The radii of curvature of the surfaces of a double convex lens are `20 cm and 30 cm`. What will be its focal length and power in air and water respectively ? Refractive indices for glass and water are `3//2 and 4//3` respectively.

Answer

To find the focal length and power of a double convex lens in air and water, we can follow these steps:

### Step 1: Identify the given values
- Radii of curvature:
- R1 = +20 cm (for the first surface)
- R2 = -30 cm (for the second surface, negative because it is on the opposite side)
- Refractive indices:
- Refractive index of glass (lens), μ = 3/2
- Refractive index of water, μ_water = 4/3

### Step 2: Calculate the focal length in air using the lens maker's formula
The lens maker's formula is given by:
\[ \frac{1}{F} = (μ - 1) \left( \frac{1}{R1} - \frac{1}{R2} \right) \]

Substituting the values for air:
\[ \frac{1}{F} = \left( \frac{3}{2} - 1 \right) \left( \frac{1}{20} - \frac{1}{(-30)} \right) \]

Calculating the terms:
- \( μ - 1 = \frac{3}{2} - 1 = \frac{1}{2} \)
- \( \frac{1}{20} - \frac{1}{(-30)} = \frac{1}{20} + \frac{1}{30} \)

Finding a common denominator (LCM of 20 and 30 is 60):
\[ \frac{1}{20} = \frac{3}{60}, \quad \frac{1}{30} = \frac{2}{60} \]
So,
\[ \frac{1}{20} + \frac{1}{(-30)} = \frac{3}{60} + \frac{2}{60} = \frac{5}{60} = \frac{1}{12} \]

Now substituting back into the lens maker's formula:
\[ \frac{1}{F} = \frac{1}{2} \cdot \frac{1}{12} = \frac{1}{24} \]
Thus,
\[ F = 24 \text{ cm} \]

### Step 3: Calculate the power of the lens in air
Power (P) is given by:
\[ P = \frac{1}{F} \quad \text{(in meters)} \]
Converting focal length to meters:
\[ F = 0.24 \text{ m} \]
Then,
\[ P = \frac{1}{0.24} \approx 4.17 \text{ diopters} \]

### Step 4: Calculate the focal length in water
First, we need to find the effective refractive index of the lens in water:
\[ μ_{effective} = \frac{μ_{lens}}{μ_{water}} = \frac{3/2}{4/3} = \frac{3}{2} \cdot \frac{3}{4} = \frac{9}{8} \approx 1.125 \]

Now, applying the lens maker's formula again in water:
\[ \frac{1}{F} = (μ_{effective} - 1) \left( \frac{1}{R1} - \frac{1}{R2} \right) \]
Substituting the values:
\[ \frac{1}{F} = (1.125 - 1) \left( \frac{1}{20} - \frac{1}{(-30)} \right) = 0.125 \cdot \frac{1}{12} \]
Calculating:
\[ \frac{1}{F} = 0.125 \cdot \frac{1}{12} = \frac{0.125}{12} = \frac{1}{96} \]
Thus,
\[ F = 96 \text{ cm} \]

### Step 5: Calculate the power of the lens in water
Convert focal length to meters:
\[ F = 0.96 \text{ m} \]
Then,
\[ P = \frac{1}{0.96} \approx 1.04 \text{ diopters} \]

### Final Answers:
- Focal length in air: 24 cm, Power in air: 4.17 diopters
- Focal length in water: 96 cm, Power in water: 1.04 diopters