To find the focal length of the plano-convex lens, we need to use the lens maker's formula. Let's break down the problem step by step.

Step 1: Understand the Given Data

Diameter of lens, D = 6 cm, so radius R = D/2 = 3 cm.

Thickness at center is small (3 mm) and can be ignored for thin lens approximation.

Speed of light in material, v = 2 × 10⁸ m/s.

Speed of light in vacuum, c = 3 × 10⁸ m/s.

Step 2: Find Refractive Index (n)

Refractive index n = c / v

So, n = (3 × 10⁸) / (2 × 10⁸) = 1.5

Step 3: Lens Maker's Formula

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

For plano-convex lens: one side is flat (plane), so radius of curvature R₁ = ∞ (infinity). The convex side has radius R₂ = -R (by sign convention, if light comes from left, convex surface towards right has negative radius).

So, R₁ = ∞, R₂ = -3 cm.

Step 4: Apply the Formula

$\frac{1}{f}=(1.5-1)\left(\frac{1}{\infty }-\frac{1}{-3}\right)=0.5\times \left(0+\frac{1}{3}\right)=0.5\times \frac{1}{3}=\frac{1}{6}$

Therefore, f = 6 cm? But wait, units: radius was in cm, so focal length f = 6 cm. However, options are in cm and 6 cm is not listed. Check sign convention.

Step 5: Correct Sign Convention

Standard convention: For a plano-convex lens with convex side towards right, R₁ = +∞ (for plane surface) and R₂ = -R (since center of curvature is on the same side as incoming light).

So, $\frac{1}{f}=(n-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)=(1.5-1)\left(\frac{1}{\infty }-\frac{1}{-R}\right)=0.5\times \left(0+\frac{1}{R}\right)=\frac{0.5}{R}$

Given R = 3 cm, so $\frac{1}{f}=\frac{0.5}{3}=\frac{1}{6}$, thus f = 6 cm. But 6 cm is not in options. Perhaps R is not 3 cm? Diameter is 6 cm, but for a plano-convex lens, the radius of curvature R is not necessarily half the diameter. Actually, the diameter given is likely the physical diameter, not the radius of curvature. We need to find R from geometry.

Step 6: Find Radius of Curvature (R) Using Geometry

For a spherical surface, from geometry: $R^2={(R-t)}^2+{\left(\frac{D}{2}\right)}^2$, where t = thickness at center = 3 mm = 0.3 cm, D = 6 cm.

So, $R^2=R^2-2Rt+t^2+{\left(\frac{D}{2}\right)}^2$

Cancel R²: $0=-2Rt+t^2+{\left(\frac{D}{2}\right)}^2$

So, $2Rt=t^2+{\left(\frac{D}{2}\right)}^2$

Since t << D, t² is very small, neglect it: $2Rt\approx {\left(\frac{D}{2}\right)}^2$

So, $R\approx \frac{D^2}{8t}=\frac{6^2}{8\times 0.3}=\frac{36}{2.4}=15\text{cm}$

Step 7: Now Apply Lens Maker's Formula

n = 1.5, R₁ = ∞ (plane surface), R₂ = -R = -15 cm (by convention).

$\frac{1}{f}=(n-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)=(1.5-1)\left(\frac{1}{\infty }-\frac{1}{-15}\right)=0.5\times \left(0+\frac{1}{15}\right)=\frac{0.5}{15}=\frac{1}{30}$

So, focal length f = 30 cm.

Final Answer

The focal length of the lens is 30 cm.

Related Topics and Formulae

Lens Maker's Formula: Used to calculate focal length based on refractive index and radii of curvature. For a thin lens: $\frac{1}{f}=(n-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$

Refractive Index: n = c / v, where c is speed in vacuum, v in medium.

Geometry of Spherical Surface: To find radius of curvature from chord length and sagitta (thickness): $R\approx \frac{D^2}{8t}$ for small t.