An object of height [tex]\( 9.0 \, \text{cm} \)[/tex] is placed [tex]\( 3.0 \, \text{cm} \)[/tex] from a lens with a focal length of [tex]\( -12.0 \, \text{cm} \)[/tex].

a) What is the distance of the image from the lens? [tex]\(\square \, \text{cm}\)[/tex]
b) What is the height of the image? [tex]\(\square \, \text{cm}\)[/tex]
c) What type of lens created this image? [tex]\(\square\)[/tex]



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.