"Optical Instruments"

About

The human eye has limits — we cannot see objects smaller than about 0.1 mm or resolve distant celestial bodies. Optical instruments extend our vision by magnifying tiny objects and bringing distant stars closer. This chapter covers microscopes and telescopes, their principles, and magnifying power.

Key Concepts

23.1 Simple Microscope (Magnifying Glass)

A single convex lens of short focal length.

Image: Virtual, erect, magnified — on the same side as the object.

Magnifying power (image at near point D = 25 cm): $M = 1 + \frac{D}{f}$

For relaxed eye (image at infinity): $M = D / f$

23.2 Compound Microscope

Two convex lenses — objective (short $f_{o}$ ) and eyepiece (short $f_{e}$ ).

Objective: Forms a real, inverted, enlarged intermediate image
Eyepiece: Acts as a simple magnifier on the intermediate image
Final image: Virtual, inverted, highly magnified

Magnifying power: $M = M_{o} \times M_{e} = \frac{L}{f _{o}} (1 + \frac{D}{f _{e}})$

Where $L$ = tube length.

Both objective and eyepiece have short focal lengths — this gives:

Large magnification without an impractically long tube
Compact design
Smaller, cheaper lenses

Eyepiece choice: Between 100 cm and 4 cm lenses → choose 4 cm for higher magnification ( $M_{e} = 1 + 25/4 = 7.25$ vs $1.25$ ).

23.3 Magnification vs Magnifying Power

Magnification	Magnifying Power
Ratio of image height to object height ( $h_{i} / h_{o}$ )	Ratio of angular size with instrument to angular size without it
Linear measure	Angular measure
Tells how big the image IS	Tells how big the object LOOKS

23.4 Astronomical Telescope

Refracting telescope: Objective (long $f_{o}$ ) + eyepiece (short $f_{e}$ ).

Magnifying power (normal adjustment): $M = \frac{f _{o}}{f _{e}}$

Tube length = $f_{o} + f_{e}$

Reflecting telescope: Uses a concave mirror as objective — no chromatic aberration, larger aperture possible.

INTEXT QUESTIONS 23.1

Q1. What is the nature of images formed by (i) simple microscope, (ii) compound microscope?

Ans: (i) Simple microscope: Virtual, erect, and magnified. (ii) Compound microscope: Virtual, inverted (w.r.t. original object), and highly magnified.

Q2. Differentiate between magnifying power and magnification.

Ans: Magnification = image height/object height (linear measure). Magnifying power = angular size with instrument/angular size without (angular measure) — it indicates how much larger the object APPEARS.

Q3. The magnifying power of a simple microscope is 11. What is its focal length?

Ans: $M = 1 + D / f = 11$ , $D = 25$ cm → $25/ f = 10$ → $f = 2.5$ cm.

Q4. Two lenses of focal lengths 100 cm and 4 cm. Which would you choose as the eyepiece of a compound microscope and why?

Ans: Choose the 4 cm lens. Shorter focal length → higher angular magnification ( $M_{e} = 1 + 25/4 = 7.25$ vs $1.25$ for 100 cm). Also keeps the tube compact without excessive aberrations.

Q5. Why should both objective and eyepiece of a compound microscope have short focal lengths?

Ans: Short $f_{o}$ increases objective magnification ( $L / f_{o}$ ). Short $f_{e}$ increases eyepiece magnification ( $1 + D / f_{e}$ ). Together they yield high overall magnification without an impractically long tube. Shorter focal lengths also allow smaller, less expensive lenses.

INTEXT QUESTIONS 23.2

Q1. How would the magnification of a telescope be affected by increasing the focal length of the objective?

Ans: Magnifying power $M = f_{o} / f_{e}$ . Increasing $f_{o}$ increases $M$ proportionally — a longer objective gives higher magnification (but also a longer tube).

Terminal Exercise

Explain the principle, construction, and working of a simple microscope. Derive $M = 1 + D / f$ .
Describe a compound microscope with a labelled diagram. Derive its magnifying power.
Why must both lenses in a compound microscope have short focal lengths?
A compound microscope has objective of focal length 1 cm and eyepiece of focal length 5 cm, separated by 20 cm. Find the magnifying power. (D = 25 cm)
Describe an astronomical telescope with a ray diagram. Derive $M = f_{o} / f_{e}$ for normal adjustment.
Compare reflecting and refracting telescopes. State two advantages of reflecting telescopes.
A telescope has an objective of focal length 100 cm and eyepiece of 5 cm. Find: (a) magnifying power, (b) tube length in normal adjustment.
Why can't we increase the magnifying power of a telescope indefinitely by decreasing $f_{e}$ ?
Define resolving power of a telescope. On what factors does it depend?
A person uses spectacles of power +2 D. What is the focal length? Is the person near-sighted or far-sighted?

Quick Revision

Instrument	Magnifying Power
Simple microscope	$M = 1 + D / f$
Compound microscope	$M = \frac{L}{f _{o}} (1 + \frac{D}{f _{e}})$
Astronomical telescope	$M = f_{o} / f_{e}$
Least distance of distinct vision (D)	25 cm

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