Animated Solution for Physics - Optics: The thickness at the centre of a plano convex lens is $3\text{ mm}$ and the diameter is $6\text{ cm}$. If the speed of light in the material of the lens is $2\times 10^8{\text{ ms}}^{-1}$, then the focal length of the lens is

Imagine a plano-convex lens not just as a piece of glass, but as a small, precise slice cut from a massive glass sphere. To truly understand its geometry, we must look at the cross-section. Let the radius of this imaginary sphere be $R$. The lens itself has a circular aperture with a diameter of $6\text{ cm}$, which means its aperture radius $r$ is $3\text{ cm}$. The maximum thickness at the center of the lens, denoted by $t$, is given as $3\text{ mm}$.

By drawing a line from the center of curvature $C$ to the edge of the lens, and another to the center of the flat surface, we form a perfect right-angled triangle. The hypotenuse is the sphere's radius $R$, the base is the distance from the center to the flat surface $(R-t)$, and the height is the aperture radius $r$.

Before we dive into the geometry, we need to know how light behaves inside this specific glass. The problem gives us the speed of light in the lens material as $v=2\times 10^8{\text{ ms}}^{-1}$.

The refractive index $\mu$ is the fundamental property that dictates how much a lens bends light. It is defined as the ratio of the speed of light in a vacuum $c$ to the speed of light in the medium $v$:

Substituting the standard speed of light in a vacuum ($3\times 10^8{\text{ ms}}^{-1}$) and the given speed in the medium:

This is a very standard refractive index for glass, confirming we are on the right track.

Now, let's return to our right-angled triangle and apply the Pythagorean theorem:

Expanding the squared binomial on the right side gives:

The $R^2$ terms on both sides cancel out beautifully, leaving us with:

Here is where physical intuition saves us from messy algebra. The thickness of the lens $t$ is just $3\text{ mm}$, which is extremely small compared to its radius. When you square a very small number, it becomes negligibly tiny. Therefore, we can safely approximate $t^2\approx 0$. This simplifies our equation dramatically:

Lens Maker's Formula

With the radius of curvature $R$ expressed in terms of known quantities, we bring in the heavy artillery: the Lens Maker's Formula. For a thin lens in air, it states:

For a plano-convex lens, the first surface is curved with radius $R_1=R$, and the second surface is perfectly flat, meaning its radius of curvature is infinite ($R_2=\infty$). Since $\frac{1}{\infty }=0$, the formula collapses to:

Substituting our derived expression for $R$:

Now, we just need to plug in the numbers. Crucially, we must convert all units to meters to avoid disastrous powers-of-ten errors.

- $r=3\text{ cm}=3\times 10^{-2}\text{ m}$ - $t=3\text{ mm}=3\times 10^{-3}\text{ m}$ - $\mu =1.5$

Substituting these into our focal length equation:

Converting back to centimeters, we get our final, elegant answer: