Given the problem, we have the following details:

• Focal length of the convex mirror, $f = 60 \, \text{cm}$
• The magnification, $m = \frac{1}{2}$, because the image is half the size of the object.
• Position of the image, $v = 30 \, \text{cm}$

Now, we use the mirror formula and the magnification formula.

Mirror formula:

$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$ Where:

• $f$ is the focal length,
• $v$ is the image distance (which is given as 30 cm),
• $u$ is the object distance (which we need to find).

Magnification formula:

$m = \frac{-v}{u}$ Given that $m = \frac{1}{2}$, we can substitute it into the magnification formula:

$\frac{1}{2} = \frac{-v}{u}$

Substitute $v = 30 \, \text{cm}$ into the equation:

$\frac{1}{2} = \frac{-30}{u}$

Solve for $u$: $u = -60 \, \text{cm}$

The object is placed 60 cm in front of the mirror.

Checking using the mirror formula:

We know that: $\frac{1}{60} = \frac{1}{30} + \frac{1}{u}$ This simplifies to: $\frac{1}{60} = \frac{1}{30} + \frac{1}{u}$

Rearranging: $\frac{1}{u} = \frac{1}{60} - \frac{1}{30} = \frac{1}{60} - \frac{2}{60} = \frac{-1}{60}$

Thus, $u = -60 \, \text{cm}$.

So, the position of the object is 60 cm, and the image is at 30 cm as given.

Do you need further details or have any questions?

Here are 5 related questions for further exploration:

1. How does the image size change if the object moves closer to the convex mirror?
2. What happens to the image distance if the object is placed farther from the mirror?
3. How does a convex mirror differ from a concave mirror in image formation?
4. Can the focal length of a convex mirror ever be negative?
5. What is the relationship between focal length and radius of curvature for mirrors?

Tip: In convex mirrors, the image is always virtual, diminished, and upright.