The focal length of a biconvex lens is 20 cm and its refractive index is 1.5. If the radii of curvatures of two surfaces of lens are in the ratio 1:2, then the larger radius of curvature is (in cm)

Options

A. 10
B. 15
C. 20
D. 30

Explanation

To solve the problem, we will use the Lensmaker's formula, which relates the focal length of a lens to its radii of curvature and refractive index. The formula is given by:

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

Where:
- \( f \) is the focal length of the lens,
- \( n \) is the refractive index of the lens,
- \( R_1 \) 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
- Focal length \( f = 20 \, \text{cm} \)
- Refractive index \( n = 1.5 \)
- The ratio of the radii of curvature \( R_1 : R_2 = 1 : 2 \)

### Step 2: Express the radii of curvature in terms of a variable
Let \( R_1 = r \) and \( R_2 = 2r \) (since \( R_2 \) is twice \( R_1 \)).

### Step 3: Substitute the values into the Lensmaker's formula
Substituting \( f \), \( n \), \( R_1 \), and \( R_2 \) into the Lensmaker's formula:

\[ \frac{1}{20} = (1.5 - 1) \left( \frac{1}{r} - \frac{1}{2r} \right) \]

### Step 4: Simplify the equation
Calculating \( n - 1 \):

\[ 1.5 - 1 = 0.5 \]

Now substitute this back into the equation:

\[ \frac{1}{20} = 0.5 \left( \frac{1}{r} - \frac{1}{2r} \right) \]

The term inside the parentheses simplifies:

\[ \frac{1}{r} - \frac{1}{2r} = \frac{2}{2r} - \frac{1}{2r} = \frac{1}{2r} \]

So now the equation becomes:

\[ \frac{1}{20} = 0.5 \cdot \frac{1}{2r} \]

This simplifies to:

\[ \frac{1}{20} = \frac{0.5}{2r} = \frac{0.25}{r} \]

### Step 5: Solve for \( r \)
Cross-multiplying gives:

\[ 1 = 20 \cdot 0.25 \cdot r \]

\[ 1 = 5r \]

Thus,

\[ r = \frac{1}{5} = 0.2 \, \text{cm} \]

### Step 6: Calculate the larger radius of curvature
Since \( R_2 = 2r \):

\[ R_2 = 2 \cdot 0.2 = 0.4 \, \text{cm} \]

However, we need to correct this step. We should have:

\[ \frac{1}{20} = \frac{0.25}{r} \implies r = 5 \, \text{cm} \]

Thus, the larger radius of curvature \( R_2 = 2r = 2 \cdot 5 = 10 \, \text{cm} \).

### Final Answer
The larger radius of curvature is \( R_2 = 30 \, \text{cm} \).