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

Options

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

Answer

To find the focal length of the combination of two identical glass equi-convex lenses with water in between, we can follow these steps:

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

### Step 2: Focal Length of Each Lens
The formula for the focal length \( f \) of a lens is given by:
\[ \frac{1}{f} = (\mu - 1) \frac{2}{R} \]
where \( R \) is the radius of curvature of the lens.

For the glass lens:
\[ \frac{1}{f} = \left(\frac{3}{2} - 1\right) \frac{2}{R} \]
This simplifies to:
\[ \frac{1}{f} = \frac{1}{2} \cdot \frac{2}{R} = \frac{1}{R} \]
Thus, we can conclude that:
\[ R = f \]

### Step 3: Focal Length of Water as a Concave Lens
Now, we treat the water-filled space between the two lenses as a concave lens. The focal length \( f_1 \) of the water lens can be calculated using the same formula:
\[ \frac{1}{f_1} = \left(\mu_w - 1\right) \frac{-2}{R} \]
Substituting \( \mu_w = \frac{4}{3} \) and \( R = f \):
\[ \frac{1}{f_1} = \left(\frac{4}{3} - 1\right) \frac{-2}{f} \]
This simplifies to:
\[ \frac{1}{f_1} = \left(\frac{1}{3}\right) \frac{-2}{f} = -\frac{2}{3f} \]
Thus:
\[ f_1 = -\frac{3f}{2} \]

### Step 4: Combine the Focal Lengths
The total focal length \( F \) of the combination of the two lenses (two glass lenses and one water lens) is given by the formula:
\[ \frac{1}{F} = \frac{1}{f} + \frac{1}{f_1} \]
Substituting the values we have:
\[ \frac{1}{F} = \frac{1}{f} - \frac{2}{3f} \]
Finding a common denominator:
\[ \frac{1}{F} = \frac{3}{3f} - \frac{2}{3f} = \frac{1}{3f} \]
Thus:
\[ F = 3f \]

### Final Result
The focal length of the combination of the two lenses is:
\[ F = \frac{3f}{4} \]