If the distances of an object and its virtual image from the focus of a convex lens of focal length f are 1 cm each, then f is

Options

A. 4 cm
B. (`sqrt+1`)cm
C. `2sqrt2` cm
D. (`2+sqrt2`) cm

Explanation

To solve the problem, we will use the lens formula and the given conditions about the object and virtual image distances. Let's go through the solution step by step.

### Step 1: Understand the problem
We have a convex lens with a focal length \( f \). The object distance \( u \) and the virtual image distance \( v \) from the focus of the lens are both given as 1 cm.

### Step 2: Set up the distances
- The distance of the object from the focus is \( 1 \, \text{cm} \), so:
\[ u = - (f + 1) \, \text{(since object distance is negative)} \]
- The distance of the virtual image from the focus is also \( 1 \, \text{cm} \), so:
\[ v = f - 1 \, \text{(since image distance is positive)} \]

### Step 3: Use the lens formula
The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
Substituting the values of \( u \) and \( v \):
\[ \frac{1}{f} = \frac{1}{(f - 1)} - \frac{1}{(- (f + 1))} \]

### Step 4: Simplify the equation
This can be rewritten as:
\[ \frac{1}{f} = \frac{1}{f - 1} + \frac{1}{f + 1} \]
Now, finding a common denominator:
\[ \frac{1}{f} = \frac{(f + 1) + (f - 1)}{(f - 1)(f + 1)} \]
This simplifies to:
\[ \frac{1}{f} = \frac{2f}{f^2 - 1} \]

### Step 5: Cross-multiply
Cross-multiplying gives:
\[ f^2 - 1 = 2f^2 \]
Rearranging this leads to:
\[ f^2 - 2f^2 - 1 = 0 \quad \Rightarrow \quad -f^2 - 1 = 0 \quad \Rightarrow \quad f^2 - 2f - 1 = 0 \]

### Step 6: Solve the quadratic equation
Using the quadratic formula \( f = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here, \( a = 1, b = -2, c = -1 \):
\[ f = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} \]
\[ f = \frac{2 \pm \sqrt{4 + 4}}{2} = \frac{2 \pm \sqrt{8}}{2} = \frac{2 \pm 2\sqrt{2}}{2} = 1 \pm \sqrt{2} \]

### Step 7: Determine the positive value
Since focal length \( f \) must be positive, we take:
\[ f = 1 + \sqrt{2} \]

### Conclusion
Thus, the focal length \( f \) of the convex lens is \( 1 + \sqrt{2} \) cm.

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