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Sagot :
Sure, let's tackle this step by step.
### 1. Position of the Image
To find the image position created by a convex mirror, we use the mirror formula:
[tex]\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \][/tex]
Here:
- \( f \) is the focal length of the mirror
- \( v \) is the image distance from the mirror
- \( u \) is the object distance from the mirror
For a convex mirror, the focal length \( f \) is taken as negative, so \( f = -15 \, \text{cm} \), and the object distance \( u \) is also considered with a negative sign since it is in front of the mirror, so \( u = -30 \, \text{cm} \).
Now, substituting these values into the mirror formula:
[tex]\[ \frac{1}{-15} = \frac{1}{v} + \frac{1}{-30} \][/tex]
Rearranging to solve for \( \frac{1}{v} \):
[tex]\[ \frac{1}{v} = \frac{1}{-15} - \frac{1}{-30} = -\frac{1}{15} + \frac{1}{30} = -\frac{2}{30} + \frac{1}{30} = -\frac{1}{30} \][/tex]
So, \(\frac{1}{v} = -\frac{1}{30}\), which gives:
[tex]\[ v = -30 \, \text{cm} \][/tex]
Therefore, the image distance \( v \) from the mirror is \(-30 \, \text{cm}\). The negative sign indicates that the image forms on the same side of the mirror as the object, which is a characteristic of a convex mirror.
### 2. Magnification of the Image
The magnification \( m \) produced by the mirror is given by the formula:
[tex]\[ m = -\frac{v}{u} \][/tex]
Substituting in the values for \( v \) and \( u \):
[tex]\[ m = -\frac{-30 \, \text{cm}}{-30 \, \text{cm}} = -1 \][/tex]
### 3. Size of the Image
The magnification \( m \) also relates the height of the image \( h_i \) to the height of the object \( h_o \) through:
[tex]\[ m = \frac{h_i}{h_o} \][/tex]
Given that the height of the object \( h_o \) is \( 5 \, \text{cm} \), and knowing the magnification \( m \):
[tex]\[ h_i = m \times h_o = -1 \times 5 \, \text{cm} = -5 \, \text{cm} \][/tex]
The negative sign indicates that the image is inverted relative to the object. However, in a convex mirror, the image is actually virtual and upright, so the interpretation here is that the image has the same orientation as the object but is smaller.
### Summary
- Position of the Image: The image is located \( 30 \, \text{cm} \) behind the mirror.
- Size of the Image: The height of the image is \( 5 \, \text{cm} \).
- Nature of the Image: The image is virtual, upright, and of the same size as the object (since the object distance and focal length were equal).
Hence, according to the calculations, the convex mirror forms an image at the same distance behind the mirror as the object is in front, with the same size but seemingly inverted based on magnification outcome analysis—keep in mind the upright nature characteristic of a virtual image produced by convex mirrors.
### 1. Position of the Image
To find the image position created by a convex mirror, we use the mirror formula:
[tex]\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \][/tex]
Here:
- \( f \) is the focal length of the mirror
- \( v \) is the image distance from the mirror
- \( u \) is the object distance from the mirror
For a convex mirror, the focal length \( f \) is taken as negative, so \( f = -15 \, \text{cm} \), and the object distance \( u \) is also considered with a negative sign since it is in front of the mirror, so \( u = -30 \, \text{cm} \).
Now, substituting these values into the mirror formula:
[tex]\[ \frac{1}{-15} = \frac{1}{v} + \frac{1}{-30} \][/tex]
Rearranging to solve for \( \frac{1}{v} \):
[tex]\[ \frac{1}{v} = \frac{1}{-15} - \frac{1}{-30} = -\frac{1}{15} + \frac{1}{30} = -\frac{2}{30} + \frac{1}{30} = -\frac{1}{30} \][/tex]
So, \(\frac{1}{v} = -\frac{1}{30}\), which gives:
[tex]\[ v = -30 \, \text{cm} \][/tex]
Therefore, the image distance \( v \) from the mirror is \(-30 \, \text{cm}\). The negative sign indicates that the image forms on the same side of the mirror as the object, which is a characteristic of a convex mirror.
### 2. Magnification of the Image
The magnification \( m \) produced by the mirror is given by the formula:
[tex]\[ m = -\frac{v}{u} \][/tex]
Substituting in the values for \( v \) and \( u \):
[tex]\[ m = -\frac{-30 \, \text{cm}}{-30 \, \text{cm}} = -1 \][/tex]
### 3. Size of the Image
The magnification \( m \) also relates the height of the image \( h_i \) to the height of the object \( h_o \) through:
[tex]\[ m = \frac{h_i}{h_o} \][/tex]
Given that the height of the object \( h_o \) is \( 5 \, \text{cm} \), and knowing the magnification \( m \):
[tex]\[ h_i = m \times h_o = -1 \times 5 \, \text{cm} = -5 \, \text{cm} \][/tex]
The negative sign indicates that the image is inverted relative to the object. However, in a convex mirror, the image is actually virtual and upright, so the interpretation here is that the image has the same orientation as the object but is smaller.
### Summary
- Position of the Image: The image is located \( 30 \, \text{cm} \) behind the mirror.
- Size of the Image: The height of the image is \( 5 \, \text{cm} \).
- Nature of the Image: The image is virtual, upright, and of the same size as the object (since the object distance and focal length were equal).
Hence, according to the calculations, the convex mirror forms an image at the same distance behind the mirror as the object is in front, with the same size but seemingly inverted based on magnification outcome analysis—keep in mind the upright nature characteristic of a virtual image produced by convex mirrors.
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