Diameter of a plano-convex lens is 6 cm and thickness at the centre is 3 mm. If the speed of light in the material of lens is $$2 \times 10^8$$ m s$$^{-1}$$, the focal length of the lens is:

We first note the given data. The plano-convex lens has a diameter $$D = 6\ \text{cm}$$, so its radius or semi-aperture is $$a = \dfrac{D}{2} = 3\ \text{cm}$$. The thickness at the centre is given as $$t = 3\ \text{mm} = 0.3\ \text{cm}$$. The speed of light in the lens material is $$v = 2 \times 10^{8}\ \text{m s}^{-1}$$, while the speed of light in vacuum is $$c = 3 \times 10^{8}\ \text{m s}^{-1}$$.

We find the refractive index $$n$$ of the material by the definition $$n = \dfrac{c}{v}$$. Substituting the values, we obtain

$$n = \dfrac{3 \times 10^{8}}{2 \times 10^{8}} = 1.5.$$

Next we determine the radius of curvature $$R$$ of the convex surface. For a spherical surface, the sagitta (central thickness in excess over the edge) is related to the radius $$R$$ and the semi-aperture $$a$$ by the formula

$$s = R - \sqrt{R^{2} - a^{2}}.$$

For a plano-convex lens whose plane face is taken as reference, this sagitta equals the given central thickness $$t$$. Thus we have

$$t = R - \sqrt{R^{2} - a^{2}}.$$

Substituting $$t = 0.3\ \text{cm}$$ and $$a = 3\ \text{cm}$$, we write

$$0.3 = R - \sqrt{R^{2} - 3^{2}}.$$

Let us denote $$\sqrt{R^{2} - 9}$$ by $$x$$, so that $$x = R - 0.3$$. Squaring both sides gives

$$x^{2} = (R - 0.3)^{2} = R^{2} - 0.6R + 0.09.$$

But by definition $$x^{2} = R^{2} - 9$$, therefore

$$R^{2} - 0.6R + 0.09 = R^{2} - 9.$$ Cancelling $$R^{2}$$ from both sides, we get

$$-0.6R + 0.09 = -9.$$

Rearranging,

$$-0.6R = -9.09 \quad \Longrightarrow \quad R = \dfrac{9.09}{0.6} \approx 15.15\ \text{cm}.$$

We now employ the Lens-Maker’s formula for a thin lens in air, which states

$$\frac{1}{f} = (n - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right).$$

For a plano-convex lens the plane side has infinite radius, so we set $$R_1 = R$$ (convex side, taken positive) and $$R_2 = \infty$$. Hence

$$\frac{1}{f} = (n - 1)\left(\frac{1}{R} - 0\right) = \frac{n - 1}{R}.$$

Substituting $$n = 1.5$$ and $$R \approx 15.15\ \text{cm}$$:

$$f = \frac{R}{n - 1} = \frac{15.15}{1.5 - 1} = \frac{15.15}{0.5} \approx 30.3\ \text{cm}.$$

The value rounds off neatly to $$30\ \text{cm}$$, matching the option provided.

Hence, the correct answer is Option A.