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

Answer

To find the focal length of the combination of two identical equiconvex lenses in contact, with the space between them 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 the lenses is filled with water, which has a refractive index \( \mu_w = \frac{4}{3} \).

### Step 2: Focal Length of Each Lens
For a single lens, the formula for the focal length \( f \) is given by:
\[ \frac{1}{f} = \left( \mu - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
For an equiconvex lens, we have:
- \( R_1 = R \) (positive for the first surface)
- \( R_2 = -R \) (negative for the second surface)

Thus, the equation becomes:
\[ \frac{1}{f} = \left( \frac{3}{2} - 1 \right) \left( \frac{1}{R} - \left(-\frac{1}{R}\right) \right) \]
\[ \frac{1}{f} = \left( \frac{1}{2} \right) \left( \frac{2}{R} \right) = \frac{1}{R} \]
This implies:
\[ R = f \]

### Step 3: Focal Length of the Water Layer
Now, we consider the water layer between the two lenses. The effective focal length \( f' \) of the water layer can be calculated using the same lens formula:
\[ \frac{1}{f'} = \left( \mu_w - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
Here, since the water behaves like an equiconcave lens:
- \( R_1 = R \) (positive)
- \( R_2 = -R \) (negative)

Thus, the equation becomes:
\[ \frac{1}{f'} = \left( \frac{4}{3} - 1 \right) \left( \frac{1}{R} - \left(-\frac{1}{R}\right) \right) \]
\[ \frac{1}{f'} = \left( \frac{1}{3} \right) \left( \frac{2}{R} \right) = \frac{2}{3R} \]

### Step 4: Combine the Focal Lengths
Now, we need to find the total focal length \( F \) of the combination of the two lenses and the water layer. The formula for the combination of lenses in contact is:
\[ \frac{1}{F} = \frac{1}{f} + \frac{1}{f'} \]
Substituting the values we found:
\[ \frac{1}{F} = \frac{1}{f} + \frac{3}{2} \cdot \frac{2}{3f} = \frac{1}{f} + \frac{2}{3f} \]
Finding a common denominator:
\[ \frac{1}{F} = \frac{3}{3f} + \frac{2}{3f} = \frac{5}{3f} \]
Thus, the total focal length \( F \) is:
\[ F = \frac{3f}{5} \]

### Final Answer
The focal length of the combination is:
\[ F = \frac{3f}{5} \]