A double convex lens is made of glass of refractive index `1.55` with both faces of same radius of curvature. Find the radius of curvature required, if focal length is `20 cm`.

Options

A. 22 cm
B. 44 cm
C. 11 cm
D. 33 cm

Explanation

To solve the problem of finding the radius of curvature required for a double convex lens made of glass with a refractive index of 1.55 and a focal length of 20 cm, we will use the lens maker's formula. Here are the steps to arrive at the solution:

### Step 1: Understand the Lens Maker's Formula
The lens maker's 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 material,
- \( R_1 \) and \( R_2 \) are the radii of curvature of the two surfaces of the lens.

### Step 2: Assign Values
For a double convex lens where both faces have the same radius of curvature:
- Let \( R_1 = R_0 \) (positive for the first surface),
- Let \( R_2 = -R_0 \) (negative for the second surface).

Given:
- \( n = 1.55 \),
- \( f = 20 \, \text{cm} \).

### Step 3: Substitute Values into the Formula
Substituting the values into the lens maker's formula:
\[ \frac{1}{20} = (1.55 - 1) \left( \frac{1}{R_0} - \frac{1}{-R_0} \right) \]

### Step 4: Simplify the Equation
This simplifies to:
\[ \frac{1}{20} = 0.55 \left( \frac{1}{R_0} + \frac{1}{R_0} \right) \]
\[ \frac{1}{20} = 0.55 \left( \frac{2}{R_0} \right) \]
\[ \frac{1}{20} = \frac{1.1}{R_0} \]

### Step 5: Solve for \( R_0 \)
Cross-multiplying gives:
\[ R_0 = 1.1 \times 20 \]
\[ R_0 = 22 \, \text{cm} \]

### Conclusion
Thus, the required radius of curvature \( R_0 \) is \( 22 \, \text{cm} \).

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