Question

An object placed $40\text{ cm}$ in front of a thin convex lens is moved to $60\text{ cm}$ from the lens. If the focal length of the lens is $30\text{ cm}$, the ratio of magnification of the image at the initial position to the final position is:

• A. $1:3$
• B. $3:1$
• C. $2:3$
• D. $3:2$

Hint:

Using the $m = \frac{f}{f+u}$ formula avoids the multi-step process of finding the image position $v$ first via $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$, saving valuable time during exams.

Answer 1

The Correct Option is B

Concept: The linear magnification ($m$) produced by a thin lens can be written directly in terms of its focal length ($f$) and the object distance ($u$) using the lens formula: \[ m = \frac{f}{f + u} \] By standard Cartesian sign convention for a real object tracking incoming rays, the object distance $u$ is treated as negative, while the focal length $f$ for a converging convex lens is positive.

Step 1: Calculate initial magnification ($m_1$).
Given $f = +30\text{ cm}$ and initial distance $u_1 = -40\text{ cm}$: \[ m_1 = \frac{30}{30 + (-40)} = \frac{30}{-10} = -3 \]

Step 2: Calculate final magnification ($m_2$).
Given $f = +30\text{ cm}$ and final distance $u_2 = -60\text{ cm}$: \[ m_2 = \frac{30}{30 + (-60)} = \frac{30}{-30} = -1 \]

Step 3: Determine the ratio of their magnitudes.
\[ \text{Ratio} = \frac{|m_1|}{|m_2|} = \frac{|-3|}{|-1|} = \frac{3}{1} \implies 3:1 \]