Q.The photograph of a house occupies an area of $1.75\ \text{cm}^2$ on a 35 mm slide. The slide is projected on to a screen, and the area of the house on the screen is $1.55\ \text{m}^2$. What is the linear magnification of the projector-screen arrangement?

The key idea is that linear magnification $m$ is the square root of the areal magnification, because area scales as the square of the linear dimension.

1. Areal magnification $M_A$ is the ratio of the image area to the object area:

Convert $1.75\ \text{cm}^2$ to $\text{m}^2$: $1.75 \times 10^{-4}\ \text{m}^2$.

2. So, …

Why linear magnification from area?

When a slide is projected, every linear dimension of the image is magnified by the same factor $m$. That means if the original object has length $L$, the image has length $m L$. Area, being length × width, scales as $m^2$ — because both dimensions get multiplied by $m$.

So if you know the area of the object on the slide and the area of the image on the screen, you can find $m$ by taking the square root of the area ratio. This works because the shape is preserved (the house looks the same, just bigger).

Step-by-step solution

1. Write down what’s given

   Area on slide (object): $A_o = 1.75\ \text{cm}^2$

   Area on screen (image): $A_i = 1.55\ \text{m}^2$

2. Convert units so they match

   The slide area is in $\text{cm}^2$, the screen area in $\text{m}^2$. Linear magnification is a pure ratio, so we need both areas in the same unit.

   $1\ \text{m} = 100\ \text{cm}$, so $1\ \text{m}^2 = (100)^2\ \text{cm}^2 = 10^4\ \text{cm}^2$.

   Therefore:

3. Relate area magnification to linear magnification For a simple projection (no distortion), the area magnification $M_A$ is the square of the linear magnification $m$: $M_A = \frac{A_i}{A_o} = m^2$ …

Method: Area-to-Linear Magnification Conversion

This method uses the fact that area magnification equals the square of linear magnification for a projector or lens system.

Steps

1. Identify the given areas

   • Object area (on slide): $A_o = 1.75\ \text{cm}^2$
   • Image area (on screen): $A_i = 1.55\ \text{m}^2$

2. Convert to consistent units

   Since $1\ \text{m} = 100\ \text{cm}$,

   $1\ \text{m}^2 = 10^4\ \text{cm}^2$

   So,

3. Relate area magnification to linear magnification Area magnification $M_{\text{area}} = \dfrac{A_i}{A_o}$ Linear magnification $m$ satisfies:

4. Calculate linear magnification

$m^2 = \frac{15500}{1.75} = 8857.14\ (\text{approx})$ …

🧠 The Core Concept

Linear magnification ($m$) is the ratio of the linear dimensions (height or width) of the image to the object.

But here, you are given areas, not lengths.

Area magnification = $m^2$ (since area scales as the square of linear dimensions).

So the correct approach is:

1. Convert areas to the same units.
2. Find area magnification.
3. Take the square root to get linear magnification.

✗ Common Mistake #1: Treating area ratio as linear magnification

What students do:

They directly divide the areas:

Why it’s wrong:

Area scales as $(\text{linear factor})^2$. Dividing areas gives area magnification, not linear magnification.

✓ How to avoid:

Always ask: “Am I comparing lengths or areas?”

If areas are given, remember:

So:

✗ Common Mistake #2: Forgetting unit conversion

What students do:

They plug in $1.75\ \text{cm}^2$ and $1.55\ \text{m}^2$ without converting.

Why it’s wrong:

Magnification is a pure ratio — units must match.

$1\ \text{m}^2 = 10^4\ \text{cm}^2$, so $1.55\ \text{m}^2 = 1.55 \times 10^4\ \text{cm}^2$.

✓ How to avoid:

Convert both areas to the same unit (usually $\text{cm}^2$ or $\text{m}^2$) before forming the ratio.

✓ Correct Step-by-Step Solution

Step 1: Convert to same units

$A_{\text{object}} = 1.75\ \text{cm}^2$

$A_{\text{image}} = 1.55\ \text{m}^2 = 1.55 \times 10^4\ \text{cm}^2$ …