A 2.5 cm candle is placed 12 cm away from a convex mirror of focal length 30 cm. Give the location of the image and the magnification.

Explanation

To solve the problem of finding the location of the image and the magnification produced by a convex mirror, we will follow these steps:

### Step 1: Identify the given values
- Height of the object (candle), \( h_o = 2.5 \, \text{cm} \)
- Object distance, \( u = -12 \, \text{cm} \) (negative as per sign convention)
- Focal length of the convex mirror, \( f = +30 \, \text{cm} \) (positive for convex mirrors)

### Step 2: Use the mirror formula
The mirror formula is given by:
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]
We can rearrange this to find \( v \):
\[ \frac{1}{v} = \frac{1}{f} - \frac{1}{u} \]

### Step 3: Substitute the values into the formula
Substituting the values we have:
\[ \frac{1}{v} = \frac{1}{30} - \frac{1}{-12} \]
This simplifies to:
\[ \frac{1}{v} = \frac{1}{30} + \frac{1}{12} \]

### Step 4: Find a common denominator and solve for \( v \)
The least common multiple of 30 and 12 is 60. We can rewrite the fractions:
\[ \frac{1}{30} = \frac{2}{60}, \quad \frac{1}{12} = \frac{5}{60} \]
Now substituting these values:
\[ \frac{1}{v} = \frac{2}{60} + \frac{5}{60} = \frac{7}{60} \]
Now, taking the reciprocal to find \( v \):
\[ v = \frac{60}{7} \approx 8.57 \, \text{cm} \]

### Step 5: Determine the nature of the image
Since \( v \) is positive, the image formed is virtual and located on the same side as the object.

### Step 6: Calculate the magnification
The magnification \( m \) is given by:
\[ m = -\frac{v}{u} \]
Substituting the values:
\[ m = -\frac{60/7}{-12} = \frac{60/7}{12} = \frac{60}{84} = \frac{5}{7} \approx 0.71 \]

### Conclusion
- The location of the image is approximately \( 8.57 \, \text{cm} \) from the mirror (virtual image).
- The magnification is approximately \( 0.71 \).