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Sagot :
To solve this problem, we need to determine the distance of the object from a concave mirror given its radius of curvature and the magnification of the image.
1. Understand the given values:
- The radius of curvature [tex]\( R \)[/tex] of the concave mirror is 30.0 cm.
- The magnification [tex]\( m \)[/tex] of the image is 2.0 times the size of the object.
2. Determine the focal length [tex]\( f \)[/tex]:
- For a concave mirror, the focal length is half of the radius of curvature.
[tex]\[ f = \frac{R}{2} \][/tex]
Substituting the given radius of curvature:
[tex]\[ f = \frac{30.0 \, \text{cm}}{2} = 15.0 \, \text{cm} \][/tex]
3. Relate magnification to image distance [tex]\( d_i \)[/tex] and object distance [tex]\( d_o \)[/tex]:
- The magnification [tex]\( m \)[/tex] of a mirror is given by the ratio of the image distance to the object distance.
[tex]\[ m = \frac{d_i}{d_o} \][/tex]
Given that [tex]\( m = 2.0 \)[/tex]:
[tex]\[ \frac{d_i}{d_o} = 2.0 \implies d_i = 2 \cdot d_o \][/tex]
4. Use the mirror equation:
- The mirror equation relates the focal length to the object distance and the image distance.
[tex]\[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \][/tex]
Substituting [tex]\( d_i = 2 \cdot d_o \)[/tex]:
[tex]\[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{2 \cdot d_o} \][/tex]
5. Solve for [tex]\( d_o \)[/tex]:
- Combine the terms on the right side of the mirror equation:
[tex]\[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{2 \cdot d_o} = \frac{2}{2 \cdot d_o} + \frac{1}{2 \cdot d_o} = \frac{3}{2 \cdot d_o} \][/tex]
- Invert both sides to solve for [tex]\( d_o \)[/tex]:
[tex]\[ \frac{2 \cdot d_o}{3} = f \][/tex]
- Substituting the given focal length [tex]\( f = 15.0 \, \text{cm} \)[/tex]:
[tex]\[ \frac{2 \cdot d_o}{3} = 15.0 \, \text{cm} \][/tex]
- Solve for [tex]\( d_o \)[/tex]:
[tex]\[ 2 \cdot d_o = 3 \cdot 15.0 \, \text{cm} = 45.0 \, \text{cm} \][/tex]
[tex]\[ d_o = \frac{45.0 \, \text{cm}}{2} = 22.5 \, \text{cm} \][/tex]
Therefore, the distance of the object from the mirror is:
[tex]\[ \boxed{22.5 \, \text{cm}} \][/tex]
The correct answer is C.
1. Understand the given values:
- The radius of curvature [tex]\( R \)[/tex] of the concave mirror is 30.0 cm.
- The magnification [tex]\( m \)[/tex] of the image is 2.0 times the size of the object.
2. Determine the focal length [tex]\( f \)[/tex]:
- For a concave mirror, the focal length is half of the radius of curvature.
[tex]\[ f = \frac{R}{2} \][/tex]
Substituting the given radius of curvature:
[tex]\[ f = \frac{30.0 \, \text{cm}}{2} = 15.0 \, \text{cm} \][/tex]
3. Relate magnification to image distance [tex]\( d_i \)[/tex] and object distance [tex]\( d_o \)[/tex]:
- The magnification [tex]\( m \)[/tex] of a mirror is given by the ratio of the image distance to the object distance.
[tex]\[ m = \frac{d_i}{d_o} \][/tex]
Given that [tex]\( m = 2.0 \)[/tex]:
[tex]\[ \frac{d_i}{d_o} = 2.0 \implies d_i = 2 \cdot d_o \][/tex]
4. Use the mirror equation:
- The mirror equation relates the focal length to the object distance and the image distance.
[tex]\[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \][/tex]
Substituting [tex]\( d_i = 2 \cdot d_o \)[/tex]:
[tex]\[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{2 \cdot d_o} \][/tex]
5. Solve for [tex]\( d_o \)[/tex]:
- Combine the terms on the right side of the mirror equation:
[tex]\[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{2 \cdot d_o} = \frac{2}{2 \cdot d_o} + \frac{1}{2 \cdot d_o} = \frac{3}{2 \cdot d_o} \][/tex]
- Invert both sides to solve for [tex]\( d_o \)[/tex]:
[tex]\[ \frac{2 \cdot d_o}{3} = f \][/tex]
- Substituting the given focal length [tex]\( f = 15.0 \, \text{cm} \)[/tex]:
[tex]\[ \frac{2 \cdot d_o}{3} = 15.0 \, \text{cm} \][/tex]
- Solve for [tex]\( d_o \)[/tex]:
[tex]\[ 2 \cdot d_o = 3 \cdot 15.0 \, \text{cm} = 45.0 \, \text{cm} \][/tex]
[tex]\[ d_o = \frac{45.0 \, \text{cm}}{2} = 22.5 \, \text{cm} \][/tex]
Therefore, the distance of the object from the mirror is:
[tex]\[ \boxed{22.5 \, \text{cm}} \][/tex]
The correct answer is C.
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