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Sagot :
Sure, let's go through the calculations step-by-step to solve the given problem.
### 1. Finding the distance of the image from the lens
Given:
- Object height ([tex]\( h_o \)[/tex]) = 9.0 cm
- Object distance ([tex]\( d_o \)[/tex]) = 3.0 cm
- Focal length ([tex]\( f \)[/tex]) = -12.0 cm
We need to find the image distance ([tex]\( d_i \)[/tex]). We can use the lens formula:
[tex]\[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \][/tex]
Rearranging to solve for [tex]\( \frac{1}{d_i} \)[/tex]:
[tex]\[ \frac{1}{d_i} = \frac{1}{f} - \frac{1}{d_o} \][/tex]
Substituting the known values:
[tex]\[ \frac{1}{d_i} = \frac{1}{-12.0 \, \text{cm}} - \frac{1}{3.0 \, \text{cm}} \][/tex]
Calculating the right-hand side:
[tex]\[ \frac{1}{d_i} = \frac{3.0 - (-12.0)}{3.0 \times (-12.0)} = \frac{3.0 + 12.0}{-36.0} = \frac{15.0}{-36.0} \][/tex]
Simplifying the fraction:
[tex]\[ \frac{15.0}{-36.0} = -\frac{5}{12} \][/tex]
Therefore:
[tex]\[ d_i = -\frac{12}{5} \, \text{cm} = -2.4 \, \text{cm} \][/tex]
So, the distance of the image from the lens is:
[tex]\[ d_i = -2.4 \, \text{cm} \][/tex]
### 2. Finding the height of the image
Next, we need to determine the height of the image ([tex]\( h_i \)[/tex]). The magnification ([tex]\( m \)[/tex]) is given by:
[tex]\[ m = -\frac{d_i}{d_o} \][/tex]
Substituting the known values:
[tex]\[ m = -\frac{-2.4 \, \text{cm}}{3.0 \, \text{cm}} = \frac{2.4}{3.0} = 0.8 \][/tex]
The height of the image is then given by:
[tex]\[ h_i = m \times h_o \][/tex]
Substituting the values for magnification and object height:
[tex]\[ h_i = 0.8 \times 9.0 \, \text{cm} = 7.2 \, \text{cm} \][/tex]
So, the height of the image is:
[tex]\[ h_i = 7.2 \, \text{cm} \][/tex]
### 3. Determining the type of lens
Given that the focal length is negative, we can identify the type of lens. A lens with a negative focal length is a diverging lens.
So, the type of lens is:
[tex]\[ \text{Diverging lens} \][/tex]
### Final Answer
1. The distance of the image from the lens is:
[tex]\[ d_i = -2.4 \, \text{cm} \][/tex]
2. The height of the image is:
[tex]\[ h_i = 7.2 \, \text{cm} \][/tex]
3. The type of lens is:
[tex]\[ \text{Diverging lens} \][/tex]
### 1. Finding the distance of the image from the lens
Given:
- Object height ([tex]\( h_o \)[/tex]) = 9.0 cm
- Object distance ([tex]\( d_o \)[/tex]) = 3.0 cm
- Focal length ([tex]\( f \)[/tex]) = -12.0 cm
We need to find the image distance ([tex]\( d_i \)[/tex]). We can use the lens formula:
[tex]\[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \][/tex]
Rearranging to solve for [tex]\( \frac{1}{d_i} \)[/tex]:
[tex]\[ \frac{1}{d_i} = \frac{1}{f} - \frac{1}{d_o} \][/tex]
Substituting the known values:
[tex]\[ \frac{1}{d_i} = \frac{1}{-12.0 \, \text{cm}} - \frac{1}{3.0 \, \text{cm}} \][/tex]
Calculating the right-hand side:
[tex]\[ \frac{1}{d_i} = \frac{3.0 - (-12.0)}{3.0 \times (-12.0)} = \frac{3.0 + 12.0}{-36.0} = \frac{15.0}{-36.0} \][/tex]
Simplifying the fraction:
[tex]\[ \frac{15.0}{-36.0} = -\frac{5}{12} \][/tex]
Therefore:
[tex]\[ d_i = -\frac{12}{5} \, \text{cm} = -2.4 \, \text{cm} \][/tex]
So, the distance of the image from the lens is:
[tex]\[ d_i = -2.4 \, \text{cm} \][/tex]
### 2. Finding the height of the image
Next, we need to determine the height of the image ([tex]\( h_i \)[/tex]). The magnification ([tex]\( m \)[/tex]) is given by:
[tex]\[ m = -\frac{d_i}{d_o} \][/tex]
Substituting the known values:
[tex]\[ m = -\frac{-2.4 \, \text{cm}}{3.0 \, \text{cm}} = \frac{2.4}{3.0} = 0.8 \][/tex]
The height of the image is then given by:
[tex]\[ h_i = m \times h_o \][/tex]
Substituting the values for magnification and object height:
[tex]\[ h_i = 0.8 \times 9.0 \, \text{cm} = 7.2 \, \text{cm} \][/tex]
So, the height of the image is:
[tex]\[ h_i = 7.2 \, \text{cm} \][/tex]
### 3. Determining the type of lens
Given that the focal length is negative, we can identify the type of lens. A lens with a negative focal length is a diverging lens.
So, the type of lens is:
[tex]\[ \text{Diverging lens} \][/tex]
### Final Answer
1. The distance of the image from the lens is:
[tex]\[ d_i = -2.4 \, \text{cm} \][/tex]
2. The height of the image is:
[tex]\[ h_i = 7.2 \, \text{cm} \][/tex]
3. The type of lens is:
[tex]\[ \text{Diverging lens} \][/tex]
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