The radii of curvature of the surfaces of a double convex lens are `20 cm and 40 cm` respectively, and its focal length is `20 cm`. What is the refractive index of the material of the lens ?

Options

A. `5/2`
B. `4/3`
C. `5/3`
D. `4/5`

Explanation

To find the refractive index of the material of the lens, we can use the lens maker's formula, which 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,
- \( R_1 \) 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
- \( R_1 = 20 \, \text{cm} \)
- \( R_2 = -40 \, \text{cm} \) (the radius of curvature for the second surface is taken as negative for a convex lens)
- \( f = 20 \, \text{cm} \)

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

\[ \frac{1}{20} = (\mu - 1) \left( \frac{1}{20} - \frac{1}{-40} \right) \]

### Step 3: Simplify the right side
Calculate \( \frac{1}{20} - \frac{1}{-40} \):

\[ \frac{1}{20} - \left(-\frac{1}{40}\right) = \frac{1}{20} + \frac{1}{40} \]

To add these fractions, find a common denominator:

\[ \frac{1}{20} = \frac{2}{40} \]

Thus,

\[ \frac{1}{20} + \frac{1}{40} = \frac{2}{40} + \frac{1}{40} = \frac{3}{40} \]

### Step 4: Substitute back into the equation
Now substitute this back into the equation:

\[ \frac{1}{20} = (\mu - 1) \left( \frac{3}{40} \right) \]

### Step 5: Solve for \( \mu - 1 \)
Rearranging gives:

\[ \mu - 1 = \frac{1/20}{3/40} \]

This can be simplified as follows:

\[ \mu - 1 = \frac{1}{20} \cdot \frac{40}{3} = \frac{40}{60} = \frac{2}{3} \]

### Step 6: Solve for \( \mu \)
Now, add 1 to both sides:

\[ \mu = 1 + \frac{2}{3} = \frac{3}{3} + \frac{2}{3} = \frac{5}{3} \]

### Final Answer
Thus, the refractive index of the material of the lens is:

\[ \mu = \frac{5}{3} \]