A converging lens of refractive index `1.5` and of focal length `15 cm` in air, has the same radii of curvature for both sides. If it is immersed in a liquid of refractive index `1.7`, find the focal length of the lens in the liquid.

Explanation

To find the focal length of a converging lens when it is immersed in a liquid, we can use the lens maker's formula. The formula for the focal length \( f \) of a lens in a medium is given by:

\[ \frac{1}{f} = \left( \mu_{relative} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]

Where:
- \( \mu_{relative} \) is the relative refractive index of the lens with respect to the medium.
- \( R_1 \) and \( R_2 \) are the radii of curvature of the lens surfaces.

### Step 1: Determine the radii of curvature
Since the lens has the same radii of curvature for both sides, we can denote them as \( R \). The focal length of the lens in air is given as \( f = 15 \, cm \).

### Step 2: Calculate the relative refractive index in air
The refractive index of the lens \( \mu_{lens} = 1.5 \) and the refractive index of air \( \mu_{air} = 1 \). Therefore, the relative refractive index in air is:

\[ \mu_{relative, air} = \frac{\mu_{lens}}{\mu_{air}} = \frac{1.5}{1} = 1.5 \]

### Step 3: Use the lens maker's formula in air
Using the lens maker's formula in air, we have:

\[ \frac{1}{f} = \left( 1.5 - 1 \right) \left( \frac{1}{R} - \frac{1}{R} \right) = \left( 0.5 \right) \left( \frac{1}{R} - \frac{1}{R} \right) \]

Since \( R_1 = R \) and \( R_2 = -R \) (for a converging lens), we can write:

\[ \frac{1}{f} = \left( 0.5 \right) \left( \frac{1}{R} + \frac{1}{R} \right) = \left( 0.5 \right) \left( \frac{2}{R} \right) = \frac{1}{R} \]

From this, we find:

\[ R = f = 15 \, cm \]

### Step 4: Calculate the relative refractive index in the liquid
Now, when the lens is immersed in a liquid with refractive index \( \mu_{liquid} = 1.7 \), the relative refractive index becomes:

\[ \mu_{relative, liquid} = \frac{\mu_{lens}}{\mu_{liquid}} = \frac{1.5}{1.7} \]

### Step 5: Substitute into the lens maker's formula for the liquid
Now we substitute into the lens maker's formula:

\[ \frac{1}{f} = \left( \frac{1.5}{1.7} - 1 \right) \left( \frac{1}{R} + \frac{1}{R} \right) \]

Calculating \( \frac{1.5}{1.7} - 1 \):

\[ \frac{1.5}{1.7} - 1 = \frac{1.5 - 1.7}{1.7} = \frac{-0.2}{1.7} \]

Now substituting \( R = 15 \, cm \):

\[ \frac{1}{f} = \left( \frac{-0.2}{1.7} \right) \left( \frac{2}{15} \right) \]

### Step 6: Simplify to find \( f \)
\[ \frac{1}{f} = \frac{-0.4}{1.7 \times 15} \]

Calculating \( 1.7 \times 15 = 25.5 \):

\[ \frac{1}{f} = \frac{-0.4}{25.5} \]

Taking the reciprocal to find \( f \):

\[ f = \frac{25.5}{-0.4} = -63.75 \, cm \]

### Final Answer
The focal length of the lens in the liquid is \( -63.75 \, cm \).

---