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. `11 cm`
B. `22 cm`
C. `7 cm`
D. `6 cm`

Explanation

To find 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 can use the lens maker's formula. The steps are as follows:

### Step 1: Understand the Lens Maker's Formula
The lens maker's formula for a thin lens is given by:
\[ \frac{1}{F} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
where:
- \( F \) is the focal length of the lens,
- \( \mu \) is the refractive index of the lens material,
- \( R_1 \) and \( R_2 \) are the radii of curvature of the two lens surfaces.

### Step 2: Identify the Parameters
For a double convex lens:
- The refractive index \( \mu = 1.55 \)
- The focal length \( F = 20 \, \text{cm} \)
- Since both faces have the same radius of curvature, let \( R_1 = R \) and \( R_2 = -R \) (the negative sign is because the second surface is concave with respect to the incoming light).

### Step 3: Substitute Values into the Formula
Substituting the values into the lens maker's formula:
\[ \frac{1}{F} = (\mu - 1) \left( \frac{1}{R} - \frac{1}{-R} \right) \]
This simplifies to:
\[ \frac{1}{F} = (\mu - 1) \left( \frac{1}{R} + \frac{1}{R} \right) = (\mu - 1) \left( \frac{2}{R} \right) \]

### Step 4: Plug in the Known Values
Now substituting \( \mu = 1.55 \) and \( F = 20 \, \text{cm} \):
\[ \frac{1}{20} = (1.55 - 1) \left( \frac{2}{R} \right) \]
\[ \frac{1}{20} = 0.55 \left( \frac{2}{R} \right) \]

### Step 5: Solve for \( R \)
Rearranging the equation:
\[ \frac{1}{20} = \frac{1.1}{R} \]
Cross-multiplying gives:
\[ R = 20 \times 1.1 = 22 \, \text{cm} \]

### Final Answer
The radius of curvature required is:
\[ R = 22 \, \text{cm} \]

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