If the radii of curvature of the two surfaces of a concave lens are 10 cm and 20 cm respectively and the refractive index is 1.5. What is its focal length ?

Options

A. `-10` cm
B. `-12` cm
C. `-13.33` cm
D. `-16.25` cm

Explanation

To find the focal length of the concave lens, we will 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,
- \( R_2 \) is the radius of curvature of the second surface.

### Step 1: Identify the values of \( R_1 \) and \( R_2 \)

Given:
- The radius of curvature of the first surface \( R_1 = -10 \, \text{cm} \) (negative for concave surface),
- The radius of curvature of the second surface \( R_2 = +20 \, \text{cm} \) (positive for convex surface).

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

We know the refractive index \( n = 1.5 \).

Now, substituting the values into the formula:

\[ \frac{1}{f} = (1.5 - 1) \left( \frac{1}{-10} - \frac{1}{20} \right) \]

### Step 3: Simplify the equation

Calculate \( n - 1 \):

\[ n - 1 = 1.5 - 1 = 0.5 \]

Now calculate \( \frac{1}{R_1} - \frac{1}{R_2} \):

\[ \frac{1}{-10} - \frac{1}{20} = -\frac{1}{10} - \frac{1}{20} \]

Finding a common denominator (which is 20):

\[ -\frac{2}{20} - \frac{1}{20} = -\frac{3}{20} \]

### Step 4: Substitute back into the equation

Now substitute back into the equation:

\[ \frac{1}{f} = 0.5 \left( -\frac{3}{20} \right) \]

Calculating this gives:

\[ \frac{1}{f} = -\frac{3}{40} \]

### Step 5: Find the focal length \( f \)

Taking the reciprocal to find \( f \):

\[ f = -\frac{40}{3} \approx -13.33 \, \text{cm} \]

Thus, the focal length of the concave lens is approximately \( -13.33 \, \text{cm} \).

### Final Answer:
The focal length of the concave lens is \( -13.33 \, \text{cm} \).

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