Microscopes and Telescopes

What Microscope and Telescope Magnification Means

Both microscopes and telescopes are two-lens systems that combine an objective lens (the one near the object or the sky) and an eyepiece (the one near the eye). Their job is to make the angular size of the image larger than the angular size we’d see with the naked eye. That’s what “magnification” means in optics — angular magnification, not just “image is bigger than object.”

This topic carries 4-5 marks in CBSE boards every year and shows up reliably in NEET and JEE Main. The good news: there are only a handful of formulas, and they all flow from one idea. The bad news: students lose marks because they confuse “near point” vs “normal” adjustments, or mix up the magnification of the objective with the system’s total magnification.

We’ll walk through both instruments, derive the magnification formulas, and lock down the cases.

Key Terms & Definitions

Angular magnification ( $M$ ): ratio of the angle subtended at the eye by the final image to the angle subtended by the object placed at the near point ( $D = 25\text{ cm}$ ). This is what “magnification” means for these instruments.

Near point ( $D$ ): the closest distance at which a normal eye can focus comfortably, conventionally $25\text{ cm}$ .

Tube length ( $L$ ): the distance between the objective and the eyepiece in a compound microscope. In a telescope, the analogous quantity is $f_o + f_e$ (when in normal adjustment).

Normal adjustment: the final image is at infinity. Eye is fully relaxed. Used for telescopes mostly.

Near-point adjustment: the final image is at the near point ( $25\text{ cm}$ ). Magnification is slightly higher but the eye has to accommodate.

Compound Microscope

A compound microscope has two converging lenses. The object sits just outside the focal point of the objective, which forms a real, inverted, magnified image. This image acts as the object for the eyepiece, which works as a simple magnifier — producing a virtual, inverted, further-magnified image.

$M = \frac{L}{f_o}\left(1 + \frac{D}{f_e}\right)$

Here $L$ is the distance between the objective’s image and the eyepiece’s first focal point (often approximated as the tube length).

$M = \frac{L}{f_o} \cdot \frac{D}{f_e}$

For high magnification: small $f_o$ , small $f_e$ , large $L$ . That’s why microscope objectives have very short focal lengths (a few mm).

Astronomical Telescope (Refracting)

A telescope’s objective is a long-focal-length converging lens that forms a real image of a distant object near its focal plane. The eyepiece (short $f_e$ ) magnifies this image.

$M = \frac{f_o}{f_e}$

Tube length: $L = f_o + f_e$ .

$M = \frac{f_o}{f_e}\left(1 + \frac{f_e}{D}\right)$

For high magnification: large $f_o$ , small $f_e$ . That’s why astronomical telescopes are long.

Why the Formulas Differ

Microscope: object is close, so it sits near the objective’s focal point, forming a magnified real image. The microscope cares about linear size on the object side and angular size on the eye side.

Telescope: object is at infinity, so the objective forms an image right at its focal plane. There’s no linear magnification by the objective in the usual sense — it just shrinks the angular size by a factor $f_o$ . The eyepiece then expands the angular size by $1/f_e$ . Net: ratio $f_o/f_e$ .

For both instruments, the eyepiece works as a simple magnifying glass with magnification $D/f_e$ (image at infinity) or $1 + D/f_e$ (image at near point). Memorise this once and apply twice.

Solved Examples

Example 1 (CBSE)

A telescope has $f_o = 100\text{ cm}$ and $f_e = 5\text{ cm}$ . Find magnification in normal adjustment and the tube length.

$M = \frac{f_o}{f_e} = \frac{100}{5} = 20$

$L = f_o + f_e = 105\text{ cm}$

Example 2 (CBSE)

A compound microscope has $f_o = 1\text{ cm}$ , $f_e = 5\text{ cm}$ , and tube length $20\text{ cm}$ . Find the magnification when the final image is at infinity. Take $D = 25\text{ cm}$ .

$M = \frac{L}{f_o} \cdot \frac{D}{f_e} = \frac{20}{1} \cdot \frac{25}{5} = 20 \times 5 = 100$

Example 3 (JEE Main)

A telescope has $f_o = 144\text{ cm}$ and $f_e = 6\text{ cm}$ . Find magnification when the final image is at the near point ( $D = 25\text{ cm}$ ).

$M = \frac{f_o}{f_e}\left(1 + \frac{f_e}{D}\right) = \frac{144}{6}\left(1 + \frac{6}{25}\right) = 24 \times \frac{31}{25} = 29.76$

Example 4 (JEE Advanced)

The objective and eyepiece of a compound microscope have focal lengths $1.25\text{ cm}$ and $5\text{ cm}$ . The object is at $1.5\text{ cm}$ from the objective. Find the position of the final image and total magnification when the image is at the near point.

Step 1: Objective. Use lens formula $1/v - 1/u = 1/f$ : $1/v - 1/(-1.5) = 1/1.25$ , so $1/v = 1/1.25 - 1/1.5 = 0.8 - 0.667 = 0.133$ , $v = 7.5\text{ cm}$ .

Step 2: Magnification of objective $m_o = v/u = 7.5/(-1.5) = -5$ .

Step 3: Eyepiece in near-point mode. Image distance $= -25\text{ cm}$ , so $1/(-25) - 1/u_e = 1/5$ , giving $u_e = -25/6 \approx -4.17\text{ cm}$ . Magnification of eyepiece $m_e = 1 + D/f_e = 1 + 5 = 6$ .

Total magnification: $|M| = |m_o \cdot m_e| = 5 \times 6 = 30$ .

Exam-Specific Tips

CBSE: Both microscope and telescope derivations are board favourites. Practice ray diagrams — they often carry 1-2 marks separately.

JEE Main: Numerical with normal vs near-point adjustment. Memorise both versions.

NEET: Often asks “which type of telescope inverts the image” — answer: simple astronomical telescope (inverts), terrestrial telescope (uses extra lens to re-invert).

JEE Advanced: Combines microscope numerics with sign conventions. Be careful with negative magnifications.

Common Mistakes to Avoid

Mixing up tube length conventions. Some textbooks use $L$ as objective-to-eyepiece distance; others use the distance from the objective’s image to the eyepiece’s focal point. For exam answers, follow NCERT.
Forgetting the negative sign. Magnifications are usually negative (image is inverted) but exam questions usually want $|M|$ .
Confusing $f_o$ and $f_e$ in telescope formulas. Telescope: large/small. Microscope: small/small. Always identify which one is the objective.
Using telescope formulas for microscopes or vice versa. Different geometry, different formulas.
Forgetting that resolving power and magnification are different things. A bigger telescope (larger aperture) has higher resolving power but not necessarily higher magnification.

Practice Questions

Q1. A telescope with $f_o = 50\text{ cm}, f_e = 2\text{ cm}$ . Magnification in normal adjustment?

$M = 50/2 = 25$ .

Q2. A compound microscope has tube length $15\text{ cm}$ , $f_o = 0.5\text{ cm}$ , $f_e = 2\text{ cm}$ . Magnification at infinity adjustment?

$M = (15/0.5)(25/2) = 30 \times 12.5 = 375$ .

Q3. What is the ratio of $f_o$ to $f_e$ for a telescope of magnification 50 in normal adjustment?

$f_o/f_e = 50$ .

Q4. Why do telescopes use a long objective?

Long $f_o$ gives high magnification (since $M = f_o/f_e$ ) and also collects more light if combined with a large aperture.

Q5. A microscope is to be made with magnification 200. If $D = 25\text{ cm}$ , $f_e = 5\text{ cm}$ , $L = 20\text{ cm}$ , find $f_o$ for image at infinity.

$M = (L/f_o)(D/f_e) \implies 200 = (20/f_o)(5)$ , so $f_o = 100/200 = 0.5\text{ cm}$ .

Q6. Why is the eyepiece’s focal length always less than the objective’s in a microscope?

Both are short, but the eyepiece is slightly larger than the objective in a typical microscope. The eyepiece needs $f_e$ small enough to give large $D/f_e$ , but the objective is even smaller because it sits very close to the object.

Q7. What happens to magnification of a telescope if $f_o$ is doubled?

Magnification doubles ( $M \propto f_o$ ). Tube length also increases.

Q8. Distinguish near-point adjustment from normal adjustment.

Near point: final image at $25\text{ cm}$ , eye accommodates, slightly higher $M$ . Normal: final image at infinity, eye relaxed, used for prolonged viewing.

FAQs

Q: Why use two lenses instead of one? A single lens (simple magnifier) gives at most $M = 1 + D/f$ . Two lenses multiply magnifications, so we get much higher overall $M$ .

Q: Can a telescope work without an eyepiece? The objective alone gives a real image but no extra angular magnification beyond what the image’s size at $f_o$ provides. The eyepiece is needed for visual use.

Q: What’s the difference between a refracting and reflecting telescope? Refracting uses a lens for the objective; reflecting uses a curved mirror. Reflecting telescopes avoid chromatic aberration and can be made much larger. Most modern research telescopes are reflecting.

Q: Why do astronomical telescopes give inverted images? The geometry of a two-converging-lens setup naturally inverts. For terrestrial use, an extra erecting lens or a prism (binoculars) restores the correct orientation.

Q: How does aperture relate to magnification? Aperture (lens diameter) doesn’t change $M$ but does affect resolving power and brightness. A bigger aperture sees finer detail and gathers more light.

Q: Why is microscope objective so small in focal length? To get large $L/f_o$ , since $L$ is fixed by tube length. Small $f_o$ also lets us place the object very close, enabling near-field imaging.