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(2)+1)cm`
C. `2sqrt(2)cm`
D. `(2+sqrt(2)) cm`

Answer

To solve the problem, we need to find the focal length \( f \) of a convex lens given that the distances of an object and its virtual image from the focus are both 1 cm.

### Step-by-step Solution:

1. **Understanding the Setup**:
- We have a convex lens with focal length \( f \).
- The object is located between the optical center and the focus of the lens.
- The distances of the object and the virtual image from the focus are both given as 1 cm.

2. **Defining Distances**:
- Let the distance of the object from the focus \( F \) be \( 1 \, \text{cm} \).
- Therefore, the object distance \( u \) from the lens is \( u = f - 1 \) (since the object is on the left side of the lens).
- The virtual image is also 1 cm from the focus, so the image distance \( v \) from the lens is \( v = -(f + 1) \) (the negative sign indicates that the image is virtual and on the same side as the object).

3. **Using the Lens Formula**:
- The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
- Substituting the values of \( u \) and \( v \):
\[ \frac{1}{f} = \frac{1}{-(f + 1)} - \frac{1}{(f - 1)} \]

4. **Finding a Common Denominator**:
- The common denominator for the right side is \( -(f + 1)(f - 1) \):
\[ \frac{1}{f} = \frac{-(f - 1) - (f + 1)}{-(f + 1)(f - 1)} \]
- Simplifying the numerator:
\[ -(f - 1) - (f + 1) = -f + 1 - f - 1 = -2f \]
- Thus, we have:
\[ \frac{1}{f} = \frac{-2f}{-(f + 1)(f - 1)} = \frac{2f}{(f + 1)(f - 1)} \]

5. **Cross-Multiplying**:
- Cross-multiplying gives:
\[ (f + 1)(f - 1) = 2f^2 \]
- Expanding the left side:
\[ f^2 - 1 = 2f^2 \]
- Rearranging gives:
\[ 0 = 2f^2 - f^2 + 1 \implies f^2 - 2f - 1 = 0 \]

6. **Solving the Quadratic Equation**:
- Using the quadratic formula \( f = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ f = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} \]
- This simplifies to:
\[ f = \frac{2 \pm \sqrt{4 + 4}}{2} = \frac{2 \pm \sqrt{8}}{2} = \frac{2 \pm 2\sqrt{2}}{2} = 1 \pm \sqrt{2} \]

7. **Choosing the Positive Value**:
- Since focal length cannot be negative, we take:
\[ f = 1 + \sqrt{2} \, \text{cm} \]

### Final Answer:
The focal length \( f \) of the convex lens is \( 1 + \sqrt{2} \, \text{cm} \).

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