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Step-by-Step Solution
Step 1: Identify given quantities
• Thickness of the lens at the center, $t = 3\,\text{mm} = 0.3\,\text{cm}$.
• Diameter of the lens, $D = 6\,\text{cm} \implies \text{Radius}, a = 3\,\text{cm}$.
• Speed of light in the material, $v = 2 \times 10^8 \,\mathrm{m\,s^{-1}}$.
• Speed of light in vacuum, $c = 3 \times 10^8 \,\mathrm{m\,s^{-1}}$.
Step 2: Calculate the refractive index
The refractive index $n$ of the lens material is given by:
$$
n = \frac{c}{v} = \frac{3 \times 10^8}{2 \times 10^8} = 1.5.
$$
Step 3: Relate the thickness to the radius of curvature
For a plane-convex lens, the center thickness $t$ (measured from the plane side to the highest point of the convex surface) is related to its radius of curvature $R$ by the relation:
$$
t = R - \sqrt{R^2 - a^2},
$$
where $a$ is the lens' radius (half its diameter).
Step 4: Solve for the radius of curvature $R$
Substitute $t = 0.3\,\text{cm}$ and $a = 3\,\text{cm}$:
$$
0.3 = R - \sqrt{R^2 - 9}.
$$
Rearrange to get:
$$
\sqrt{R^2 - 9} = R - 0.3.
$$
Square both sides:
$$
R^2 - 9 = (R - 0.3)^2 = R^2 - 0.6R + 0.09.
$$
Simplify:
$$
-9 = -0.6R + 0.09 \quad\Longrightarrow\quad -0.6R = -9 - 0.09 = -9.09.
$$
Hence,
$$
R = \frac{9.09}{0.6} \approx 15.15\,\text{cm}.
$$
Step 5: Use the lens formula for a plane-convex surface
For a plane-convex lens, the focal length $f$ is given by:
$$
f = \frac{R}{n - 1},
$$
where $R$ is the radius of curvature (convex side) and $n$ is the refractive index.
Step 6: Compute the focal length
Substituting $R = 15.15\,\text{cm}$ and $n = 1.5$:
$$
f = \frac{15.15}{1.5 - 1} = \frac{15.15}{0.5} = 30.3\,\text{cm} \approx 30\,\text{cm}.
$$
The focal length of the lens is therefore approximately 30 cm.
Reference Image
