Sure, let's go through each part of the problem step-by-step:
### Given
- Object height ([tex]\( h_o \)[/tex]) = 1.0 cm
- Object distance ([tex]\( u \)[/tex]) = 12 cm
- Focal length ([tex]\( f \)[/tex]) = 16 cm
### (a) Find the position of the image.
To find the image distance ([tex]\( v \)[/tex]), we can use the lens formula:
[tex]\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \][/tex]
Rearranging this formula to solve for [tex]\( v \)[/tex]:
[tex]\[ \frac{1}{v} = \frac{1}{f} + \frac{1}{u} \][/tex]
Plugging in the given values:
[tex]\[ \frac{1}{v} = \frac{1}{16} + \frac{1}{-12} \][/tex]
Calculating this:
[tex]\[ \frac{1}{v} = \frac{1}{16} - \frac{1}{12} \][/tex]
[tex]\[ \frac{1}{v} = \frac{3 - 4}{48} \][/tex]
[tex]\[ \frac{1}{v} = -\frac{1}{48} \][/tex]
Therefore:
[tex]\[ v = -48 \text{ cm} \][/tex]
So, the position of the image is at -48 cm.
### (b) Is the image real or virtual?
If the image distance [tex]\( v \)[/tex] is negative, it indicates that the image is virtual.
Since [tex]\( v = -48 \)[/tex] cm, the image is virtual.
### (c) Find the size of the image.
To find the size of the image ([tex]\( h_i \)[/tex]), we use the magnification formula:
[tex]\[ m = \frac{h_i}{h_o} = -\frac{v}{u} \][/tex]
Rearranging this to solve for [tex]\( h_i \)[/tex]:
[tex]\[ h_i = m \cdot h_o \][/tex]
First, calculate magnification ([tex]\( m \)[/tex]):
[tex]\[ m = -\frac{v}{u} \][/tex]
[tex]\[ m = -\frac{-48}{-12} \][/tex]
[tex]\[ m = 4 \][/tex]
Now, calculate the image height ([tex]\( h_i \)[/tex]):
[tex]\[ h_i = m \cdot h_o \][/tex]
[tex]\[ h_i = 4 \cdot 1.0 \][/tex]
[tex]\[ h_i = -4.0 \text{ cm} \][/tex]
### (d) Is the image erect or inverted?
The orientation of the image is determined by the magnification sign. If the magnification is positive, the image is erect. If the magnification is negative, the image is inverted.
Given the magnification [tex]\( m = 4 \)[/tex], which is positive, the image is inverted.
### Summary
- (a) The position of the image is -48 cm.
- (b) The image is virtual.
- (c) The size of the image is -4.0 cm.
- (d) The image is inverted.
