MHT-CET Physics · Optics (Ray)

The Microscope and the Telescope

A compound microscope magnifies by (v₀/u₀)(D/fₑ) with a short-focus objective near the object, a telescope in normal adjustment by f₀/fₑ with a tube f₀ + fₑ long, and a telescope's large objective aperture is for resolving fine detail.

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Why this matters

6 PYQs, one HARD: the objective's object distance from a microscope's length and magnification, a telescope's eyepiece from its length, why a telescope needs a large aperture, what a microscope objective looks like, and how a simple magnifier changes from blue light to red. One card.

Concept 1 of 1: Magnifying Power and Resolution

A simple magnifier gives D/f (relaxed eye); red light has a lower index, a longer focal length and so less magnification than blue. A compound microscope's objective has a short focal length and a small aperture; for a relaxed eye its magnification is (v₀/u₀)(D/fₑ) and its length is v₀ + fₑ. An astronomical telescope in normal adjustment magnifies f₀/fₑ and is f₀ + fₑ long. A large objective aperture gathers more light and, above all, resolves finer detail — resolving power grows with the aperture.

Definition

Simple magnifier: $M = \dfrac{D}{f}$ (D = 25 cm); red light ⇒ smaller M than blue.
Compound microscope, relaxed eye: $M = \dfrac{v_0}{u_0}\cdot\dfrac{D}{f_e}$ , length $L = v_0 + f_e$ (L = 15, fₑ = 6, M = 25 ⇒ $u_0 = 1.5$ cm).
Telescope, normal adjustment: $M = \dfrac{f_0}{f_e}$ , $L = f_0 + f_e$ (1.5 m, 1.56 m ⇒ $f_e = 0.06$ m).
Large telescope aperture: higher resolution (RP ∝ aperture/λ).

Magnifying power

M_{\text{micro}} = \frac{v_0}{u_0}\cdot\frac{D}{f_e}, \qquad M_{\text{tele}} = \frac{f_0}{f_e}

Worked example

A compound microscope is 20 cm long with an eyepiece of focal length 5 cm, relaxed eye. Its objective forms the image 15 cm away from an object 1 cm in front. Magnifying power?

Practice this conceptself-check · 2 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 20 April Shift I · Q49Hard

Example 1 · Optics (Ray) · Optical Instruments — Microscope and Telescope

The length of the compound microscope is 15 cm . The magnifying power for relaxed eye is 25. If the focal length of eye lens is 6 cm then the object distance for objective lens will be

Thinking a large aperture raises the magnification

Magnification is set by focal lengths. A large aperture collects more light and resolves finer detail.

Using the full tube length as the objective's image distance

For a relaxed eye the intermediate image sits at the eyepiece's focus, so v₀ = L − fₑ.

Drill 5 more on magnifying power and resolution

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (1)

Magnifying Power and Resolution
Magnifying power
$M_{\text{micro}} = \frac{v_0}{u_0}\cdot\frac{D}{f_e}, \qquad M_{\text{tele}} = \frac{f_0}{f_e}$

Watch out for (2)

Thinking a large aperture raises the magnification
→ Magnifying Power and Resolution
Using the full tube length as the objective's image distance
→ Magnifying Power and Resolution

Test yourself on Optics (Ray)

20 past MHT-CET questions from this chapter, timed at 18 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

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