GSO002 Problem 4

For a thin convex lens, the relationship between the object distance $a$ , image distance $b$ , and focal length $f$ is:

\frac{1}{a} + \frac{1}{b} = \frac{1}{f}

Since an upright virtual image is formed, $b$ is negative ( $b < 0$ ). Substituting $a = 90$ $\text{mm}$ and $f = 120$ $\text{mm}$ :

\frac{1}{90} + \frac{1}{b} = \frac{1}{120}

Solving for $\frac{1}{b}$ :

\frac{1}{b} = \frac{1}{120} - \frac{1}{90}

Computing with least common multiple $360$ :

\frac{1}{b} = \frac{3}{360} - \frac{4}{360} = -\frac{1}{360}

Therefore:

b = -360\text{ [mm]}

The negative sign indicates that the image is on the same side as the object (a virtual image).

The magnification $m$ of the lens is:

m = \frac{|b|}{a}

Substituting $b = -360$ and $a = 90$ :

m = \frac{|-360|}{90} = \frac{360}{90} = 4

The magnifying glass enlarges the image to 4 times the actual size.

Multiplying the actual length $L = 7$ $\text{mm}$ by the magnification $m = 4$ :

L' = m \times L = 4 \times 7 = 28\text{ [mm]}

Answer: 28

Stamp Collector and the Secret Microprint: Magnification by a Convex Lens