The focal length of a combination of two lenses is doubled if the separation between them is doubled. If the separation is increased to 4 times, the magnitude of focal length is

Options

A. doubled
B. quadrupled
C. halved
D. same

Explanation

To solve the problem step by step, we will analyze the given conditions and derive the equivalent focal length of the lens combination.

### Step 1: Understand the initial conditions
We have two lenses with focal lengths \( f_1 \) and \( f_2 \) separated by a distance \( d \). The formula for the equivalent focal length \( f \) of the combination of two lenses is given by:

\[ \frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2} - \frac{d}{f_1 f_2} \]

### Step 2: Analyze the first condition
According to the problem, when the separation \( d \) is doubled (i.e., \( 2d \)), the equivalent focal length is also doubled (i.e., \( 2f \)). Thus, we can write:

\[ \frac{1}{2f} = \frac{1}{f_1} + \frac{1}{f_2} - \frac{2d}{f_1 f_2} \]

### Step 3: Set up equations
Now we have two equations:
1. \( \frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2} - \frac{d}{f_1 f_2} \) (Equation 1)
2. \( \frac{1}{2f} = \frac{1}{f_1} + \frac{1}{f_2} - \frac{2d}{f_1 f_2} \) (Equation 2)

### Step 4: Subtract the equations
Subtract Equation 1 from Equation 2:

\[ \frac{1}{2f} - \frac{1}{f} = -\frac{2d}{f_1 f_2} + \frac{d}{f_1 f_2} \]

This simplifies to:

\[ -\frac{1}{2f} = -\frac{d}{f_1 f_2} \]

Thus, we can express \( d \) in terms of \( f \):

\[ \frac{d}{f_1 f_2} = \frac{1}{2f} \quad \Rightarrow \quad d = \frac{f_1 f_2}{2f} \]

### Step 5: Analyze the second condition
Now, we need to find the equivalent focal length when the separation is increased to \( 4d \):

\[ \frac{1}{f_{eq}} = \frac{1}{f_1} + \frac{1}{f_2} - \frac{4d}{f_1 f_2} \]

### Step 6: Substitute the value of \( d \)
Substituting \( d = \frac{f_1 f_2}{2f} \) into the equation:

\[ \frac{1}{f_{eq}} = \frac{1}{f_1} + \frac{1}{f_2} - \frac{4 \cdot \frac{f_1 f_2}{2f}}{f_1 f_2} \]

This simplifies to:

\[ \frac{1}{f_{eq}} = \frac{1}{f_1} + \frac{1}{f_2} - \frac{2}{f} \]

### Step 7: Substitute \( \frac{1}{f} \)
From Equation 1, we know:

\[ \frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2} - \frac{d}{f_1 f_2} \]

Substituting this into our equation gives:

\[ \frac{1}{f_{eq}} = \frac{1}{f} + \frac{d}{f_1 f_2} - \frac{2}{f} \]

### Step 8: Simplify
Using \( d = \frac{f_1 f_2}{2f} \):

\[ \frac{1}{f_{eq}} = \frac{1}{f} + \frac{1}{2f} - \frac{2}{f} \]

This simplifies to:

\[ \frac{1}{f_{eq}} = -\frac{1}{2f} \]

### Step 9: Find \( f_{eq} \)
Taking the reciprocal gives:

\[ f_{eq} = -2f \]

### Final Answer
The magnitude of the equivalent focal length when the separation is increased to four times is:

\[ |f_{eq}| = 2f \]