The image of an object, formed by a plano-convex lens at a distance of 8 m behind the lens, is real

Q: Why is the magnification taken as $1/3$ and not $-1/3$?

For a real image formed by a single lens, the image is always inverted, so the actual magnification is $m = -1/3$. However, using the magnitude $|m| = |v/u|$ is often simpler to find the distances, after which we apply the proper Cartesian sign conventions ($v = +8\text{ m}$, $u = -24\text{ m}$).

Q: Why is the radius of the flat surface taken as infinity?

A perfectly flat plane can be thought of as a sphere with an infinitely large radius. Therefore, its curvature is zero, and $1/R = 1/\infty = 0$.

Q: Does it matter if the curved surface or the flat surface faces the object?

No, for a thin lens, the focal length remains exactly the same regardless of its orientation. The Lens Maker's formula will yield the same $f$ whether $R_1 = R, R_2 = \infty$ or $R_1 = \infty, R_2 = -R$.

Q: How do we know the image distance $v$ is positive?

The problem states the image is formed 'behind the lens'. According to the standard Cartesian sign convention, the direction of incident light (front to back) is positive, making distances behind the lens positive.

Q: Why does the wavelength change inside the lens?

When light enters a denser medium like glass, its speed decreases while its frequency remains constant. Since speed equals frequency times wavelength ($v = f\lambda$), the wavelength must decrease proportionally.

Question

The image of an object, formed by a plano-convex lens at a distance of $8	ext{ m}$ behind the lens, is real and is one-third the size of the object. The wavelength of light inside the lens is $2/3$ times the wavelength in free space. The radius of the curved surface of the lens is

Options:

(A) $1	ext{ m}$

(B) $2	ext{ m}$

(C) $3	ext{ m}$

(D) $6	ext{ m}$

Sigma Percentile · Accepted Answer

Imagine you are setting up an optical experiment. You have a plano-convex lens, and you place an object in front of it. A crisp, real image forms on a screen 8 meters behind the lens, but it's shrunk to exactly one-third the size of your original object. Your task? Find the radius of curvature of that curved lens surface.

### Decoding the Refractive Index
Before we even think about the lens's shape, we need to know what it's made of. The problem gives us a fascinating clue: the wavelength of light inside the lens is $2/3$ of its wavelength in free space.

Why does this matter? Because the refractive index $\mu$ of a medium is directly tied to how much it "slows down" light, which in turn compresses its wavelength. The relationship is beautifully simple:
$$ \mu = \frac{\lambda_{	ext{vacuum}}}{\lambda_{	ext{medium}}} $$

Substituting the given ratio, we find:
$$ \mu = \frac{1}{2/3} = \frac{3}{2} = 1.5 $$

So, our lens is made of a standard glass with a refractive index of 1.5.

### Pinpointing the Object
Next, let's figure out exactly where our object is sitting. We know the image is real and one-third the size of the object. In optics, the magnification $m$ relates the image distance $v$ to the object distance $u$:
$$ |m| = \left|\frac{v}{u}ight| = \frac{1}{3} $$

Since the real image is formed 8 meters behind the lens, $v = +8	ext{ m}$. Plugging this in:
$$ \frac{1}{3} = \frac{8}{|u|} \implies |u| = 24	ext{ m} $$

Following the Cartesian sign convention, the object is placed in front of the lens, so $u = -24	ext{ m}$.

### Unveiling the Focal Length
With both $u$ and $v$ in hand, the lens's focal length $f$ is just a thin lens formula away:
$$ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} $$

Let's plug in our coordinates:
$$ \frac{1}{f} = \frac{1}{8} - \frac{1}{-24} = \frac{1}{8} + \frac{1}{24} $$

Finding a common denominator:
$$ \frac{1}{f} = \frac{3}{24} + \frac{1}{24} = \frac{4}{24} = \frac{1}{6} $$

This tells us the focal length $f$ is exactly $+6	ext{ m}$.

### The Grand Finale: Lens Maker's Formula
Now for the final piece of the puzzle. How does the focal length relate to the physical shape of the lens? Enter the Lens Maker's formula:
$$ \frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}ight) $$

For a plano-convex lens, one surface is flat, meaning its radius of curvature is infinite ($R_2 = \infty$). The other surface has the radius $R$ we are looking for.
$$ \frac{1}{6} = \left(\frac{3}{2} - 1ight)\left(\frac{1}{R} - \frac{1}{\infty}ight) $$

Since $1/\infty = 0$, the equation simplifies beautifully:
$$ \frac{1}{6} = \left(\frac{1}{2}ight)\left(\frac{1}{R}ight) $$

Multiplying both sides by 2:
$$ \frac{1}{3} = \frac{1}{R} \implies R = 3	ext{ m} $$

And there we have it! The radius of the curved surface is exactly **$3	ext{ m}$**. By systematically breaking down the light's wavelength, the image's magnification, and the lens's geometry, we've completely decoded the optical system.

Select Answer:

Visualized Solution

$μ = \frac{λ _{air}}{λ _{medium}}$

$∣ m ∣ = \frac{v}{u}$

$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$

$\frac{1}{f} = (μ - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}})$

The Sigma Insight: Lens

Decoding the Refractive Index Before we even think about the lens's shape, we need to know what it's made of

Pinpointing the Object Next, let's figure out exactly where our object is sitting

Unveiling the Focal Length

The Grand Finale

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