1. Given data

• Diameter of plano-convex lens: $D = 6\,\text{cm}$
• Radius of aperture: $a = D/2 = 3\,\text{cm} = 0.03\,\text{m}$
• Thickness at center: $t = 3\,\text{mm} = 0.3\,\text{cm} = 0.003\,\text{m}$
• Speed of light in lens material: $v = 2 \times 10^8\,\text{m/s}$

2. Find refractive index of lens material

Using $n = \frac{c}{v} = \frac{3 \times 10^8}{2 \times 10^8} = 1.5$

3. Find radius of curvature of convex surface

For a plano-convex lens, the central thickness above the plane face comes from the sagitta of the spherical surface.

If $R$ is the radius of curvature and $a$ is the semi-aperture, then $t = R - \sqrt{R^2 - a^2}$

Here, $0.3 = R - \sqrt{R^2 - 9}$ when all lengths are in cm.

So, $\sqrt{R^2 - 9} = R - 0.3$

Squaring both sides, $R^2 - 9 = R^2 - 0.6R + 0.09$

$-9 = -0.6R + 0.09$

$-9.09 = -0.6R$

$R = \frac{9.09}{0.6} = 15.15\,\text{cm}$

Approximately, $R \approx 15\,\text{cm}$

4. Use lens maker formula

For a thin plano-convex lens in air, $\frac{1}{f} = (n-1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$

One surface is plane, so $R_2 = \infty$, hence $\frac{1}{f} = (n-1)\frac{1}{R}$

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

Therefore, $f \approx 30\,\text{cm}$

5. Check options

• A: $15\,\text{cm}$ ❌
• B: $20\,\text{cm}$ ❌
• C: $30\,\text{cm}$ ✅
• D: $10\,\text{cm}$ ❌

So the correct option is C.