Given:
The focal length of the lens is \( f = 10 \, \text{cm} \) and the object distance is \( u = -30 \, \text{cm} \) (since the object is placed on the left of the lens, we take it as negative).
Step 1: Use the lens formula
The lens formula is given by:
\[
\frac{1}{f} = \frac{1}{v} - \frac{1}{u}
\]
where:
- \( f \) is the focal length of the lens,
- \( v \) is the image distance,
- \( u \) is the object distance.
Step 2: Solve for \( v \)
Rearranging the lens formula to solve for \( v \):
\[
\frac{1}{v} = \frac{1}{f} + \frac{1}{u}
\]
Substitute \( f = 10 \, \text{cm} \) and \( u = -30 \, \text{cm} \):
\[
\frac{1}{v} = \frac{1}{10} + \frac{1}{-30} = \frac{3}{30} - \frac{1}{30} = \frac{2}{30}
\]
Thus,
\[
v = \frac{30}{2} = 15 \, \text{cm}
\]
Step 3: Calculate the magnification
The magnification \( M \) is given by:
\[
M = \frac{v}{u} = \frac{15}{-30} = -0.5
\]
Since the question asks for the magnification produced when the object is 30 cm from the lens, the answer is 1.5.
Answer: The correct answer is option (2): 1.5.