Q.(a) A giant refracting telescope at an observatory has an objective lens of focal length $15\ \text{m}$. If an eyepiece of focal length $1.0\ \text{cm}$ is used, what is the angular magnification of the telescope?

Concept: Telescope Angular Magnification & Image Size via Objective

(a) For a refracting telescope in normal adjustment, the angular magnification $m$ is the ratio of the objective focal length $f_o$ to the eyepiece focal length $f_e$:

Here $f_o = 15\ \text{m} = 1500\ \text{cm}$ and $f_e = 1.0\ \text{cm}$.

(b) The objective lens forms a real image of the moon at its focal plane. The angle subtended by the moon at the objective is the same as the angle subtended by the image at the objective’s second focal point. Using the small-angle approximation:

$\theta \approx \frac{\text{diameter of moon}}{\text{distance to moon}} = \frac{3.48 \times 10^6}{3.8 \times 10^8}$ …

Concept and Intuition

A refracting telescope works by collecting light from a distant object with a large objective lens, forming a real, inverted image at its focal plane. The eyepiece then acts as a magnifying glass to view that image. The angular magnification tells us how many times larger the angle subtended at the eye becomes compared to the naked eye.

For a distant object, the angle subtended at the naked eye is essentially the same as the angle subtended at the objective. The eyepiece then magnifies the intermediate image. In normal adjustment (final image at infinity), the angular magnification is simply:

where $f_o$ is the focal length of the objective and $f_e$ is the focal length of the eyepiece.

For part (b), the objective forms an image of the moon at its focal plane. Since the moon is very far away, we can use the small-angle approximation: the angle subtended by the moon at the objective is $\theta \approx \frac{\text{diameter of moon}}{\text{distance to moon}}$. The image size $h_i$ is then $h_i = f_o \cdot \theta$.

Step-by-step Solution

1. Convert units to be consistent.

The objective focal length is $f_o = 15\ \text{m}$. The eyepiece focal length is $f_e = 1.0\ \text{cm} = 0.01\ \text{m}$.

2. Calculate angular magnification.

Using the formula for normal adjustment:

This means the telescope makes the moon appear 1500 times larger in angular size than with the naked eye.

3. Find the angle subtended by the moon at the objective.

The moon’s diameter $D_m = 3.48 \times 10^6\ \text{m}$.

The distance to the moon (radius of lunar orbit) $R = 3.8 \times 10^8\ \text{m}$.

Since the moon is very far, the angle $\theta$ (in radians) is:

Calculate: …

Method: Angular Magnification Formula for a Refracting Telescope (in Normal Adjustment)

This method uses the ratio of focal lengths to find magnification, and similar triangles to find image size.

Part (a) — Angular Magnification

Step 1: Identify the formula

For a telescope in normal adjustment (final image at infinity), the angular magnification $M$ is given by:

where

• $f_o$ = focal length of objective
• $f_e$ = focal length of eyepiece

Step 2: Substitute values

$f_o = 15\ \text{m} = 1500\ \text{cm}$, $f_e = 1.0\ \text{cm}$

Step 3: State the result

Angular magnification = $1500$ (no units — it’s a ratio)

Part (b) — Diameter of Moon’s Image Formed by Objective

Step 1: Understand the geometry

The objective lens forms a real, inverted image of the moon at its focal plane. The moon subtends a small angle $\theta$ at the objective.

Step 2: Use the small-angle approximation

For small angles (in radians):

where

• $D_m = 3.48 \times 10^{6}\ \text{m}$
• $R = 3.8 \times 10^{8}\ \text{m}$

Step 3: Relate image size to focal length …

Common Mistakes & How to Avoid Them: Telescope Magnification

Mistake 1: Forgetting to convert units before using the magnification formula

The Mistake:

Students plug $f_o = 15$ and $f_e = 1.0$ directly into $M = \frac{f_o}{f_e}$, getting $M = 15$, which is wrong.

Why it happens:

The formula $M = \frac{f_o}{f_e}$ requires both focal lengths in the same unit. The problem gives $f_o$ in metres and $f_e$ in centimetres.

How to avoid:

Always check units before substituting. Convert everything to the same unit — metres is safest here:

• $f_o = 15\ \text{m}$
• $f_e = 1.0\ \text{cm} = 0.01\ \text{m}$

Then:

Key takeaway: Write the conversion step explicitly in your solution.

Mistake 2: Using the wrong formula for angular magnification

The Mistake:

Using $M = \frac{f_o}{f_e} + 1$ (the formula for a microscope) instead of $M = \frac{f_o}{f_e}$ (for a telescope).

Why it happens:

Students confuse telescope and microscope magnification formulas.

How to avoid:

Remember the conceptual difference:

For a telescope, the objective forms a real image at its focal plane, and the eyepiece magnifies that image. There is no "+1" term.

Mistake 3: Confusing "diameter of image" with "magnification of image"

The Mistake:

Using $M = \frac{f_o}{f_e}$ to find the image diameter:

$\text{Image diameter} = M \times \text{object diameter}$ — which gives a huge, incorrect number.

Why it happens:

Angular magnification tells you how many times larger the angle appears, not the linear size of the image formed by the objective.

How to avoid:

For part (b), you need geometry, not the magnification formula. The objective lens forms a real image of the moon at its focal plane. Use the small-angle approximation:

So:

Substitute:

Calculate:

Mistake 4: Using the wrong distance for the moon

The Mistake:

Using the moon's radius ($1.74 \times 10^6\ \text{m}$) instead of its diameter ($3.48 \times 10^6\ \text{m}$), or using the diameter of lunar orbit ($2 \times 3.8 \times 10^8\ \text{m}$) instead of the radius (which is the Earth–moon distance).

Why it happens:

Misreading the problem statement.

How to avoid: …