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

Options

A. 25 cm
B. 125 cm
C. 250 cm
D. `-407.5 cm`

Answer

To solve the problem, we will use the Lensmaker's formula, which is given by:

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

Where:
- \( f \) is the focal length of the lens,
- \( \mu_l \) is the refractive index of the lens,
- \( \mu_m \) is the refractive index of the medium,
- \( R_1 \) is the radius of curvature of the first surface,
- \( R_2 \) is the radius of curvature of the second surface.

### Step 1: Identify the given values
- Refractive index of the lens, \( \mu_l = 1.6 \)
- Refractive index of the medium, \( \mu_m = 1.63 \)
- Radius of curvature for both surfaces, \( R_1 = 15 \, \text{cm} \) and \( R_2 = -15 \, \text{cm} \) (since \( R_2 \) is on the opposite side).

### Step 2: Substitute the values into the Lensmaker's formula
Substituting the values into the formula, we have:

\[ \frac{1}{f} = \left( \frac{1.6}{1.63} - 1 \right) \left( \frac{1}{15} - \frac{1}{-15} \right) \]

### Step 3: Calculate \( \frac{1}{R_1} - \frac{1}{R_2} \)
Calculating \( \frac{1}{R_1} - \frac{1}{R_2} \):

\[ \frac{1}{15} - \left(-\frac{1}{15}\right) = \frac{1}{15} + \frac{1}{15} = \frac{2}{15} \]

### Step 4: Calculate \( \frac{\mu_l}{\mu_m} - 1 \)
Calculating \( \frac{\mu_l}{\mu_m} - 1 \):

\[ \frac{1.6}{1.63} - 1 = \frac{1.6 - 1.63}{1.63} = \frac{-0.03}{1.63} \]

### Step 5: Substitute back into the formula
Now substituting back into the formula:

\[ \frac{1}{f} = \left( \frac{-0.03}{1.63} \right) \left( \frac{2}{15} \right) \]

### Step 6: Simplify the expression
Calculating this gives:

\[ \frac{1}{f} = \frac{-0.03 \times 2}{1.63 \times 15} \]

Calculating \( 1.63 \times 15 = 24.45 \):

\[ \frac{1}{f} = \frac{-0.06}{24.45} \]

### Step 7: Calculate \( f \)
Now, calculating \( f \):

\[ f = \frac{24.45}{-0.06} \approx -407.5 \, \text{cm} \]

### Final Answer
Thus, the focal length of the lens when immersed in the fluid is:

\[ f \approx -407.5 \, \text{cm} \]