To solve this problem, let's break it into steps and derive the necessary quantities.
### Step 1: Given Information
1. Height of the object ([tex]\( h_{object} \)[/tex]) = 5 cm
2. Focal length of the convex mirror ([tex]\( f \)[/tex]) = 10 cm
3. Image distance ([tex]\( v \)[/tex]) = -15 cm (negative because the image is virtual for a convex mirror).
### Step 2: Mirror Equation
The mirror equation is given by:
[tex]\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \][/tex]
We need to find the object distance ([tex]\( u \)[/tex]). Rearrange the mirror equation to solve for [tex]\( \frac{1}{u} \)[/tex]:
[tex]\[ \frac{1}{u} = \frac{1}{f} - \frac{1}{v} \][/tex]
### Step 3: Calculations
Substitute the known values into the rearranged mirror equation:
[tex]\[ \frac{1}{u} = \frac{1}{10} - \frac{1}{-15} \][/tex]
Convert each term to a common denominator:
[tex]\[ \frac{1}{u} = \frac{1}{10} + \frac{1}{15} = \frac{3}{30} + \frac{2}{30} = \frac{5}{30} \][/tex]
Therefore:
[tex]\[ \frac{1}{u} = \frac{1}{6} \][/tex]
Taking the reciprocal to find [tex]\( u \)[/tex]:
[tex]\[ u = 6 \text{ cm} \][/tex]
### Step 4: Magnification
The magnification ([tex]\( m \)[/tex]) for mirrors is given by the ratio of the image distance to the object distance:
[tex]\[ m = -\frac{v}{u} \][/tex]
Substitute the known values:
[tex]\[ m = -\frac{-15}{6} = 2.5 \][/tex]
### Step 5: Height of the Image
The height of the image ([tex]\( h_{image} \)[/tex]) can be calculated using the magnification:
[tex]\[ h_{image} = m \cdot h_{object} \][/tex]
Substitute the known values:
[tex]\[ h_{image} = 2.5 \cdot 5 \text{ cm} = 12.5 \text{ cm} \][/tex]
### Final Results
- The object distance ([tex]\( u \)[/tex]) is 6 cm.
- The magnification ([tex]\( m \)[/tex]) is 2.5.
- The height of the image ([tex]\( h_{image} \)[/tex]) is 12.5 cm.
Thus, the image distance ([tex]\( v \)[/tex]) is -15 cm (already given), and the height of the image is 12.5 cm.
