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 in contact with a space filled with water, we can follow these steps:

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

### Step 2: Use the Lensmaker's Formula
The lensmaker's formula is given by:
\[ \frac{1}{F} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
For a convex lens, \( R_1 \) is positive and \( R_2 \) is negative (for the second surface). Since both lenses are identical, we can denote the radius of curvature as \( R \).

### Step 3: Calculate the Focal Length of the Glass Lenses
For each glass lens:
\[ \frac{1}{f} = \left(\frac{3}{2} - 1\right) \left( \frac{1}{R} - \frac{1}{-R} \right) \]
This simplifies to:
\[ \frac{1}{f} = \left(\frac{1}{2}\right) \left( \frac{2}{R} \right) = \frac{1}{R} \]
Thus, we confirm that:
\[ f = R \]

### Step 4: Calculate the Focal Length of the Water Lens
Now, we treat the water-filled space as a lens. Using the lensmaker's formula for the water lens:
\[ \frac{1}{F_w} = \left(\frac{4}{3} - 1\right) \left( \frac{1}{R} - \frac{1}{-R} \right) \]
This simplifies to:
\[ \frac{1}{F_w} = \left(\frac{1}{3}\right) \left( \frac{2}{R} \right) = \frac{2}{3R} \]
Thus, the focal length of the water lens is:
\[ F_w = \frac{3R}{2} \]

### Step 5: Combine the Focal Lengths
Now we need to find the equivalent focal length \( F_{eq} \) of the combination of the two glass lenses and the water lens. The formula for the combination of lenses in contact is:
\[ \frac{1}{F_{eq}} = \frac{1}{F_1} + \frac{1}{F_2} + \frac{1}{F_w} \]
Since both glass lenses have the same focal length \( f \):
\[ \frac{1}{F_{eq}} = \frac{1}{f} + \frac{1}{f} + \frac{1}{F_w} \]
Substituting \( F_w = \frac{3R}{2} \) and \( f = R \):
\[ \frac{1}{F_{eq}} = \frac{1}{R} + \frac{1}{R} + \frac{2}{3R} \]
This simplifies to:
\[ \frac{1}{F_{eq}} = \frac{2}{R} + \frac{2}{3R} = \frac{6}{3R} + \frac{2}{3R} = \frac{8}{3R} \]
Thus:
\[ F_{eq} = \frac{3R}{8} \]

### Step 6: Substitute \( R \) with \( f \)
Since \( R = f \):
\[ F_{eq} = \frac{3f}{8} \]

### Final Answer
The focal length of the combination of the two lenses is:
\[ F_{eq} = \frac{3f}{8} \]

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