1. Given data

• Thickness at centre: $t = 3\text{ mm} = 0.3\text{ cm}$
• Diameter of lens: $6\text{ cm}$, so radius (aperture radius) is $a = 3\text{ cm}$
• Speed of light in lens material: $v = 2\times 10^8\text{ m s}^{-1}$
• Speed of light in vacuum: $c = 3\times 10^8\text{ m s}^{-1}$

So refractive index of lens material is $n = \frac{c}{v} = \frac{3\times 10^8}{2\times 10^8} = \frac{3}{2}$

2. Find radius of curvature of the convex surface

For a plane-convex lens, the plane side has infinite radius, and the curved side is part of a sphere.

Let $R$ be the radius of curvature of the spherical surface.

The sagitta formula relates the central thickness (rise) $t$, aperture radius $a$, and radius of curvature $R$: $R^2 = a^2 + (R-t)^2$

Expanding, $R^2 = a^2 + R^2 - 2Rt + t^2$ $2Rt = a^2 + t^2$ $R = \frac{a^2+t^2}{2t}$

Substitute values: $R = \frac{3^2 + 0.3^2}{2(0.3)} = \frac{9 + 0.09}{0.6} = \frac{9.09}{0.6} = 15.15\text{ cm}$

Since $t \ll a$, this is essentially $R \approx 15\text{ cm}$

3. Use lens maker formula

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

For a plane-convex lens:

• one surface is plane, so $R_2 = \infty$
• one surface is convex, so $R_1 = R$

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

Now, $n-1 = \frac{3}{2}-1 = \frac{1}{2}$

Hence, $f = \frac{15.15}{1/2} = 30.3\text{ cm}$

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

4. Match with options

The correct option is:

B: $30\text{ cm}$