When the gap between two identical concave thin lenses `(mu=(3)/(2), f=10cm)` placed in contact is filled with certain liquid, the image of an object placed at 15 cm from lens combination shifts away from the lens by `5//4` cm. If the refractive index of liquid is `mu`, then write `3mu` as your answer.

Answer

To solve the problem step by step, we will follow the logical sequence of calculations based on the given information about the lenses and the liquid.

### Step 1: Understand the Lenses
We have two identical concave lenses with:
- Refractive index, \( \mu = \frac{3}{2} \)
- Focal length, \( f = -10 \, \text{cm} \) (since it is a concave lens).

### Step 2: Calculate the Equivalent Focal Length of the Lens Combination
For two lenses in contact, the equivalent focal length \( f_{eq} \) is given by:
\[ \frac{1}{f_{eq}} = \frac{1}{f_1} + \frac{1}{f_2} \]
Since both lenses are identical:
\[ \frac{1}{f_{eq}} = \frac{1}{-10} + \frac{1}{-10} = -\frac{2}{10} = -\frac{1}{5} \]
Thus, the equivalent focal length is:
\[ f_{eq} = -5 \, \text{cm} \]

### Step 3: Determine the Image Distance Without Liquid
Using the lens formula:
\[ \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \]
Where:
- \( u = -15 \, \text{cm} \) (object distance, negative as per sign convention)
- \( f = -5 \, \text{cm} \)

Substituting the values:
\[ \frac{1}{v} - \frac{1}{-15} = \frac{1}{-5} \]
This simplifies to:
\[ \frac{1}{v} + \frac{1}{15} = -\frac{1}{5} \]
\[ \frac{1}{v} = -\frac{1}{5} - \frac{1}{15} \]
Finding a common denominator (15):
\[ \frac{1}{v} = -\frac{3}{15} - \frac{1}{15} = -\frac{4}{15} \]
Thus:
\[ v = -\frac{15}{4} \, \text{cm} \]

### Step 4: Calculate the New Image Distance with Liquid
The image shifts away from the lens by \( \frac{5}{4} \, \text{cm} \), so the new image distance \( v' \) becomes:
\[ v' = -\frac{15}{4} + \frac{5}{4} = -\frac{10}{4} = -\frac{5}{2} \, \text{cm} \]

### Step 5: Find the New Equivalent Focal Length with Liquid
Using the new image distance, we apply the lens formula again:
\[ \frac{1}{v'} - \frac{1}{u} = \frac{1}{f_{liquid}} \]
Substituting the known values:
\[ \frac{1}{-\frac{5}{2}} - \frac{1}{-15} = \frac{1}{f_{liquid}} \]
This simplifies to:
\[ -\frac{2}{5} + \frac{1}{15} = \frac{1}{f_{liquid}} \]
Finding a common denominator (15):
\[ -\frac{6}{15} + \frac{1}{15} = -\frac{5}{15} = -\frac{1}{3} \]
Thus:
\[ f_{liquid} = -3 \, \text{cm} \]

### Step 6: Relate Focal Length to Refractive Index
Using the lens maker's formula for the liquid lens:
\[ \frac{1}{f} = (\mu - 1) \left(\frac{1}{R_1} - \frac{1}{R_2}\right) \]
For our case:
- \( f = -3 \, \text{cm} \)
- \( R_1 = 10 \, \text{cm} \) and \( R_2 = -10 \, \text{cm} \)

Substituting:
\[ \frac{1}{-3} = (\mu - 1) \left(\frac{1}{10} - \left(-\frac{1}{10}\right)\right) \]
This simplifies to:
\[ \frac{1}{-3} = (\mu - 1) \left(\frac{2}{10}\right) = \frac{(\mu - 1)}{5} \]
Cross-multiplying gives:
\[ -5 = 3(\mu - 1) \]
Solving for \( \mu \):
\[ -5 = 3\mu - 3 \implies 3\mu = -5 + 3 = -2 \implies \mu = \frac{-2}{3} \]

### Step 7: Final Answer
The question asks for \( 3\mu \):
\[ 3\mu = 3 \times \frac{-2}{3} = -2 \]

Thus, the final answer is:
\[ \boxed{-2} \]