Two thin symmetrical lenses of different nature and of different material have equal radii of curvature `R = 15 cm`. The lenses are put close together and immersed in water `(mu_(w) = (4)/(3))`. The focal length of the system in water is 30 cm. The difference between refractive indices of the two lenses is

Options

A. `(1)/(2)`
B. `(1)/(4)`
C. `(1)/(3)`
D. `(3)/(4)`

Answer

To solve the problem, we will use the lens maker's formula and the information provided about the system of two lenses immersed in water.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Radius of curvature, \( R = 15 \, \text{cm} \)
- Focal length of the system in water, \( f = 30 \, \text{cm} \)
- Refractive index of water, \( \mu_w = \frac{4}{3} \)

2. **Use the Lens Maker's Formula:**
The lens maker's formula for a thin lens in a medium is given by:
\[ \frac{1}{f} = \left( \mu - \mu_w \right) \frac{1}{R} \]
For two lenses, we can write:
\[ \frac{1}{f} = \left( \mu_1 - \mu_w \right) \frac{1}{R} + \left( \mu_2 - \mu_w \right) \frac{1}{R} \]
Simplifying this, we have:
\[ \frac{1}{f} = \frac{1}{R} \left( (\mu_1 - \mu_w) + (\mu_2 - \mu_w) \right) \]
This can be rewritten as:
\[ \frac{1}{f} = \frac{1}{R} \left( \mu_1 + \mu_2 - 2\mu_w \right) \]

3. **Substitute Known Values:**
Substitute \( \mu_w = \frac{4}{3} \), \( R = 15 \, \text{cm} \), and \( f = 30 \, \text{cm} \):
\[ \frac{1}{30} = \frac{1}{15} \left( \mu_1 + \mu_2 - 2 \cdot \frac{4}{3} \right) \]

4. **Simplify the Equation:**
Multiply both sides by 15:
\[ \frac{15}{30} = \mu_1 + \mu_2 - \frac{8}{3} \]
This simplifies to:
\[ \frac{1}{2} = \mu_1 + \mu_2 - \frac{8}{3} \]

5. **Rearranging the Equation:**
Rearranging gives:
\[ \mu_1 + \mu_2 = \frac{1}{2} + \frac{8}{3} \]
To add these fractions, convert \( \frac{1}{2} \) to a fraction with a denominator of 6:
\[ \frac{1}{2} = \frac{3}{6} \quad \text{and} \quad \frac{8}{3} = \frac{16}{6} \]
Therefore:
\[ \mu_1 + \mu_2 = \frac{3}{6} + \frac{16}{6} = \frac{19}{6} \]

6. **Finding the Difference in Refractive Indices:**
We know that:
\[ \Delta \mu = \mu_2 - \mu_1 \]
We can express \( \mu_2 \) in terms of \( \mu_1 \):
\[ \mu_2 = \frac{19}{6} - \mu_1 \]
Therefore:
\[ \Delta \mu = \left( \frac{19}{6} - \mu_1 \right) - \mu_1 = \frac{19}{6} - 2\mu_1 \]

7. **Using the Focal Length Relation:**
From the earlier step, we also have:
\[ \Delta \mu = \frac{R}{f} = \frac{15}{30} = \frac{1}{2} \]

8. **Setting Up the Equation:**
So we have:
\[ \frac{19}{6} - 2\mu_1 = \frac{1}{2} \]

9. **Solving for \( \mu_1 \):**
Convert \( \frac{1}{2} \) to sixths:
\[ \frac{1}{2} = \frac{3}{6} \]
Thus:
\[ \frac{19}{6} - 2\mu_1 = \frac{3}{6} \]
Rearranging gives:
\[ 2\mu_1 = \frac{19}{6} - \frac{3}{6} = \frac{16}{6} \]
Therefore:
\[ \mu_1 = \frac{8}{6} = \frac{4}{3} \]

10. **Finding \( \mu_2 \):**
Substitute \( \mu_1 \) back to find \( \mu_2 \):
\[ \mu_2 = \frac{19}{6} - \frac{4}{3} = \frac{19}{6} - \frac{8}{6} = \frac{11}{6} \]

11. **Calculating the Difference:**
Finally, the difference in refractive indices is:
\[ \Delta \mu = \mu_2 - \mu_1 = \frac{11}{6} - \frac{4}{3} = \frac{11}{6} - \frac{8}{6} = \frac{3}{6} = \frac{1}{2} \]

### Final Answer:
The difference between the refractive indices of the two lenses is \( \frac{1}{2} \).