Q.A card sheet divided into squares each of size $1\ \text{mm}^2$ is being viewed at a distance of $9\ \text{cm}$ through a magnifying glass (a converging lens of focal length $9\ \text{cm}$) held close to the eye.

Concept: Microscope Magnification — The linear magnification $m$ of a lens is $v/u$, while angular magnification $M$ is the ratio of the angle subtended at the eye by the image to that by the object at the near point (25 cm).

Reasoning

1. Given: $f = 9\ \text{cm}$, object distance $u = -9\ \text{cm}$ (lens close to eye, object at focus). Using lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$:

The image is at infinity — the lens is being used in the normal adjustment for relaxed eye viewing.

2. (a) Linear magnification $m = v/u$ is infinite ($\infty$) because $v \to \infty$. Each square of area $1\ \text{mm}^2$ therefore appears to have infinite area in the virtual image — practically, the image is formed at infinity and cannot be assigned a finite linear size.

3. (b) Angular magnification (magnifying power) for normal adjustment: …

(a) Linear magnification and image area

The lens has $f = +9\ \text{cm}$ and the object is at $u = -9\ \text{cm}$. From the lens equation:

The image is formed at infinity (emergent rays are parallel). The linear magnification is

so the size — and hence the area — of each square in the virtual image is not finite. An image at infinity has no well-defined linear size; its area is effectively infinite. In practice one describes such an image by the angle it subtends, not by a length or area.

(b) Angular magnification (magnifying power)

With the image at infinity and the lens close to the eye, the angle subtended by the image equals the angle the object subtends at the lens, $\theta' \approx h/f$, while unaided the object would be held at the near point, subtending $\theta \approx h/D$. Hence

(c) Are they equal? …

Method: Lens Formula & Magnification Approach

This problem uses two distinct concepts — linear magnification (image size) and angular magnification (magnifying power). We solve step-by-step.

(a) Linear magnification and image area

Step 1: Identify given data

• Focal length, $f = 9\ \text{cm}$
• Object distance, $u = -9\ \text{cm}$ (negative by sign convention)
• Object is at the focus of the lens.

Step 2: Apply lens formula

Substitute $u = -9\ \text{cm}$, $f = 9\ \text{cm}$:

Step 3: Linear magnification

So the linear magnification is infinite — the image is formed at infinity.

Step 4: Area of each square in the virtual image

• Since the image is at infinity, the area is not defined in the usual sense.
• However, the angular size of each square is what matters (see part b).
• Answer: The concept of area in the image is meaningless here; the image is at infinity.

(b) Angular magnification (magnifying power)

Step 1: Formula for magnifying power when image is at infinity

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

Step 2: Substitute values

Step 3: Interpretation

• The angular magnification is 2.78 — the angle subtended by the image is 2.78 times that of the object placed at $D$.

(c) Comparison and explanation

🔍 Mistake 1: Confusing Linear Magnification with Angular Magnification

The error:

Students often think the magnification in part (a) and part (b) are the same quantity, so they give the same answer for both.

Why it’s wrong:

• Linear magnification ($m$) is the ratio of image height to object height.
• Angular magnification ($M$) is the ratio of the angle subtended by the image at the eye to the angle subtended by the object at the near point (usually $D = 25\ \text{cm}$).

How to avoid:

Always check what the question asks:

• If it says “magnification produced by the lens” → use $m = v/u$.
• If it says “angular magnification” or “magnifying power” → use $M = D/f$ (for relaxed eye) or $M = 1 + D/f$ (for image at near point).

🔍 Mistake 2: Using the Wrong Formula for Angular Magnification

The error:

Using $M = \frac{D}{f}$ when the image is at the near point, or using $M = 1 + \frac{D}{f}$ when the eye is relaxed.

Why it’s wrong:

• For a relaxed eye (image at infinity): $M = \frac{D}{f}$.
• For image at the near point (least distance of distinct vision, $D = 25\ \text{cm}$): $M = 1 + \frac{D}{f}$.

How to avoid:

Read the problem carefully. Here, the lens is held close to the eye and the object is at $9\ \text{cm}$ (focal length also $9\ \text{cm}$). That means the image is at infinity → relaxed eye → use $M = D/f$.

🔍 Mistake 3: Forgetting to Convert Units

The error:

Using $f = 9\ \text{cm}$ and $D = 25\ \text{cm}$ without converting to meters, or mixing cm and mm.

Why it’s wrong:

Formulas like $M = D/f$ are unitless ratios, so units must match. If $f$ is in cm, $D$ must also be in cm.

How to avoid:

Always write units beside every number. Convert everything to the same unit before plugging in.

🔍 Mistake 4: Calculating Area Magnification Incorrectly

The error:

Thinking area magnification = linear magnification $m$.

Why it’s wrong:

Area scales as the square of linear magnification.

If $m$ is the linear magnification, then:

How to avoid:

• First find $m$ using $m = v/u$.
• Then multiply the original area by $m^2$.

🔍 Mistake 5: Assuming $v$ is Always Positive

The error:

Using $v$ as positive without checking sign conventions.

Why it’s wrong:

For a virtual image (as in a simple microscope), $v$ is negative in the Cartesian sign convention (object on left, image on same side as object).

How to avoid:

Use the lens formula with proper signs:

$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ …