Discover new information and insights with the help of IDNLearn.com. Discover prompt and accurate responses from our experts, ensuring you get the information you need quickly.
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]
We greatly appreciate every question and answer you provide. Keep engaging and finding the best solutions. This community is the perfect place to learn and grow together. IDNLearn.com provides the answers you need. Thank you for visiting, and see you next time for more valuable insights.