An extended object of size 2mm is placed on the principal axis of a converging lens of focal length 10cm. It is found that when the object is placed perpendicular to the principal axis the image formed is 4mm in size. The size of image when it is placed along the principal axis is `"____________"` mm.

Explanation

To solve the problem step by step, we will use the concepts of magnification and the properties of lenses.

### Step 1: Identify the given data
- Size of the object (O) = 2 mm
- Focal length of the lens (f) = 10 cm = 100 mm (since we need to keep the units consistent)
- Size of the image when the object is placed perpendicular to the principal axis (I_perpendicular) = 4 mm

### Step 2: Calculate the magnification when the object is placed perpendicular to the principal axis
The magnification (M) when the object is placed perpendicular to the principal axis is given by the formula:
\[ M = \frac{I}{O} \]
Where:
- I = size of the image
- O = size of the object

Substituting the known values:
\[ M = \frac{4 \text{ mm}}{2 \text{ mm}} = 2 \]

### Step 3: Determine the magnification when the object is placed along the principal axis
When the object is placed along the principal axis, the magnification is given by:
\[ M^2 = \frac{I}{O} \]
Since we already calculated M = 2, we can substitute this into the equation:
\[ M^2 = 2^2 = 4 \]

### Step 4: Calculate the size of the image when the object is placed along the principal axis
Using the magnification formula:
\[ M^2 = \frac{I}{O} \]
We can rearrange it to find the size of the image (I):
\[ I = M^2 \cdot O \]
Substituting the values we have:
\[ I = 4 \cdot 2 \text{ mm} = 8 \text{ mm} \]

### Conclusion
The size of the image when the object is placed along the principal axis is **8 mm**.

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