from distant faint objects

The telescope's large objective also improves resolving power — the ability to distinguish two close objects.

Why This Works

Both instruments use two converging lenses, but their goals differ fundamentally:

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The human eye is itself an optical instrument with a variable focal length lens (accommodation). Defects arise when the focal length or eye shape changes: myopia (far point becomes finite) and hypermetropia (near point moves beyond 25 cm).

Alternative Method — Compare by Ray Diagrams

Draw the ray diagram for each: in a microscope, the object is between $f_o$ and $2f_o$ (producing a magnified real image), while in a telescope, parallel rays from infinity converge at $f_o$ (producing a small real image at the focal plane). The eyepiece then magnifies this intermediate image in both cases.

💡 Expert Tip

For NEET: the sign of magnification tells you the image orientation. Negative $M$ means inverted. Both instruments give inverted final images in normal use. An erecting lens can make the image upright (terrestrial telescope) but this is rarely tested.

Common Mistake

⚠️ Common Mistake

Students swap the magnification formulas — using $f_o/f_e$ for the microscope and $L/(f_o f_e)$ for the telescope. The telescope formula is simpler ( $M = f_o/f_e$ ) because the object is at infinity. The microscope formula involves tube length $L$ because the intermediate image size depends on how far the objective projects the image. If in doubt, remember: telescope has the simpler formula.

Question

Solution — Step by Step

Why This Works

Alternative Method — Compare by Ray Diagrams

Common Mistake

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