Q.A small telescope has an objective lens of focal length $144\ \text{cm}$ and an eyepiece of focal length $6.0\ \text{cm}$. What is the magnifying power of the telescope? What is the separation between the objective and the eyepiece?

For a telescope in normal adjustment, $M=-\frac{f_o}{f_e}$ and separation $L=f_o+f_e$.

With $f_o=144\ \text{cm}$, $f_e=6.0\ \text{cm}$:

$M = -\frac{144}{6.0} = -24, \qquad L = 144+6.0 = 150\ \text{cm}$ …

Setting up

The objective (large focal length) forms a real, inverted image of a distant object at its focal plane; the eyepiece then magnifies that image as a simple magnifier. In normal adjustment, the eyepiece's first focal point coincides with the objective's focal plane, so the final image forms at infinity and the eye is fully relaxed.

The negative sign in $M$ records that the final image is inverted relative to the object — the objective's real image is already inverted, and the eyepiece (a single simple lens) does not flip it back.

Step 1 — magnifying power

The telescope magnifies the object's angular size 24 times, and the image is inverted.

Step 2 — separation between objective and eyepiece

$L = f_o+f_e = 144+6.0 = 150\ \text{cm}$ …

Method: Normal Adjustment (Relaxed Eye) for a Telescope

In this method, the final image is formed at infinity — the eye is relaxed and least strained. This is the standard approach for telescope problems in Indian board exams.

Step 1: Recall the formula for magnifying power

For a telescope in normal adjustment:

Where:

• $f_o$ = focal length of objective = 144 cm
• $f_e$ = focal length of eyepiece = 6.0 cm

The negative sign indicates an inverted image (which is fine for astronomical telescopes).

Step 2: Calculate magnifying power

Magnifying power = 24 (magnitude). The image is inverted.

Step 3: Find the separation between lenses

In normal adjustment, the distance between objective and eyepiece equals the sum of their focal lengths:

$L = 144 + 6.0 = 150\ \text{cm}$ …

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

1. Using the Wrong Formula for Magnifying Power

Mistake:

Students often confuse the formula for a microscope with that for a telescope.

• Microscope: $M = \frac{L}{f_o} \cdot \frac{D}{f_e}$ (where $L$ is tube length, $D$ is near point)
• Telescope (normal adjustment): $M = \frac{f_o}{f_e}$

Here, the problem is about a telescope, so the correct formula is:

How to avoid:

• Read the problem carefully — it says telescope, not microscope.
• Memorise the key distinction:
  • Telescope: $M = \frac{f_o}{f_e}$ (for distant objects)
  • Microscope: $M = \frac{L}{f_o} \cdot \frac{D}{f_e}$

2. Forgetting the Sign Convention (or Using It Incorrectly)

Mistake:

Some students insert a negative sign unnecessarily. In most Indian board exams (CBSE, ICSE, state boards), magnifying power for a telescope is asked as a positive number (the absolute value). The formula $M = \frac{f_o}{f_e}$ gives the magnitude.

How to avoid:

• Unless the question explicitly asks for sign (e.g., “with sign”), give the positive value.
• For normal adjustment, the image is inverted, but magnifying power is reported as $|M|$.

3. Incorrect Separation Between Objective and Eyepiece

Mistake:

Students sometimes add the focal lengths incorrectly or use the tube length formula from microscopes.

Correct approach:

For a telescope in normal adjustment (final image at infinity), the separation between objective and eyepiece is:

How to avoid:

• Remember: In a telescope, the objective forms a real image at its focus, and the eyepiece is placed so that this image lies at its first focus. Hence, the distance between the two lenses is sum of focal lengths.
• Do not subtract or use $f_o - f_e$.

4. Unit Errors or Missing Units

Mistake:

• Using cm and m interchangeably without conversion.
• Forgetting to write units in the final answer.

How to avoid:

• Both focal lengths are given in cm — keep them in cm.
• Write the final answer as: …