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 reasoning outlined in the video transcript.

### Step 1: Understand the Lens System
We have two identical concave lenses with a refractive index \( \mu = \frac{3}{2} \) and a focal length \( f = -10 \, \text{cm} \) (negative because they are concave).

### Step 2: Find the Radius of Curvature
Using the lens maker's formula:
\[ \frac{1}{f} = (\mu - 1) \left( \frac{1}{r_1} - \frac{1}{r_2} \right) \]
For our two identical lenses, \( r_1 = -r \) and \( r_2 = r \). Thus,
\[ \frac{1}{-10} = \left( \frac{3}{2} - 1 \right) \left( \frac{1}{-r} - \frac{1}{r} \right) \]
This simplifies to:
\[ \frac{1}{-10} = \frac{1}{2} \left( \frac{-2}{r} \right) \]
\[ \frac{1}{-10} = \frac{-1}{r} \]
From this, we find:
\[ r = 10 \, \text{cm} \]

### Step 3: Find the Combined Focal Length of the Lens System
The combined focal length \( F \) of two lenses in contact is given by:
\[ \frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} \]
Since both lenses are identical:
\[ \frac{1}{F} = \frac{1}{-10} + \frac{1}{-10} = \frac{-2}{10} = \frac{-1}{5} \]
Thus, the combined focal length is:
\[ F = -5 \, \text{cm} \]

### Step 4: Use the Lens Formula to Find the Image Position
Using the lens formula:
\[ \frac{1}{v} - \frac{1}{u} = \frac{1}{F} \]
Where \( u = -15 \, \text{cm} \) (object distance is negative in lens convention):
\[ \frac{1}{v} - \frac{1}{-15} = \frac{1}{-5} \]
This simplifies to:
\[ \frac{1}{v} + \frac{1}{15} = \frac{-1}{5} \]
Finding a common denominator:
\[ \frac{1}{v} = \frac{-3}{15} - \frac{1}{15} = \frac{-4}{15} \]
Thus:
\[ v = -\frac{15}{4} \, \text{cm} \]

### Step 5: Determine the New Image Position with Liquid
When the liquid is added, the image shifts away by \( \frac{5}{4} \, \text{cm} \). The new image position \( v' \) is:
\[ v' = -\frac{15}{4} - \frac{5}{4} = -5 \, \text{cm} \]

### Step 6: Find the New Focal Length with the Liquid
Using the lens formula again for the new image position:
\[ \frac{1}{F'} = \frac{1}{v'} - \frac{1}{u} \]
Substituting \( v' = -5 \, \text{cm} \) and \( u = -15 \, \text{cm} \):
\[ \frac{1}{F'} = \frac{1}{-5} - \frac{1}{-15} \]
Finding a common denominator:
\[ \frac{1}{F'} = \frac{-3 + 1}{15} = \frac{-2}{15} \]
Thus:
\[ F' = -\frac{15}{2} \, \text{cm} \]

### Step 7: Find the Focal Length of the Liquid Lens
Using the lens maker's formula for the liquid lens:
\[ \frac{1}{F'} = \frac{1}{F_1} + \frac{1}{F_2} + \frac{1}{F_3} \]
Where \( F_1 = -10 \, \text{cm} \) and \( F_3 = -10 \, \text{cm} \):
\[ \frac{1}{F'} = \frac{1}{-10} + \frac{1}{F_2} + \frac{1}{-10} \]
Thus:
\[ \frac{1}{F'} = \frac{-2}{10} + \frac{1}{F_2} \]
Substituting \( F' = -\frac{15}{2} \):
\[ \frac{-2}{15} = \frac{-2}{10} + \frac{1}{F_2} \]
Solving for \( \frac{1}{F_2} \):
\[ \frac{1}{F_2} = \frac{-2}{15} + \frac{3}{15} = \frac{1}{15} \]
Thus:
\[ F_2 = 15 \, \text{cm} \]

### Step 8: Find the Refractive Index of the Liquid
Using the lens maker's formula again:
\[ \frac{1}{F_2} = (\mu - 1) \left( \frac{1}{r} - \frac{1}{-r} \right) \]
Substituting \( F_2 = 15 \, \text{cm} \) and \( r = 10 \, \text{cm} \):
\[ \frac{1}{15} = (\mu - 1) \left( \frac{1}{10} + \frac{1}{10} \right) \]
\[ \frac{1}{15} = (\mu - 1) \left( \frac{2}{10} \right) \]
\[ \frac{1}{15} = \frac{(\mu - 1)}{5} \]
Thus:
\[ \mu - 1 = \frac{1}{3} \implies \mu = \frac{4}{3} \]

### Step 9: Calculate \( 3\mu \)
Finally, we need to find \( 3\mu \):
\[ 3\mu = 3 \times \frac{4}{3} = 4 \]

### Final Answer
The final answer is:
\[ \boxed{4} \]