Sure, let's solve this problem step-by-step.
### Step 1: Determine the Focal Length
For a convex mirror, the focal length (f) is related to the radius of curvature (R) by the formula:
[tex]\[ f = \frac{R}{2} \][/tex]
Since the mirror is convex, the focal length is taken as negative:
[tex]\[ f = -\frac{R}{2} \][/tex]
Given that [tex]\( R = 60 \, \text{cm} \)[/tex], we get:
[tex]\[ f = -\frac{60}{2} = -30 \, \text{cm} \][/tex]
### Step 2: Use the Mirror Formula to Find Image Distance
The mirror formula is:
[tex]\[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \][/tex]
where [tex]\( d_o \)[/tex] is the object distance and [tex]\( d_i \)[/tex] is the image distance. We need to solve for [tex]\( d_i \)[/tex].
Rearranging the mirror formula to solve for [tex]\( \frac{1}{d_i} \)[/tex]:
[tex]\[ \frac{1}{d_i} = \frac{1}{f} - \frac{1}{d_o} \][/tex]
Substitute the known values [tex]\( f = -30 \, \text{cm} \)[/tex] and [tex]\( d_o = 10 \, \text{cm} \)[/tex]:
[tex]\[ \frac{1}{d_i} = \frac{1}{-30} - \frac{1}{10} \][/tex]
[tex]\[ \frac{1}{d_i} = -\frac{1}{30} - \frac{1}{10} \][/tex]
[tex]\[ \frac{1}{d_i} = -\frac{1}{30} - \frac{3}{30} \][/tex]
[tex]\[ \frac{1}{d_i} = -\frac{4}{30} \][/tex]
[tex]\[ \frac{1}{d_i} = -\frac{2}{15} \][/tex]
Taking the reciprocal to find [tex]\( d_i \)[/tex]:
[tex]\[ d_i = -\frac{15}{2} \][/tex]
[tex]\[ d_i = -7.5 \, \text{cm} \][/tex]
The negative sign indicates that the image is formed on the same side as the object, which is a characteristic of convex mirrors.
### Step 3: Calculate the Magnification
The magnification (m) for mirrors is given by:
[tex]\[ m = -\frac{d_i}{d_o} \][/tex]
Substituting the known values [tex]\( d_i = -7.5 \, \text{cm} \)[/tex] and [tex]\( d_o = 10 \, \text{cm} \)[/tex]:
[tex]\[ m = -\left( \frac{-7.5}{10} \right) \][/tex]
[tex]\[ m = \frac{7.5}{10} \][/tex]
[tex]\[ m = 0.75 \][/tex]
### Conclusion
- The image position ([tex]\( d_i \)[/tex]) is [tex]\( -7.5 \, \text{cm} \)[/tex]. This indicates that the image is virtual and located 7.5 cm behind the mirror.
- The magnification ([tex]\( m \)[/tex]) is [tex]\( 0.75 \)[/tex], which means the image is smaller than the object and 75% of its size.
Thus, the image is positioned at [tex]\( -7.5 \, \text{cm} \)[/tex] with a magnification of [tex]\( 0.75 \)[/tex].
