An image Y is formed of point object X by a lens whose optic axis is AB as shown in figure. Draw a r

Q: Why does the line $XY$ pass through the optical centre of the lens?

By definition, the optical centre is the point on the principal axis where a light ray passes through without suffering any angular deviation. Since $Y$ is the image of $X$, the straight line connecting them represents this undeviated ray.

Q: How did we find the pole of the mirror using $Y'$?

The pole $O$ acts like a plane mirror for paraxial rays. The angle of incidence equals the angle of reflection. By reflecting $Y$ across the principal axis to $Y'$, the line $XY'$ represents the path of the ray if it hadn't been reflected. Its intersection with the axis gives the exact point of reflection, which is the pole.

Q: Why does the line $XY$ pass through the centre of curvature $C$ for the mirror?

A ray directed towards the centre of curvature strikes the mirror normally (at $90^\circ$) and retraces its path. Therefore, the object $X$, the centre of curvature $C$, and the image $Y$ must all lie on this straight line.

Q: Can this method be used for any position of $X$ and $Y$?

Yes, as long as $X$ and $Y$ represent a valid real object-image pair for the respective optical device, these geometric constructions will accurately locate the device and its cardinal points.

Question

An image $Y$ is formed of point object $X$ by a lens whose optic axis is $AB$ as shown in figure. Draw a ray diagram to locate the lens and its focus. If the image $Y$ of the object $X$ is formed by a concave mirror (having the same optic axis as $AB$) instead of lens, draw another ray diagram to locate the mirror and its focus. Write down the steps of construction of the ray diagrams.

Sigma Percentile · Accepted Answer

The problem presents us with a fascinating geometric puzzle. We are given the optic axis $AB$, a point object $X$, and its real, inverted image $Y$. Our mission is to locate the optical device—first a lens, then a concave mirror—and pinpoint their respective foci using pure ray geometry.

### Decoding the Setup

Before we draw any lines, let's analyze what we know. The image $Y$ is formed on the opposite side of the principal axis relative to the object $X$. This means the image is real and inverted.

If the device is a lens, it must be a convex lens, because a concave lens always produces a virtual, erect image on the same side as the object. If the device is a mirror, it must be a concave mirror, as convex mirrors only form virtual, erect images.

### Constructing the Lens Diagram

Let's begin by locating the convex lens. We rely on a fundamental principle of thin lenses: any light ray passing through the optical centre $O$ travels straight without any angular deviation.

Since $Y$ is the image of $X$, the straight line connecting them represents this undeviated ray. By drawing a line from $X$ to $Y$, the exact point where it intersects the principal axis $AB$ is our optical centre $O$. We can now draw our convex lens at this position.

Next, we need to find the focus $F$. We use another standard ray: a ray parallel to the principal axis. We draw a line from $X$ parallel to $AB$, striking the lens at a point $L$. After refraction, this ray must pass through the focus to reach the image $Y$. By joining $L$ and $Y$, the point where this refracted ray crosses the principal axis is the focus $F$.

### Constructing the Mirror Diagram

Now, let's replace the lens with a concave mirror. Finding the pole $O$ of the mirror requires a clever geometric trick. For paraxial rays, the pole acts like a plane mirror where the angle of incidence equals the angle of reflection.

We drop a perpendicular from $Y$ to the axis and extend it equally to the other side to find a virtual point $Y'$. The line connecting $X$ to $Y'$ represents the path of the ray if it hadn't been reflected. The intersection of $XY'$ with the principal axis gives us the exact location of the pole $O$. We can now sketch our concave mirror at $O$.

To find the centre of curvature $C$, we recall that a ray passing through $C$ hits the mirror normally and retraces its path. Therefore, the straight line connecting the object $X$ and the image $Y$ must pass directly through $C$. We draw this line, and its intersection with the axis is $C$.

Finally, to locate the focus $F$ of the mirror, we draw a ray from $X$ parallel to the principal axis, striking the mirror at point $M$. After reflection, this ray must pass through the focus to reach $Y$. By joining $M$ and $Y$, the intersection with the axis gives us the focus $F$.

### The Beauty of Ray Optics

This problem beautifully demonstrates the power of geometric optics. Without knowing any focal lengths or object distances, we used the fundamental laws of refraction and reflection to completely reconstruct the optical systems. It's a perfect example of how simple geometric rules govern the complex behavior of light!

Visualized Solution

\text{Initial Setup}

\text{Optical Centre } O \text{ of the Lens}

\text{Focus } F \text{ of the Lens}

\text{Pole } O \text{ of the Concave Mirror}

\text{Centre of Curvature } C

\text{Focus } F \text{ of the Mirror}

The Sigma Insight: Lens

Decoding the Setup

Constructing the Lens Diagram

Constructing the Mirror Diagram

The Beauty of Ray Optics

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