Q.What should be the distance between the object in Exercise 9.23 and the magnifying glass if the virtual image of each square in the figure is to have an area of $6.25\ \text{mm}^2$. Would you be able to see the squares distinctly with your eyes very close to the magnifier?
[Note: Exercises 9.22 to 9.24 will help you clearly understand the difference between magnification in absolute size and the angular magnification (or magnifying power) of an instrument.]

Each small square on the card has side $1\ \text{mm}$ and the magnifying glass has focal length $f = 9\ \text{cm}$ (carried over from the previous exercise).

Required magnification. The virtual image of a square must have area $6.25\ \text{mm}^2$, so its side is $\sqrt{6.25} = 2.5\ \text{mm}$ and the linear magnification is

Object distance. For an erect virtual image $v = m\,u = 2.5u$. The lens formula $\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$ gives

so the object is $5.4\ \text{cm}$ from the lens and the image is at $v = 2.5\times(-5.4) = -13.5\ \text{cm}$. …

Data

From the earlier exercise, each square has side $1\ \text{mm}$ and the magnifying glass (converging lens) has focal length $f = 9\ \text{cm}$.

Step 1 — Magnification needed for the required image area

The virtual image of each square must have area $6.25\ \text{mm}^2$, so the image side is

The linear magnification is therefore

Step 2 — Object distance from the lens formula

For an erect virtual image the magnification is $m = v/u$, so $v = 2.5u$. Substituting into $\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$:

The object is $5.4\ \text{cm}$ in front of the lens — inside $f = 9\ \text{cm}$, which correctly gives a virtual image.

Step 3 — Image position

$v = 2.5u = 2.5 \times (-5.4\ \text{cm}) = -13.5\ \text{cm}$ …

Method: Thin Lens Formula with Magnification for a Virtual Image

This problem uses the thin lens equation combined with linear magnification to find the object distance when the image size is specified.

Steps

1. Identify given data

   • Lens focal length: $f = 10\ \text{cm}$ (from Exercise 9.23 context)
   • Original square side length: $a = 1\ \text{mm}$ (from figure in Exercise 9.23)
   • Required virtual image area: $A_i = 6.25\ \text{mm}^2$
   • Therefore, image side length: $a_i = \sqrt{6.25} = 2.5\ \text{mm}$

2. Calculate required linear magnification

   Linear magnification $m = \frac{\text{image size}}{\text{object size}} = \frac{2.5\ \text{mm}}{1\ \text{mm}} = 2.5$

3. Apply magnification formula for a lens

   For a thin lens: $m = \frac{v}{u}$

   Since the image is virtual and upright, $m$ is positive:

4. Use thin lens equation

(Note: sign convention — for virtual image, $v$ is negative if using real-is-positive; but here we use magnitudes with sign awareness)

Substituting $v = -2.5u$ (virtual image on same side as object):

$u = -14\ \text{cm}$ …

Common Mistakes: Microscope Magnification (Exercise 9.23)

Mistake 1: Confusing Linear Magnification with Angular Magnification

The error: Students often use the formula for angular magnification ($M = D/f$) directly to find the image distance or object distance, when the problem actually asks about absolute size magnification (linear magnification).

Why it's wrong: The question asks for the image to have an area of $6.25\ \text{mm}^2$. This is about linear magnification $m = \frac{h_i}{h_o}$, not angular magnification. The two are fundamentally different:

• Linear magnification $m = \frac{v}{u}$ (absolute size change)
• Angular magnification $M = \frac{D}{f}$ (apparent size change when eye is relaxed)

How to avoid: Read carefully — if the problem gives actual dimensions of the image (like area), use linear magnification. If it asks about "magnifying power" or "angular magnification," use the angle-based formula.

Mistake 2: Forgetting to Take Square Root for Area-to-Length Conversion

The error: Students treat the area $6.25\ \text{mm}^2$ as if it were a linear dimension, plugging it directly into magnification formulas.

Why it's wrong: Magnification is defined for linear dimensions (length, height), not area. If the image area is $6.25\ \text{mm}^2$, the linear magnification factor is:

How to avoid: Always convert area to linear dimension by taking the square root before using any magnification formula.

Mistake 3: Using the Wrong Sign Convention for Virtual Image

The error: Students treat the image distance $v$ as positive when using the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.

Why it's wrong: For a magnifying glass (convex lens used as a simple microscope), the image is virtual and on the same side as the object. According to the Cartesian sign convention:

• $v$ is negative for virtual images
• $u$ is negative (object on left side)

The correct lens formula becomes:

where both $u$ and $v$ are negative.

How to avoid: Draw a ray diagram first. If the image is on the same side as the object, $v$ is negative. Always write the sign convention at the top of your solution.

Mistake 4: Assuming the Image is at Infinity (Relaxed Eye)

The error: Students automatically set $v = \infty$ (image at infinity) because that's the "normal" adjustment for a magnifying glass. …