Convex lens made up of glass `(mu_(g)=1.5)` and radius of curvature R is dipped into water. Its focal length will be (Refractice index of water =4/3)

Options

A. 4R
B. 2R
C. R
D. R/2

Answer

To solve the problem of finding the focal length of a convex lens made of glass when it is dipped in water, we can use the lens maker's formula. Let's go through the steps systematically.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Refractive index of glass, \( \mu_g = 1.5 \)
- Refractive index of water, \( \mu_w = \frac{4}{3} \)
- Radius of curvature of the lens, \( R \)

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

3. **Assign Values to Radii of Curvature:**
For a convex lens:
- \( R_1 = R \) (the first surface),
- \( R_2 = -R \) (the second surface, negative because it is on the opposite side).

4. **Substitute Values into the Formula:**
\[ \frac{1}{f} = \left( 1.5 - \frac{4}{3} \right) \left( \frac{1}{R} - \left(-\frac{1}{R}\right) \right) \]

5. **Simplify the Expression:**
- Calculate \( \mu_L - \mu_{surrounding} \):
\[ 1.5 - \frac{4}{3} = \frac{3}{2} - \frac{4}{3} = \frac{9}{6} - \frac{8}{6} = \frac{1}{6} \]
- Calculate \( \frac{1}{R} - \left(-\frac{1}{R}\right) \):
\[ \frac{1}{R} + \frac{1}{R} = \frac{2}{R} \]

6. **Combine the Results:**
\[ \frac{1}{f} = \left( \frac{1}{6} \right) \left( \frac{2}{R} \right) = \frac{2}{6R} = \frac{1}{3R} \]

7. **Find the Focal Length:**
\[ f = 3R \]

### Final Answer:
The focal length of the convex lens when dipped in water is \( f = 3R \).