Step 1: Understanding the Concept:
The problem requires finding the focal length of a biconvex lens given its radii of curvature and refractive index. This can be solved using the Lens Maker's Formula.
Step 2: Key Formula or Approach:
The Lens Maker's Formula is:
\[ \frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
where:
\(f\) = focal length of the lens
\(\mu\) = refractive index of the lens material with respect to the surrounding medium (air, in this case)
\(R_1\) = radius of curvature of the first surface (where light enters)
\(R_2\) = radius of curvature of the second surface
Step 3: Detailed Explanation:
Given data:
- Type of lens: Biconvex
- Refractive index, \(\mu = 1.5\)
- Radius of curvature of each surface = 20 cm.
We must apply the Cartesian sign convention. Assume light travels from left to right.
- For the first surface (left), it is convex towards the incident light. Its center of curvature is on the right side. Thus, \(R_1 = +20\) cm.
- For the second surface (right), it is also convex, but its center of curvature is on the left side. Thus, \(R_2 = -20\) cm.
Now, substitute these values into the Lens Maker's Formula:
\[ \frac{1}{f} = (1.5 - 1) \left( \frac{1}{+20} - \frac{1}{-20} \right) \]
\[ \frac{1}{f} = (0.5) \left( \frac{1}{20} + \frac{1}{20} \right) \]
\[ \frac{1}{f} = (0.5) \left( \frac{2}{20} \right) \]
\[ \frac{1}{f} = \left(\frac{1}{2}\right) \left( \frac{1}{10} \right) \]
\[ \frac{1}{f} = \frac{1}{20} \]
Therefore, the focal length is:
\[ f = 20 \text{ cm} \]
Step 4: Final Answer:
The focal length of the lens is 20 cm. This corresponds to option (C).