Optical Instruments (Microscope & Telescope)

PHYPrism, Dispersion & Optical Instruments

Optical instruments combine two converging lenses to extend the eye's reach: a compound microscope magnifies tiny nearby objects, while an astronomical telescope magnifies distant ones. The shared core idea is two-stage magnification — the objective forms a real, inverted intermediate image that the eyepiece then views as a simple magnifier, so the total magnifying power is the product (microscope) or ratio (telescope) of the two lens contributions.

ISC examines this through short numericals on magnifying power, tube length, and "which lens is the objective," usually in normal adjustment (final image at infinity) or with the final image at the near point $D=25$ cm.

Compound microscope — magnifying power (image at infinity / normal adjustment)

M = m_o \times m_e = \dfrac{L}{f_o}\times\dfrac{D}{f_e}

m_o=\dfrac{L}{f_o}

is the linear magnification of the objective, where

L

(the "tube length") is the distance of the real intermediate image beyond the objective's focus, approximately the separation of the lenses for high magnification;

m_e=\dfrac{D}{f_e}

is the angular magnification of the eyepiece used as a magnifier with the final image at infinity;

f_o,f_e

are the objective and eyepiece focal lengths and

D=25

cm is the least distance of distinct vision. Magnitudes are used for the final numerical magnifying power.

Compound microscope — magnifying power (final image at

D

)

M = \dfrac{v_o}{|u_o|}\left(1+\dfrac{D}{f_e}\right)

v_o,u_o

are the image and object distances for the objective (found from the lens formula);

\dfrac{v_o}{|u_o|}

is the objective's linear magnification and the bracket is the eyepiece angular magnification when the final image forms at the near point

D=25

cm. Use this when the final image is at

D

, not at infinity.

Lens formula (used for the objective)

\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}

u

= object distance,

v

= image distance,

f

= focal length, all measured from the optic centre with the ISC/NCERT Cartesian sign convention (distances against the incident light are negative). Apply to the objective to obtain

v_o

, and the magnitude of the objective magnification is

|m_o|=\dfrac{v_o}{u_o}

Astronomical telescope — magnifying power (normal adjustment)

M = \dfrac{f_o}{f_e}

f_o

= objective focal length (large),

f_e

= eyepiece focal length (small); normal adjustment means the final image is at infinity, so the intermediate image lies at the common focus of both lenses. The objective is the lens of larger focal length.

Astronomical telescope — tube length and image-at-

D

magnification

L = f_o + f_e \qquad M_D = \dfrac{f_o}{f_e}\left(1+\dfrac{f_e}{D}\right)

L

= tube length in normal adjustment (separation of the lenses);

M_D

is the magnifying power when the final image is formed at the near point

D=25

cm instead of infinity, which gives a slightly larger value than

\dfrac{f_o}{f_e}

Both instruments use two convex (converging) lenses. In a microscope the objective has the SHORTER focal length; in a telescope the objective has the LONGER focal length and the eyepiece the shorter one.
Magnifying power is an ANGULAR magnification (ratio of the angle subtended at the eye by the image to that subtended by the object), not a linear size ratio — that is why the near-point distance $D=25$ cm enters the eyepiece term.
Normal adjustment = final image at infinity, so the eye is fully relaxed. Then the microscope gives $M=\dfrac{L}{f_o}\cdot\dfrac{D}{f_e}$ and the telescope gives $M=\dfrac{f_o}{f_e}$ .
Telescope tube length in normal adjustment is $L=f_o+f_e$ ; microscope tube length is the lens separation $L=v_o+|u_e|$ , where $u_e$ is the eyepiece object distance fixed by where the intermediate image sits relative to $f_e$ .
If the final image forms at the near point $D$ instead of infinity, the eyepiece term changes: the microscope eyepiece factor becomes $\left(1+\dfrac{D}{f_e}\right)$ and the telescope becomes $\dfrac{f_o}{f_e}\left(1+\dfrac{f_e}{D}\right)$ , both giving slightly larger magnification.
For the microscope objective, always use the lens formula $\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$ with the Cartesian sign convention ( $u$ negative for a real object) to obtain $v_o$ and $|m_o|=\dfrac{v_o}{u_o}$ .
A telescope objective is made wide-aperture to gather more light and improve brightness and resolving power; this is conceptually distinct from magnifying power and is a common follow-up theory question.
The final image in both the simple astronomical telescope and the compound microscope is inverted relative to the object, because a single inversion occurs at the objective and the eyepiece does not re-invert it; so quote magnifying power as a magnitude unless a sign is explicitly asked for.

	Compound microscope	Astronomical telescope
Object	Tiny, placed just beyond the objective's focus	Effectively at infinity
Objective focal length	Very short	Long
Objective aperture	Small	Large, to gather light and resolve detail
Eyepiece focal length	Short, but longer than the objective's	Short, much shorter than the objective's
Magnifying power	$M = \dfrac{v_o}{u_o}\left(1 + \dfrac{D}{f_e}\right)$ (near point)	$M = \dfrac{f_o}{f_e}$ (normal adjustment)
Tube length	$L = v_o + u_e$	$L = f_o + f_e$

Both instruments are two convex lenses in a tube; everything that differs follows from one being pointed at something tiny and near, the other at something huge and far.

Worked example · 2 marks

An astronomical telescope in normal adjustment has an objective of focal length

144\,\text{cm}

and an eyepiece of focal length

6\,\text{cm}

. Find its magnifying power and the length of the tube.

Normal adjustment means the final image is at infinity, so the intermediate image sits at the common focus of the two lenses.
Magnifying power $M = \dfrac{f_o}{f_e} = \dfrac{144}{6}$ .
Tube length $L = f_o + f_e = 144 + 6$ .

Answer

M = 24

and

L = 150\,\text{cm}

Where the marks go

Mixing up which lens is the objective: in a telescope the LARGER focal-length lens is the objective; students wrongly carry over the microscope rule (smaller $f$ = objective) and invert the answer.
Using $\dfrac{D}{f_e}$ for the eyepiece even when the final image is at the near point — the at- $D$ case requires the factor $\left(1+\dfrac{D}{f_e}\right)$ (microscope) or $\left(1+\dfrac{f_e}{D}\right)$ (telescope), which give different (larger) values.
Sign-convention slips in the objective lens formula: forgetting that the real object distance $u_o$ is negative, which corrupts $v_o$ and the whole magnification.
Approximating the MICROSCOPE tube length as $f_o+f_e$ (that formula is only for the telescope in normal adjustment); the microscope tube length must come from $v_o+|u_e|$ , with all distances kept in the same unit (cm).

How the board asks it

Numericalcompound-microscope magnifying power; lens formula for the objective; near-point eyepiece factor
A compound microscope has an objective of focal length $2\ \text{cm}$ and an eyepiece of focal length $5\ \text{cm}$ . An object is placed $2.5\ \text{cm}$ in front of the objective. Calculate the magnifying power when the final image is formed at the least distance of distinct vision $D=25\ \text{cm}$ .
Numericaltelescope magnifying power and tube length in normal adjustment3 mkAsked 2026
An astronomical telescope in normal adjustment has an objective of focal length $144\ \text{cm}$ and an eyepiece of focal length $6\ \text{cm}$ . Calculate its magnifying power and the length of the telescope tube.
Diagram / graphtwo-stage image formation; intermediate image3 mkAsked 2026
Draw a labelled ray diagram showing the formation of the final image by an astronomical telescope in normal adjustment, marking the objective, the eyepiece and the intermediate image.
Derive / proveangular magnification as the ratio of the two focal lengths
Obtain an expression for the magnifying power of an astronomical telescope in normal adjustment, and hence write its tube length in terms of $f_o$ and $f_e$ .
Distinguishwhich lens is the objective; relative focal lengths and apertures1 mkAsked 2026
Distinguish between a compound microscope and an astronomical telescope with reference to the focal lengths and apertures of their objective and eyepiece lenses.
Give reasonswide-aperture objective for light gathering and resolving power1 mkAsked 2024 · 2025
Why is the objective of an astronomical telescope made of large aperture and long focal length, while its eyepiece has a short focal length? Give reasons.
Define / state'normal use' means the final image is at the least distance of distinct vision1 mkAsked 2023
What is meant by a microscope in normal use?
Give reasonsa magnifying glass with the image at $D$ 3 mkAsked 2025
A convex lens of small focal length is used as a magnifying glass with the image at the least distance of distinct vision. Where must the object be placed, and what are two characteristics of the image?

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Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.