An object of height `1cm` is set at angles to the optical axis of a double convex lens of optical power `5D` and `25cm` away from the lens. Determine the focal length of the lens, the position of the image. The linear magnification of the lens, and the height of the image formed by it.

Explanation

To solve the problem step by step, we will determine the focal length of the lens, the position of the image, the linear magnification, and the height of the image formed by the lens.

### Step 1: Determine the Focal Length of the Lens
The power \( P \) of a lens is related to its focal length \( f \) by the formula:
\[ P = \frac{1}{f} \]
Given that the power of the lens is \( 5D \), we can rearrange the formula to find the focal length:
\[ f = \frac{1}{P} = \frac{1}{5} = 0.2 \text{ m} = 20 \text{ cm} \]

### Step 2: Determine the Position of the Image
We will use the lens formula to find the position of the image. The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
Where:
- \( f \) is the focal length (20 cm),
- \( u \) is the object distance (which is negative for real objects, so \( u = -25 \) cm),
- \( v \) is the image distance (which we need to find).

Substituting the known values into the lens formula:
\[ \frac{1}{20} = \frac{1}{v} - \frac{1}{-25} \]
This simplifies to:
\[ \frac{1}{20} = \frac{1}{v} + \frac{1}{25} \]
To combine the fractions, we find a common denominator (100):
\[ \frac{1}{20} = \frac{5}{100} \quad \text{and} \quad \frac{1}{25} = \frac{4}{100} \]
Thus, we have:
\[ \frac{5}{100} = \frac{1}{v} + \frac{4}{100} \]
Subtracting \( \frac{4}{100} \) from both sides gives:
\[ \frac{1}{100} = \frac{1}{v} \]
Taking the reciprocal of both sides, we find:
\[ v = 100 \text{ cm} = 1 \text{ m} \]

### Step 3: Determine the Linear Magnification
The linear magnification \( m \) of a lens is given by the formula:
\[ m = \frac{h_i}{h} = \frac{v}{u} \]
Where:
- \( h_i \) is the height of the image,
- \( h \) is the height of the object (1 cm),
- \( v \) is the image distance (100 cm),
- \( u \) is the object distance (-25 cm).

Substituting the known values:
\[ m = \frac{100}{-25} = -4 \]

### Step 4: Determine the Height of the Image
Using the magnification formula, we can find the height of the image:
\[ m = \frac{h_i}{h} \]
Rearranging gives:
\[ h_i = m \cdot h \]
Substituting the values:
\[ h_i = -4 \cdot 1 \text{ cm} = -4 \text{ cm} \]

### Summary of Results
1. Focal length of the lens \( f = 20 \text{ cm} \)
2. Position of the image \( v = 100 \text{ cm} \) (1 m)
3. Linear magnification \( m = -4 \)
4. Height of the image \( h_i = -4 \text{ cm} \)