Two concave lenses `L_1 and L_2` are kept in contact with each other. If the space between the two lenses is filled with a material of smaller refractive index, the magnitude of the focal length of the combination

Options

A. A. becomes undefined
B. B. remains unchanged
C. C. increases
D. D. decreases

Answer

To solve the problem of finding the magnitude of the focal length of a combination of two concave lenses \( L_1 \) and \( L_2 \) when the space between them is filled with a material of smaller refractive index, we can follow these steps:

### Step 1: Understand the Focal Length of Individual Lenses
For a concave lens, the focal length \( f \) can be calculated using the lens maker's formula:

\[ \frac{1}{f} = \frac{n - 1}{R_1} - \frac{n - 1}{R_2} \]

Where:
- \( n \) is the refractive index of the lens material,
- \( R_1 \) and \( R_2 \) are the radii of curvature of the lens surfaces.

For two concave lenses \( L_1 \) and \( L_2 \), both will have negative focal lengths.

### Step 2: Calculate Focal Lengths of Individual Lenses
Assuming both lenses have the same refractive index \( n \) and the same radius of curvature \( R \):

For lens \( L_1 \):
\[ \frac{1}{f_1} = -\frac{2(n - 1)}{R} \]

For lens \( L_2 \):
\[ \frac{1}{f_2} = -\frac{2(n - 1)}{R} \]

### Step 3: Combine the Focal Lengths
The total focal length \( f_c \) of the combination of two lenses in contact is given by the formula:

\[ \frac{1}{f_c} = \frac{1}{f_1} + \frac{1}{f_2} \]

Substituting the values:
\[ \frac{1}{f_c} = -\frac{2(n - 1)}{R} - \frac{2(n - 1)}{R} = -\frac{4(n - 1)}{R} \]

Thus, the focal length of the combination without any material is:
\[ f_c = -\frac{R}{4(n - 1)} \]

### Step 4: Introduce the Material of Smaller Refractive Index
Let the refractive index of the material filling the space between the lenses be \( n_1 \) (where \( n_1 < n \)). The focal length of the combination will change because of this material.

For the new focal length \( f_3 \) due to the material:
\[ \frac{1}{f_3} = \frac{n_1 - 1}{R} - \frac{n - 1}{R} \]
This simplifies to:
\[ \frac{1}{f_3} = \frac{n_1 - n}{R} \]

### Step 5: Calculate the New Total Focal Length
Now, the total focal length \( f_{total} \) of the combination with the material is:
\[ \frac{1}{f_{total}} = \frac{1}{f_c} + \frac{1}{f_3} \]

Substituting the values:
\[ \frac{1}{f_{total}} = -\frac{4(n - 1)}{R} + \frac{n_1 - n}{R} \]

### Step 6: Analyze the Result
Since \( n_1 < n \), the term \( \frac{n_1 - n}{R} \) is negative and will reduce the magnitude of the negative focal length. Thus, the overall focal length \( f_{total} \) will be less negative than \( f_c \), indicating that the magnitude of the focal length has decreased.

### Conclusion
The magnitude of the focal length of the combination of the two concave lenses when filled with a material of smaller refractive index decreases.