A

To find the focal length of a plano-convex lens, we use the lensmaker's formula:
$$ \frac{1}{f} = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) $$
where:
- $f$ is the focal length
- $n$ is the refractive index of the lens material
- $R_1$ is the radius of curvature of the first surface (convex side)
- $R_2$ is the radius of curvature of the second surface (plano side).
Step 1: Determine the radius of curvature. Since the diameter of the lens is 5 cm, the radius is:
$$ R = \frac{5}{2} = 2.5 ext{ cm} $$
Since the first surface is convex, we take this as positive, and the second surface is plano, hence its radius is infinite. Therefore, we have:
$$ R_1 = 2.5 ext{ cm}, \, R_2 = \infty $$
Step 2: Next, we calculate the refractive index. The speed of light in vacuum is $c = 3 \times 10^8 \text{ m/s}$. Given that the speed of light inside the lens is $v = 2 \times 10^8 \text{ m/s}$:
$$ n = \frac{c}{v} = \frac{3 \times 10^8}{2 \times 10^8} = 1.5 $$
Step 3: Substitute the values into the lensmaker's formula:
$$ \frac{1}{f} = (1.5 - 1) \left( \frac{1}{2.5} - 0 \right) = 0.5 \cdot \frac{1}{2.5} = \frac{0.5}{2.5} = \frac{1}{5} $$
Thus, we find:
$$ f = 5 ext{ cm} $$
Therefore, the focal length of the lens is 5 cm.