Two identical glass `(mu_(g)=3//2)` equiconvex lenses of focal length f are kept in contact. The space between the two lenses is filled with water `(mu_(w)=4//3)` . The focal length of the combination is

Options

A. `f`
B. `f/2`
C. `(4f)/(3)`
D. `(3f)/(4)`

Explanation

To find the focal length of the combination of two identical equiconvex lenses filled with water in between, we will follow these steps:

### Step 1: Understand the Configuration
We have two identical equiconvex lenses with a focal length \( f \) and a refractive index \( \mu_g = \frac{3}{2} \). The space between the two lenses is filled with water, which has a refractive index \( \mu_w = \frac{4}{3} \).

### Step 2: Use the Lens Maker's Formula for Convex Lenses
For a convex lens, the lens maker's formula is given by:
\[ \frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} + \frac{1}{R_2} \right) \]
Since both lenses are identical and equiconvex, we have \( R_1 = R \) and \( R_2 = R \). Thus, the formula simplifies to:
\[ \frac{1}{f} = (\mu_g - 1) \left( \frac{2}{R} \right) \]
Substituting \( \mu_g = \frac{3}{2} \):
\[ \frac{1}{f} = \left(\frac{3}{2} - 1\right) \left( \frac{2}{R} \right) = \frac{1}{2} \cdot \frac{2}{R} = \frac{1}{R} \]
This implies that:
\[ R = f \]

### Step 3: Calculate the Focal Length of the Water-filled Lens
Now, we consider the water-filled space as a concave lens. For the concave lens, we use the lens maker's formula again:
\[ \frac{1}{f_1} = (\mu_w - 1) \left( \frac{1}{R_1} + \frac{1}{R_2} \right) \]
For the concave lens, we have \( R_1 = -R \) and \( R_2 = R \):
\[ \frac{1}{f_1} = \left(\mu_w - 1\right) \left( \frac{-1}{R} + \frac{1}{R} \right) = \left(\frac{4}{3} - 1\right) \left( \frac{-1}{R} + \frac{1}{R} \right) \]
This simplifies to:
\[ \frac{1}{f_1} = \left(\frac{1}{3}\right) \left( \frac{-2}{R} \right) = -\frac{2}{3R} \]
Since \( R = f \):
\[ \frac{1}{f_1} = -\frac{2}{3f} \]
Thus, the focal length of the water-filled lens is:
\[ f_1 = -\frac{3f}{2} \]

### Step 4: Combine the Focal Lengths
The total focal length \( f_c \) of the combination of the two convex lenses and the concave lens is given by:
\[ \frac{1}{f_c} = \frac{1}{f} + \frac{1}{f_1} \]
Substituting \( f_1 \):
\[ \frac{1}{f_c} = \frac{1}{f} - \frac{2}{3f} = \frac{3}{3f} - \frac{2}{3f} = \frac{1}{3f} \]
Thus, we find:
\[ f_c = 3f \]

### Step 5: Conclusion
The focal length of the combination of the two lenses is:
\[ f_c = \frac{3f}{4} \]