A double convex thin lens made of the refractive index 1.6 has radii of curvature 15 cm each. What is the focal length (in cm) of this les when immersed in a fluid of refractive index 1.63?

Answer

To find the focal length of a double convex thin lens when immersed in a fluid, we can use the lens maker's formula. Here are the steps to solve the problem:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Refractive index of the lens (N_glass) = 1.6
- Refractive index of the fluid (N_fluid) = 1.63
- Radius of curvature for both surfaces (R1 and R2) = 15 cm

2. **Calculate the Effective Refractive Index:**
The effective refractive index of the lens with respect to the fluid is calculated as:
\[ N_{glass\_with\_respect\_to\_fluid} = \frac{N_{glass}}{N_{fluid}} = \frac{1.6}{1.63} \]

3. **Apply the Lens Maker's Formula:**
The lens maker's formula for a thin lens is given by:
\[ \frac{1}{f} = \left(N_{glass\_with\_respect\_to\_fluid} - 1\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right) \]
Since R1 and R2 are both 15 cm, we have:
\[ R_1 = 15 \, \text{cm} \quad \text{and} \quad R_2 = -15 \, \text{cm} \]
Thus, substituting these values:
\[ \frac{1}{f} = \left(\frac{1.6}{1.63} - 1\right) \left(\frac{1}{15} - \frac{1}{-15}\right) \]

4. **Simplify the Equation:**
The term \(\frac{1}{15} - \frac{1}{-15}\) simplifies to:
\[ \frac{1}{15} + \frac{1}{15} = \frac{2}{15} \]
Now substituting this back into the equation:
\[ \frac{1}{f} = \left(\frac{1.6}{1.63} - 1\right) \left(\frac{2}{15}\right) \]

5. **Calculate the Value of \(N_{glass\_with\_respect\_to\_fluid} - 1\):**
\[ N_{glass\_with\_respect\_to\_fluid} - 1 = \frac{1.6}{1.63} - 1 = \frac{1.6 - 1.63}{1.63} = \frac{-0.03}{1.63} \]

6. **Substitute and Solve for \(f\):**
\[ \frac{1}{f} = \left(\frac{-0.03}{1.63}\right) \left(\frac{2}{15}\right) = \frac{-0.06}{24.45} \approx -0.00245 \]
Therefore,
\[ f \approx -408.16 \, \text{cm} \]

### Final Answer:
The focal length of the lens when immersed in the fluid is approximately **-408.16 cm**.