Answer :
Let's solve the problem step-by-step:
### Given:
- Object height ([tex]\( h \)[/tex]): [tex]\( 9.0 \)[/tex] cm
- Object distance ([tex]\( u \)[/tex]): [tex]\( 3.0 \)[/tex] cm
- Focal length ([tex]\( f \)[/tex]): [tex]\( -12.0 \)[/tex] cm
### To Find:
1. The distance of the image from the lens ([tex]\( v \)[/tex])
2. The height of the image ([tex]\( h' \)[/tex])
3. The type of lens
### Step 1: Determine the distance of the image from the lens ([tex]\( v \)[/tex])
We can use the lens formula:
[tex]\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \][/tex]
Given [tex]\( f = -12.0 \)[/tex] cm and [tex]\( u = 3.0 \)[/tex] cm, we rearrange the formula to solve for [tex]\( v \)[/tex]:
[tex]\[ \frac{1}{v} = \frac{1}{f} + \frac{1}{u} \][/tex]
[tex]\[ \frac{1}{v} = \frac{1}{-12.0} + \frac{1}{3.0} \][/tex]
[tex]\[ \frac{1}{v} = -\frac{1}{12.0} + \frac{4}{12.0} \][/tex]
[tex]\[ \frac{1}{v} = \frac{3}{12.0} \][/tex]
[tex]\[ \frac{1}{v} = \frac{1}{4.0} \][/tex]
[tex]\[ v = 4.0 \text{ cm} \][/tex]
So, the distance of the image from the lens is [tex]\( 4.0 \)[/tex] cm.
### Step 2: Determine the height of the image ([tex]\( h' \)[/tex])
To find the height of the image, we can use the magnification ( [tex]\( m \)[/tex] ) formula:
[tex]\[ m = \frac{h'}{h} = -\frac{v}{u} \][/tex]
Let's first find the magnification:
[tex]\[ m = -\frac{v}{u} = -\frac{4.0 \text{ cm}}{3.0 \text{ cm}} = -\frac{4}{3} \][/tex]
[tex]\[ m = -\frac{4}{3} \][/tex]
Now, using the magnification to find the image height [tex]\( h' \)[/tex]:
[tex]\[ h' = m \cdot h = -\frac{4}{3} \cdot 9.0 \text{ cm} = -12.0 \text{ cm} \][/tex]
So, the height of the image is [tex]\( -12.0 \)[/tex] cm. The negative sign indicates that the image is inverted.
### Step 3: Determine the type of lens
The focal length given is [tex]\( -12.0 \)[/tex] cm, which is negative. A negative focal length indicates that the lens is a diverging lens.
So, the type of lens is a diverging lens.
### Summary:
- The distance of the image from the lens is [tex]\( 4.0 \)[/tex] cm.
- The height of the image is [tex]\( -12.0 \)[/tex] cm.
- The type of lens is a diverging lens.
### Given:
- Object height ([tex]\( h \)[/tex]): [tex]\( 9.0 \)[/tex] cm
- Object distance ([tex]\( u \)[/tex]): [tex]\( 3.0 \)[/tex] cm
- Focal length ([tex]\( f \)[/tex]): [tex]\( -12.0 \)[/tex] cm
### To Find:
1. The distance of the image from the lens ([tex]\( v \)[/tex])
2. The height of the image ([tex]\( h' \)[/tex])
3. The type of lens
### Step 1: Determine the distance of the image from the lens ([tex]\( v \)[/tex])
We can use the lens formula:
[tex]\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \][/tex]
Given [tex]\( f = -12.0 \)[/tex] cm and [tex]\( u = 3.0 \)[/tex] cm, we rearrange the formula to solve for [tex]\( v \)[/tex]:
[tex]\[ \frac{1}{v} = \frac{1}{f} + \frac{1}{u} \][/tex]
[tex]\[ \frac{1}{v} = \frac{1}{-12.0} + \frac{1}{3.0} \][/tex]
[tex]\[ \frac{1}{v} = -\frac{1}{12.0} + \frac{4}{12.0} \][/tex]
[tex]\[ \frac{1}{v} = \frac{3}{12.0} \][/tex]
[tex]\[ \frac{1}{v} = \frac{1}{4.0} \][/tex]
[tex]\[ v = 4.0 \text{ cm} \][/tex]
So, the distance of the image from the lens is [tex]\( 4.0 \)[/tex] cm.
### Step 2: Determine the height of the image ([tex]\( h' \)[/tex])
To find the height of the image, we can use the magnification ( [tex]\( m \)[/tex] ) formula:
[tex]\[ m = \frac{h'}{h} = -\frac{v}{u} \][/tex]
Let's first find the magnification:
[tex]\[ m = -\frac{v}{u} = -\frac{4.0 \text{ cm}}{3.0 \text{ cm}} = -\frac{4}{3} \][/tex]
[tex]\[ m = -\frac{4}{3} \][/tex]
Now, using the magnification to find the image height [tex]\( h' \)[/tex]:
[tex]\[ h' = m \cdot h = -\frac{4}{3} \cdot 9.0 \text{ cm} = -12.0 \text{ cm} \][/tex]
So, the height of the image is [tex]\( -12.0 \)[/tex] cm. The negative sign indicates that the image is inverted.
### Step 3: Determine the type of lens
The focal length given is [tex]\( -12.0 \)[/tex] cm, which is negative. A negative focal length indicates that the lens is a diverging lens.
So, the type of lens is a diverging lens.
### Summary:
- The distance of the image from the lens is [tex]\( 4.0 \)[/tex] cm.
- The height of the image is [tex]\( -12.0 \)[/tex] cm.
- The type of lens is a diverging lens.
