A plano-convex lens has a thickness of 4 cm. When placed on a horizontal table, with the curved surf

Q: Why do we use the apparent depth formula in the first case?

In the first case, the top surface of the lens is flat. When light travels from a denser medium to a rarer medium through a plane surface, the image appears closer to the surface. The apparent depth formula $d_{\text{app}} = \frac{d_{\text{actual}}}{\mu}$ perfectly captures this effect.

Q: Why is $u = -4 \text{ cm}$ in the second case?

The object is at the center of the plane face, which is at the bottom. The light travels upwards towards the curved surface. Measuring from the pole of the curved surface downwards (opposite to the direction of light), the object distance $u$ is $-4 \text{ cm}$.

Q: Why is $R_1 = \infty$ and $R_2 = -25 \text{ cm}$ in the Lens Maker's formula?

For a plano-convex lens, one surface is flat, meaning its radius of curvature is infinite ($R_1 = \infty$). The other surface is curved. Depending on the orientation you choose for the formula, if the light hits the flat surface first, the curved surface bulges away, making its center of curvature lie on the negative side, hence $R_2 = -25 \text{ cm}$.

Q: What if we didn't ignore the thickness of the lens?

If the thickness wasn't negligible, the standard Lens Maker's formula wouldn't be accurate. We would have to use the thick lens formula, which includes an extra term $\frac{(\mu - 1)t}{\mu R_1 R_2}$ to account for the distance the light travels inside the lens.

Q: Why is $\mu_2 = 1$ in the spherical refraction formula?

The light is traveling from inside the lens (which has a refractive index $\mu_1 = \frac{4}{3}$) out into the air. The refractive index of air is approximately $1$, so $\mu_2 = 1$.

Question

A plano-convex lens has a thickness of $4	ext{ cm}$. When placed on a horizontal table, with the curved surface in contact with it, the apparent depth of the bottom most point of the lens is found to be $3	ext{ cm}$. If the lens is inverted such that the plane face is in contact with the table, the apparent depth of the centre of the plane face is found to be $25/8	ext{ cm}$. Find the focal length of the lens. Assume thickness to be negligible while finding its focal length.

Sigma Percentile · Accepted Answer

### The Tale of Two Orientations

The problem presents a fascinating scenario involving a plano-convex lens. We are asked to analyze the lens from two different perspectives to ultimately uncover its focal length.

In the first orientation, the curved surface rests on the table, meaning the light rays originating from the bottom-most point travel upwards and refract out of the flat top surface. In the second orientation, the lens is flipped. The flat surface is on the table, and the light rays travel upwards to refract out of the curved top surface. Let's break down each case to extract the hidden parameters of the lens.

### Decoding the First Orientation

When the curved surface is on the table, the light rays emerge from the plane surface. This is a classic case of apparent depth. When light travels from a denser medium to a rarer medium through a flat boundary, the object appears closer to the surface than it actually is.

The formula for apparent depth is:
$$d_{	ext{app}} = \frac{d_{	ext{actual}}}{\mu}$$

We are given the actual thickness of the lens $d_{	ext{actual}} = 4 	ext{ cm}$ and the apparent depth $d_{	ext{app}} = 3 	ext{ cm}$. Substituting these values, we can easily find the refractive index $\mu$ of the lens material:
$$3 = \frac{4}{\mu} \implies \mu = \frac{4}{3}$$

### Unveiling the Second Orientation

Now, let's flip the lens. The flat surface is on the table, and the light rays refract out of the curved surface. This requires the spherical refraction formula:
$$\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}$$

Here, the object is at the center of the plane face (the bottom of the lens). Since we measure distances from the pole of the curved surface against the direction of incident light, the object distance $u = -4 	ext{ cm}$. The image is formed at an apparent depth of $25/8 	ext{ cm}$, so $v = -\frac{25}{8} 	ext{ cm}$. The light travels from the lens ($\mu_1 = \frac{4}{3}$) into the air ($\mu_2 = 1$).

Substituting these values into our equation:
$$\frac{1}{-25/8} - \frac{4/3}{-4} = \frac{1 - 4/3}{-R}$$

Let's simplify the algebra:
$$-\frac{8}{25} + \frac{1}{3} = \frac{-1/3}{-R}$$
$$\frac{-24 + 25}{75} = \frac{1}{3R}$$
$$\frac{1}{75} = \frac{1}{3R} \implies R = 25 	ext{ cm}$$

### The Master Equation: Lens Maker's Formula

With both the refractive index $\mu = \frac{4}{3}$ and the radius of curvature $R = 25 	ext{ cm}$ in our arsenal, we are ready to find the focal length. The Lens Maker's formula is our ultimate tool:
$$\frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}ight)$$

For a plano-convex lens, one surface is perfectly flat, meaning its radius of curvature is infinite ($R_1 = \infty$). If we assume the light enters the flat surface first, the curved surface bulges away from the incoming light, making its center of curvature lie on the negative side. Thus, $R_2 = -25 	ext{ cm}$.

### The Final Triumph

Let's plug everything into the Lens Maker's formula:
$$\frac{1}{f} = \left(\frac{4}{3} - 1ight)\left(\frac{1}{\infty} - \frac{1}{-25}ight)$$
$$\frac{1}{f} = \left(\frac{1}{3}ight)\left(0 + \frac{1}{25}ight)$$
$$\frac{1}{f} = \frac{1}{75}$$

Inverting both sides, we arrive at our final answer:
$$f = 75 	ext{ cm}$$

By analyzing the lens from two different orientations, we systematically unlocked its refractive index and radius of curvature, leading us straight to the focal length. A beautiful demonstration of how changing perspectives can solve complex problems!

Enter Numerical Value:

Visualized Solution

\text{Visualizing the Two Orientations}

\text{Refraction at a Plane Surface}

\text{Calculating Refractive Index } \mu

\text{Refraction at a Spherical Surface}

\text{Substituting Values for Case 2}

\text{Calculating Radius of Curvature } R

\text{Lens Maker's Formula}

\text{Substituting into Lens Maker's Formula}

\text{Final Calculation}

\text{The Way Forward}

The Sigma Insight: Refraction at Spherical Surface

The Tale of Two Orientations

Decoding the First Orientation

Unveiling the Second Orientation

The Master Equation

The Final Triumph

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