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

Options

A. -407 cm
B. 250 cm
C. 125 cm
D. 25cm

Answer

To find the focal length of a double convex thin lens made of glass when immersed in a liquid, we can use the lens maker's formula. The formula is given by:

\[ \frac{1}{F} = \left( \frac{\mu_{lens}}{\mu_{media}} - 1 \right) \left( \frac{1}{R_1} + \frac{1}{R_2} \right) \]

Where:
- \( F \) is the focal length of the lens,
- \( \mu_{lens} \) is the refractive index of the lens material,
- \( \mu_{media} \) is the refractive index of the medium (liquid),
- \( R_1 \) 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, \( \mu_{lens} = 1.6 \)
- Refractive index of the liquid, \( \mu_{media} = 1.63 \)
- Radius of curvature of both surfaces, \( R_1 = +15 \, \text{cm} \) (positive for the first surface) and \( R_2 = -15 \, \text{cm} \) (negative for the second surface).

### Step 2: Substitute the values into the formula
Using the lens maker's formula:

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

### Step 3: Calculate \( \frac{1.6}{1.63} - 1 \)
Calculating the first part:

\[ \frac{1.6}{1.63} \approx 0.98037 \]
Thus,
\[ 0.98037 - 1 = -0.01963 \]

### Step 4: Calculate \( \frac{1}{15} + \frac{1}{-15} \)
Calculating the second part:

\[ \frac{1}{15} + \frac{1}{-15} = 0 \]

### Step 5: Combine the results
Since the second part equals zero, we have:

\[ \frac{1}{F} = (-0.01963) \cdot 0 = 0 \]

This indicates that the focal length approaches infinity, which means that the lens does not converge light when immersed in the liquid of higher refractive index.

### Conclusion
The focal length of the lens when immersed in the liquid is effectively infinite.