The focal length of the lens used in question 118 is

Options

A. `(D^(2)+d^(2))/(2D)`
B. `(D^(2)-d^(2))/(4D)`
C. `(D^(2)-d^(2))/(2D)`
D. `(D^(2)+d^(2))/(d)`

Explanation

To find the focal length of the lens used in the problem, we can follow these steps:

### Step 1: Understand the Problem
We have a convex lens that is placed in two different positions while forming the image on the same screen. The distance between the two positions of the lens is denoted as \( d \), and the total distance between the object and the screen is denoted as \( D \).

### Step 2: Set Up the Lens Displacement Equations
From the lens displacement formula, we can establish two equations based on the positions of the lens:
1. \( V + u = D \) (for the first position)
2. \( V - u = d \) (for the second position)

Where:
- \( V \) is the image distance,
- \( u \) is the object distance.

### Step 3: Solve for \( V \) and \( u \)
We can add the two equations to eliminate \( u \):
\[ (V + u) + (V - u) = D + d \]
This simplifies to:
\[ 2V = D + d \implies V = \frac{D + d}{2} \]

Next, we subtract the second equation from the first to eliminate \( V \):
\[ (V + u) - (V - u) = D - d \]
This simplifies to:
\[ 2u = D - d \implies u = \frac{D - d}{2} \]

### Step 4: Use the Lens Formula
The lens formula relates the focal length \( f \) to the object distance \( u \) and the image distance \( V \):
\[ \frac{1}{f} = \frac{1}{V} - \frac{1}{u} \]

Substituting the values of \( V \) and \( u \):
\[ \frac{1}{f} = \frac{2}{D + d} - \frac{2}{D - d} \]

### Step 5: Find a Common Denominator
To combine the fractions, we need a common denominator:
\[ \frac{1}{f} = \frac{2(D - d) - 2(D + d)}{(D + d)(D - d)} \]
This simplifies to:
\[ \frac{1}{f} = \frac{2D - 2d - 2D - 2d}{(D + d)(D - d)} = \frac{-4d}{(D + d)(D - d)} \]

### Step 6: Solve for \( f \)
Taking the reciprocal gives us:
\[ f = \frac{(D + d)(D - d)}{-4d} = \frac{D^2 - d^2}{4d} \]

### Conclusion
The focal length \( f \) of the lens is:
\[ f = \frac{D^2 - d^2}{4d} \]