Distance of an object from the first focus of an equi-convex lens is `10 cm` and the distance of its real image from second focus is `40 cm.` The focal length of the lens is

Options

A. `25 cm`
B. `10 cm`
C. `20 cm`
D. `40 cm`

Explanation

To find the focal length of the equi-convex lens, we can use the lens formula, which is given by:

\[ \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \]

Where:
- \( v \) is the image distance from the lens,
- \( u \) is the object distance from the lens,
- \( f \) is the focal length of the lens.

### Step 1: Identify the distances
From the problem:
- The distance of the object from the first focus is \( 10 \, \text{cm} \). Therefore, the object distance \( u \) from the lens is:
\[ u = - (f + 10) \quad \text{(since object distance is taken as negative)} \]

- The distance of the real image from the second focus is \( 40 \, \text{cm} \). Therefore, the image distance \( v \) from the lens is:
\[ v = f + 40 \quad \text{(since image distance is taken as positive)} \]

### Step 2: Substitute the values into the lens formula
Using the lens formula:
\[ \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \]
Substituting the values of \( v \) and \( u \):
\[ \frac{1}{f + 40} - \frac{1}{-(f + 10)} = \frac{1}{f} \]

### Step 3: Simplify the equation
This can be rewritten as:
\[ \frac{1}{f + 40} + \frac{1}{f + 10} = \frac{1}{f} \]

### Step 4: Find a common denominator
The common denominator for the left side is \((f + 40)(f + 10)\):
\[ \frac{(f + 10) + (f + 40)}{(f + 40)(f + 10)} = \frac{1}{f} \]
This simplifies to:
\[ \frac{2f + 50}{(f + 40)(f + 10)} = \frac{1}{f} \]

### Step 5: Cross-multiply to eliminate the fraction
Cross-multiplying gives:
\[ (2f + 50)f = (f + 40)(f + 10) \]

### Step 6: Expand both sides
Expanding both sides:
\[ 2f^2 + 50f = f^2 + 10f + 40f + 400 \]
This simplifies to:
\[ 2f^2 + 50f = f^2 + 50f + 400 \]

### Step 7: Rearrange the equation
Rearranging gives:
\[ 2f^2 + 50f - f^2 - 50f - 400 = 0 \]
This simplifies to:
\[ f^2 - 400 = 0 \]

### Step 8: Solve for \( f \)
Factoring gives:
\[ (f - 20)(f + 20) = 0 \]
Thus, \( f = 20 \, \text{cm} \) (we discard the negative value as focal length cannot be negative).

### Final Answer
The focal length of the lens is:
\[ f = 20 \, \text{cm} \]