Q.An angular magnification (magnifying power) of $30\text{X}$ is desired using an objective of focal length $1.25\ \text{cm}$ and an eyepiece of focal length $5\ \text{cm}$. How will you set up the compound microscope?

Setup: form the final image at the near point $D = 25\ \text{cm}$ (the arrangement that gives maximum magnifying power).

Eyepiece magnification.

Required objective magnification.

Object placement at the objective ($|v_o| = 5|u_o|$, $f_o = 1.25\ \text{cm}$):

so $v_o = 5 \times 1.5 = 7.5\ \text{cm}$.

Eyepiece object distance ($v_e = -25\ \text{cm}$, $f_e = 5\ \text{cm}$): …

Given

$f_o = 1.25\ \text{cm}$, $f_e = 5\ \text{cm}$, desired magnifying power $M = 30$, near point $D = 25\ \text{cm}$.

Step 1 — Eyepiece magnification (image at near point)

Step 2 — Objective magnification

Since $M = m_o \times m_e$,

Step 3 — Object position at the objective

The objective forms a real, inverted image, so $|v_o| = 5|u_o|$. Using $\dfrac{1}{v_o} - \dfrac{1}{u_o} = \dfrac{1}{f_o}$ with $u_o<0$, $v_o>0$:

The object sits $1.5\ \text{cm}$ from the objective, just beyond its focus $f_o = 1.25\ \text{cm}$.

Step 4 — Eyepiece object distance …

Method: Two-Lens Ray Diagram Approach for Compound Microscope Setup

This method uses the magnification formula for a compound microscope to determine the required tube length and lens positions.

Step 1: Recall the magnification formula

For a compound microscope in normal adjustment (final image at infinity), the total angular magnification is:

Where:

• $M$ = total angular magnification (magnifying power)
• $m_o$ = linear magnification of objective
• $m_e$ = angular magnification of eyepiece
• $L$ = tube length (distance between second focal point of objective and first focal point of eyepiece)
• $f_o$ = focal length of objective = $1.25\ \text{cm}$
• $f_e$ = focal length of eyepiece = $5\ \text{cm}$
• $D$ = least distance of distinct vision = $25\ \text{cm}$ (standard value)

Step 2: Substitute known values

Given $M = 30\text{X}$:

Step 3: Solve for tube length $L$

The negative sign indicates the image formed by the objective is real and inverted (as expected). The magnitude gives:

Tube length $L = 7.5\ \text{cm}$

Step 4: Determine lens positions

The tube length $L$ is the distance between: …

Here are the common mistakes students make when solving this compound microscope problem, along with how to avoid each.

Mistake 1: Confusing Magnification for the Final Image at Infinity vs. at Near Point

The error:

Students often blindly use the formula for angular magnification when the final image is at infinity ($M = \frac{L}{f_o} \cdot \frac{D}{f_e}$) without checking the problem’s condition. Here, the desired magnification is 30X, but using the infinity formula gives a different tube length.

How to avoid:

Always check which case is implied. In most exam problems, unless stated otherwise, the final image is formed at the near point (25 cm). The correct formula for that case is:

where $D = 25\ \text{cm}$ (least distance of distinct vision).

For this problem, using $f_o = 1.25\ \text{cm}$, $f_e = 5\ \text{cm}$, and $M = 30$, you solve for $L$:

So the tube length is $6.25\ \text{cm}$.

Mistake 2: Forgetting to Add the Eyepiece Focal Length to Get Total Microscope Length

The error:

Students stop after finding $L$ (distance between the second focal point of the objective and the first focal point of the eyepiece) and report that as the total length of the microscope.

How to avoid:

The total length of the microscope is the distance between the objective and the eyepiece. This is:

For this problem:

Always draw a quick ray diagram to remind yourself: the objective’s second focal point and the eyepiece’s first focal point coincide — so the physical separation includes both focal lengths.

Mistake 3: Using the Wrong Sign Convention for Lens Formula

The error:

When verifying the image distances, students plug values into the lens formula without consistent sign convention, leading to negative distances that confuse them.

How to avoid:

Use the Cartesian sign convention (distances measured from the optical centre, positive in the direction of incident light). For the objective:

• $u_o$ is negative (object is real, on left)
• $f_o$ is positive (convex lens)
• $v_o$ is positive (real image on right)

For the eyepiece:

• $u_e$ is negative (object is real, on left)
• $f_e$ is positive
• $v_e$ is negative (final virtual image on left)

This consistency prevents sign errors.

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