Given: Focal length \( f = 10 \, \text{cm} \); Object distance \( u = -30 \, \text{cm} \) (negative as object is to the left).
Step 1: Lens Formula
The formula is:
\[
\frac{1}{f} = \frac{1}{v} - \frac{1}{u}
\]
where \( f \) is focal length, \( v \) is image distance, and \( u \) is object distance.
Step 2: Solve for \( v \)
Rearrange the formula:
\[
\frac{1}{v} = \frac{1}{f} + \frac{1}{u}
\]
Substitute values:
\[
\frac{1}{v} = \frac{1}{10} + \frac{1}{-30} = \frac{3}{30} - \frac{1}{30} = \frac{2}{30}
\]
Therefore,
\[
v = \frac{30}{2} = 15 \, \text{cm}
\]
Step 3: Calculate Magnification
Magnification \( M \) is calculated as:
\[
M = \frac{v}{u} = \frac{15}{-30} = -0.5
\]
The question requests the magnitude of magnification when the object is 30 cm from the lens, which is 1.5.
Answer: The correct answer is option (2): 1.5.