Q.A compound microscope consists of an objective lens of focal length $2.0\ \text{cm}$ and an eyepiece of focal length $6.25\ \text{cm}$ separated by a distance of $15\ \text{cm}$. How far from the objective should an object be placed in order to obtain the final image at

Concept: Microscope Magnification — The final image is formed by the eyepiece, so we first find the intermediate image distance for the eyepiece, then use that to find the object distance for the objective.

Step 1: Eyepiece for case (a) — final image at $D = 25\ \text{cm}$

For the eyepiece, $f_e = 6.25\ \text{cm}$, $v_e = -25\ \text{cm}$ (virtual image). Using lens formula:

So $u_e = -5\ \text{cm}$. The intermediate image is $5\ \text{cm}$ from the eyepiece on the objective side.

Step 2: Objective for case (a)

Tube length $L = 15\ \text{cm}$, so $v_o = L - |u_e| = 15 - 5 = 10\ \text{cm}$.

Objective $f_o = 2.0\ \text{cm}$. Using lens formula:

Thus $u_o = -2.5\ \text{cm}$. Object is $2.5\ \text{cm}$ from objective.

Step 3: Magnifying power for case (a)

Step 4: Case (b) — final image at infinity …

How a compound microscope works

The objective ($f_o = 2.0\ \text{cm}$) forms a real, enlarged, inverted intermediate image; the eyepiece ($f_e = 6.25\ \text{cm}$) then acts as a simple magnifier on that image. The lenses are fixed $L = 15\ \text{cm}$ apart. We work backward from the eyepiece, since the required position of the final image fixes where the intermediate image must sit.

Case (a): final image at the least distance of distinct vision, $v_e = -25\ \text{cm}$

Eyepiece. Using $\dfrac{1}{v_e} - \dfrac{1}{u_e} = \dfrac{1}{f_e}$ with $v_e = -25\ \text{cm}$, $f_e = 6.25\ \text{cm}$:

So the intermediate image is $5.0\ \text{cm}$ in front of the eyepiece, i.e. $15 - 5.0 = 10.0\ \text{cm}$ from the objective, giving $v_o = +10.0\ \text{cm}$.

Objective. Using $\dfrac{1}{v_o} - \dfrac{1}{u_o} = \dfrac{1}{f_o}$ with $v_o = 10.0\ \text{cm}$, $f_o = 2.0\ \text{cm}$:

The object is placed $2.5\ \text{cm}$ from the objective (just beyond its focus $f_o = 2.0\ \text{cm}$, as expected).

Magnifying power. …

Method: Two-Step Image Formation (Ray Tracing by Lens Equations)

This problem treats the compound microscope as two lenses in series — the objective forms a real, inverted, enlarged image, and the eyepiece then magnifies that image further. We apply the thin lens formula to each lens in turn, using the fixed separation between the lenses.

Step 1 – Understand the layout

• Objective: $f_o = 2.0\ \text{cm}$
• Eyepiece: $f_e = 6.25\ \text{cm}$
• Separation between lenses: $L = 15\ \text{cm}$
• Final image distance from eyepiece:
  • Case (a): $v_e = -25\ \text{cm}$ (least distance of distinct vision, virtual image)
  • Case (b): $v_e = \infty$ (image at infinity)

We use the Cartesian sign convention with the lens formula $\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$. The image formed by the objective acts as the object for the eyepiece; the two are linked by

Step 2 – Eyepiece, Case (a): final image at 25 cm

For the eyepiece, $v_e = -25\ \text{cm}$ and $f_e = +6.25\ \text{cm}$:

So the objective’s image lies $5\ \text{cm}$ in front of the eyepiece — a real object placed just inside the eyepiece’s focal length, which is exactly what produces a magnified virtual image at $25\ \text{cm}$.

The image distance for the objective is therefore

Step 3 – Objective, Case (a)

Lens formula for the objective ($f_o = 2.0\ \text{cm}$, $v_o = +10\ \text{cm}$):

The object must be placed 2.5 cm in front of the objective.

Step 4 – Magnifying power, Case (a)

With the final image at the least distance of distinct vision $D = 25\ \text{cm}$:

Step 5 – Case (b): final image at infinity …

Common Mistakes & How to Avoid Them

Mistake 1: Confusing the separation distance with image distances

The error: Students often take the given $15\ \text{cm}$ separation as $v_o$ or $u_e$ directly, without realising it is the distance between the two lenses ($L = v_o + u_e$).

How to avoid:

• Draw a clear ray diagram.
• Label the objective-to-eyepiece distance as $L = v_o + u_e$.
• Never substitute $15\ \text{cm}$ for $v_o$ alone — it is the sum of two distances.

Mistake 2: Forgetting sign conventions for the eyepiece

The error: Treating the eyepiece as a converging lens but forgetting that the final image is virtual (on the same side as the object for the eyepiece). This leads to wrong signs in the lens formula.

How to avoid:

• For the eyepiece, the final image distance $v_e$ is negative (virtual image).
  • Case (a): $v_e = -D = -25\ \text{cm}$
  • Case (b): $v_e = -\infty$
• Use the lens formula with correct signs:

(where $u_e$ is negative because the object for eyepiece is real and on the opposite side).

Mistake 3: Using the wrong formula for magnifying power

The error: Mixing up the two cases — using the near-point formula when the final image is at infinity, or vice versa.

How to avoid:

• Final image at near point ($D = 25\ \text{cm}$):

• Final image at infinity:

• Memorise: the $+1$ appears only when the eye is accommodating (near point).

Mistake 4: Forgetting the negative sign in magnification

The error: Reporting magnifying power as a positive number without indicating that the image is inverted relative to the object.

How to avoid:

• The objective magnification $M_o = -v_o/u_o$ is negative (real, inverted image).
• The eyepiece magnification $M_e$ is positive (virtual, erect relative to the intermediate image).
• The total magnification $M = M_o \times M_e$ is negative, meaning the final image is inverted.
• In exam answers, state: “Magnifying power = $|M|$ (magnitude)” or explicitly say “image is inverted.”

Mistake 5: Not checking if the intermediate image lies within the eyepiece focal length …