Q.A small telescope has an objective lens of focal length $140\ \text{cm}$ and an eyepiece of focal length $5.0\ \text{cm}$. What is the magnifying power of the telescope for viewing distant objects when

Concept: Telescope Magnification — ratio of the angle subtended by the image to the angle subtended by the object.

Reasoning:

1. For a telescope in normal adjustment, the final image is at infinity. The magnifying power is simply the ratio of the objective focal length to the eyepiece focal length:

2. When the final image is at the near point ($D = 25\ \text{cm}$), the eyepiece is adjusted so the image forms at $D$. The magnifying power becomes:

3. Substitute $f_o = 140\ \text{cm}$, $f_e = 5.0\ \text{cm}$, and $D = 25\ \text{cm}$.

Calculations:

(a) Normal adjustment:

$M = \frac{140}{5.0} = 28$ …

Why Magnification Works This Way

A telescope makes distant objects appear larger by increasing the angular size of the image compared to the object. The objective lens forms a real, inverted image of the distant object at its focal plane. The eyepiece then acts like a magnifying glass to view that intermediate image. The magnifying power $M$ is defined as:

For a distant object, the angle subtended at the unaided eye is essentially the same as the angle subtended at the objective. The trick is that the eyepiece lets you bring the intermediate image much closer to your eye, making it appear under a larger angle.

Let’s apply these directly.

Step-by-Step Solution

Given:

$f_o = 140\ \text{cm}$, $f_e = 5.0\ \text{cm}$, least distance of distinct vision $D = 25\ \text{cm}$.

1. Normal adjustment (final image at infinity)

In normal adjustment, the eyepiece is adjusted so that the intermediate image lies exactly at its focal point. The eyepiece then produces parallel rays (image at infinity), which the relaxed eye views without strain. The magnifying power is simply the ratio of focal lengths:

So the telescope magnifies 28 times.

2. Final image at the least distance of distinct vision ($D = 25\ \text{cm}$) …

Method: Magnification Formula for a Refracting Telescope

We use the standard magnification equations for a telescope in two common viewing configurations.

Key Concept

For a telescope, the magnifying power $M$ is the ratio of the angle subtended by the final image at the eye to the angle subtended by the object at the unaided eye. For distant objects, this simplifies to:

• Normal adjustment (final image at infinity):

where $f_o$ = focal length of objective, $f_e$ = focal length of eyepiece.

The negative sign indicates an inverted image (not always required in magnitude).

• Final image at near point (least distance of distinct vision $D = 25\ \text{cm}$):

Step-by-Step Solution

Given:

$f_o = 140\ \text{cm}$, $f_e = 5.0\ \text{cm}$, $D = 25\ \text{cm}$

(a) Normal adjustment (image at infinity)

Step 1: Apply the formula

Step 2: Substitute values

Step 3: Interpret

The magnifying power is 28 (magnitude). The negative sign means the image is inverted relative to the object.

Answer (a): $\boxed{28}$ (or $|M| = 28$)

(b) Final image at least distance of distinct vision ($25\ \text{cm}$)

Step 1: Apply the formula for near-point viewing

Step 2: Substitute values

Step 3: Simplify inside the bracket

Step 4: Multiply

$M = -28 \times 1.2 = -33.6$ …

Here are the common mistakes students make on this telescope magnification problem, and how to avoid each.

1. Using the wrong formula for each case

The Mistake:

Students often use the same formula for both parts. For normal adjustment (final image at infinity), the magnifying power is:

For the final image at the near point ($D = 25\ \text{cm}$), the correct formula is:

Some students incorrectly use $M = \frac{f_o}{f_e} + \frac{D}{f_e}$ or forget the $+1$ term.

How to Avoid:

• Memorise the two distinct cases:
  • Normal adjustment → $M = \frac{f_o}{f_e}$ (image at infinity, relaxed eye)
  • Near point adjustment → $M = \frac{f_o}{f_e} \left(1 + \frac{f_e}{D}\right)$ (image at $D$, maximum strain)
• Write the formula explicitly before substituting numbers.

2. Confusing objective and eyepiece focal lengths

The Mistake:

Swapping $f_o$ and $f_e$ in the formula. For a telescope, $f_o$ is large (objective) and $f_e$ is small (eyepiece). Using $f_o = 5\ \text{cm}$ and $f_e = 140\ \text{cm}$ gives a magnifying power less than 1 — physically impossible for a telescope.

How to Avoid:

• Always label: $f_o = 140\ \text{cm}$ (objective), $f_e = 5.0\ \text{cm}$ (eyepiece).
• Check: Magnification > 1 for a telescope. If your answer is < 1, you’ve swapped them.

3. Forgetting to convert units

The Mistake:

Using $f_o = 140\ \text{cm}$ and $f_e = 5.0\ \text{cm}$ directly in the formula — that’s fine here because both are in cm. But if $D$ is given in cm and $f_e$ in metres, or vice versa, students forget to convert.

How to Avoid:

• Ensure all lengths are in the same unit before substituting. Here, $D = 25\ \text{cm}$ matches $f_e = 5.0\ \text{cm}$, so no conversion needed.
• If units differ, convert everything to cm or metres consistently.

4. Misinterpreting “normal adjustment”

The Mistake:

Thinking “normal adjustment” means the final image is at the near point ($25\ \text{cm}$). Actually, normal adjustment means the final image is at infinity — the eye is relaxed.

How to Avoid:

• Memorise the definitions:
  • Normal adjustment → final image at infinity → $M = f_o / f_e$
  • Near point adjustment → final image at $D$ → $M = \frac{f_o}{f_e} \left(1 + \frac{f_e}{D}\right)$
• Read the question carefully: “when the telescope is in normal adjustment (i.e., when the final image is at infinity)” — the hint is already given.

5. Sign errors in the formula derivation

The Mistake:

Some textbooks derive magnification with a negative sign (indicating inverted image). Students sometimes include the negative sign in the final answer, or forget it when the question asks for “magnifying power” (which is usually taken as positive).

How to Avoid:

• Magnifying power is generally reported as a positive number (absolute value). The negative sign indicates inversion, but exam questions typically want the magnitude.
• If the question asks for “magnifying power,” give $|M|$.

6. Arithmetic errors in the second part

The Mistake: …