The radius of curvature of the faces of a double convex lens are `10 cm and 15 cm`. If focal length of lens of lens is `12 cm`, find the refractive index of the material of th lens.

Explanation

To find the refractive index of the material of the double convex lens, we can use the lens maker's formula, which 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,
- \( R_1 \) is the radius of curvature of the first surface (positive for convex),
- \( R_2 \) is the radius of curvature of the second surface (negative for convex).

### Step 1: Identify the values
Given:
- \( R_1 = 10 \, \text{cm} \) (for the first surface, convex)
- \( R_2 = -15 \, \text{cm} \) (for the second surface, convex)
- \( f = 12 \, \text{cm} \)

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

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

Substituting the values:

\[ \frac{1}{12} = (n - 1) \left( \frac{1}{10} - \frac{1}{-15} \right) \]

### Step 3: Calculate the terms inside the parentheses
Calculate \( \frac{1}{10} - \frac{1}{-15} \):

\[ \frac{1}{10} + \frac{1}{15} = \frac{3}{30} + \frac{2}{30} = \frac{5}{30} = \frac{1}{6} \]

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

\[ \frac{1}{12} = (n - 1) \left( \frac{1}{6} \right) \]

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

\[ n - 1 = \frac{1/12}{1/6} = \frac{1}{12} \times \frac{6}{1} = \frac{6}{12} = \frac{1}{2} \]

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

\[ n = 1 + \frac{1}{2} = \frac{3}{2} = 1.5 \]

### Final Answer
The refractive index of the material of the lens is \( n = 1.5 \).

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