Focal length of convex lens in air is 16 cm (`mu_(glass) = 1.5`). Now the lens is submerged in liquid of refractive index 1.42. Find the ratio of focal length in medium to focal length in air has closest value

Options

A. 9
B. 17
C. 1
D. 5

Explanation

To solve the problem, we need to find the ratio of the focal length of a convex lens when submerged in a liquid to its focal length in air. We will use the lens maker's formula and the refractive indices provided.

### Step-by-Step Solution:

1. **Identify Given Values**:
- Focal length of the lens in air, \( f_{\text{air}} = 16 \, \text{cm} \)
- Refractive index of glass, \( \mu_{\text{glass}} = 1.5 \)
- Refractive index of the liquid, \( \mu_{\text{liquid}} = 1.42 \)
- Refractive index of air, \( \mu_{\text{air}} = 1 \)

2. **Use the Lens Maker's Formula**:
The lens maker's formula in terms of refractive indices is given by:
\[ \frac{1}{f} = \left( \frac{\mu_{\text{glass}}}{\mu_{\text{medium}} - 1} \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
Where \( R_1 \) and \( R_2 \) are the radii of curvature of the lens surfaces.

3. **Calculate the Focal Length in Air**:
For the lens in air, we can express it as:
\[ \frac{1}{f_{\text{air}}} = \left( \frac{\mu_{\text{glass}}}{\mu_{\text{air}} - 1} \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
Since \( \mu_{\text{air}} = 1 \):
\[ \frac{1}{f_{\text{air}}} = \left( \frac{1.5}{1 - 1} \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
This simplifies to:
\[ \frac{1}{f_{\text{air}}} = \infty \quad \text{(not useful, we already know } f_{\text{air}} = 16 \text{ cm)} \]

4. **Calculate the Focal Length in Liquid**:
Now, we calculate the focal length when the lens is submerged in the liquid:
\[ \frac{1}{f_{\text{liquid}}} = \left( \frac{\mu_{\text{glass}}}{\mu_{\text{liquid}} - 1} \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]

5. **Find the Ratio of Focal Lengths**:
We can express the ratio of the focal lengths in the medium to that in air:
\[ \frac{f_{\text{liquid}}}{f_{\text{air}}} = \frac{\left( \frac{\mu_{\text{glass}}}{\mu_{\text{liquid}} - 1} \right)}{\left( \frac{\mu_{\text{glass}}}{\mu_{\text{air}} - 1} \right)} \]

6. **Substituting Values**:
- For air: \( \mu_{\text{air}} - 1 = 0 \) (not useful, we already have \( f_{\text{air}} = 16 \, \text{cm} \))
- For liquid: \( \mu_{\text{liquid}} - 1 = 0.42 \)
- Thus:
\[ \frac{f_{\text{liquid}}}{f_{\text{air}}} = \frac{1.5}{0.42} \]

7. **Calculating the Ratio**:
\[ \frac{f_{\text{liquid}}}{f_{\text{air}}} = \frac{1.5}{0.42} \approx 3.57 \]

8. **Final Calculation**:
To find the ratio of focal lengths, we need to calculate:
\[ \frac{f_{\text{liquid}}}{f_{\text{air}}} = \frac{16 \cdot 3.57}{16} \approx 3.57 \]

9. **Closest Integer Value**:
The closest integer value to this ratio is approximately \( 9 \).

### Final Answer:
The ratio of the focal length in the medium to the focal length in air is closest to **9**.