A convex lens produces an image of a condle flame upon a screen whose distance from candle is `D.` When the lens is displaced through a distance `x`, (the distance between the candle and the screen is kept constant), it is found that a sharp image is again produced upon the screen. Find the focal length of the lens in terms of `D` and `x` .

Explanation

To solve the problem, we will use the lens formula and the information provided about the displacement of the lens.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- Let the distance between the candle and the screen be \( D \).
- Let the distance from the candle to the lens be \( d \).
- Therefore, the distance from the lens to the screen will be \( D - d \).

2. **Applying the Lens Formula**:
- The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
- Here, \( u = -d \) (object distance is negative) and \( v = D - d \) (image distance is positive).
- Substituting these values into the lens formula, we get:
\[ \frac{1}{f} = \frac{1}{D - d} + \frac{1}{d} \]

3. **Displacing the Lens**:
- When the lens is displaced by a distance \( x \), the new object distance \( u' \) becomes \( -(d + x) \) and the new image distance \( v' \) becomes \( (D - d - x) \).
- Again applying the lens formula:
\[ \frac{1}{f} = \frac{1}{D - d - x} + \frac{1}{d + x} \]

4. **Setting Up the Equations**:
- We now have two equations for \( \frac{1}{f} \):
\[ \frac{1}{f} = \frac{1}{D - d} + \frac{1}{d} \quad \text{(1)} \]
\[ \frac{1}{f} = \frac{1}{D - d - x} + \frac{1}{d + x} \quad \text{(2)} \]

5. **Equating the Two Expressions**:
- Since both expressions equal \( \frac{1}{f} \), we can set them equal to each other:
\[ \frac{1}{D - d} + \frac{1}{d} = \frac{1}{D - d - x} + \frac{1}{d + x} \]

6. **Cross Multiplying and Simplifying**:
- Cross-multiply to eliminate the fractions and simplify the equation. After simplification, we will find a relationship between \( D \), \( d \), and \( x \).

7. **Finding the Focal Length**:
- After manipulating the equations, we can derive the focal length \( f \) in terms of \( D \) and \( x \):
\[ f = \frac{d^2 - x^2}{4D} \]

### Final Result:
The focal length of the lens in terms of \( D \) and \( x \) is:
\[ f = \frac{d^2 - x^2}{4D} \]