A wire is bent in the shape of a right angled triangle and is placed in front of a concave mirror of

Q: Why is the image formed behind the mirror?

Whenever an object is placed between the pole and the principal focus of a concave mirror, the reflected rays diverge. When extended backwards, they appear to meet behind the mirror, forming a virtual, erect, and magnified image.

Q: Why are the lines in the correct option drawn as dashed lines?

In optics, it is a standard convention to represent virtual images (images formed by the apparent intersection of rays behind a mirror or lens) using dashed lines, while real objects and real images are drawn with solid lines.

Q: How do we know the image extends to infinity?

One vertex of the right-angled triangle is placed exactly at the focus ($u = -f$) of the concave mirror. According to the mirror formula, as $u$ approaches $-f$, the image distance $v$ approaches infinity.

Q: Will any slanted line placed in front of a mirror become horizontal?

No, this is a very special case! The image becomes perfectly horizontal only because the extended line of the hypotenuse passes exactly through the focus. By the laws of reflection, any ray passing through the focus reflects parallel to the principal axis.

Q: Why did the $(u+f)$ term cancel out in the magnification step?

It's a beautiful consequence of the geometry. The height of the object increases linearly as you move away from the focus ($y = u+f$). At the same time, the magnification decreases inversely with the exact same factor ($m = \frac{f}{u+f}$). When you multiply them to find the image height, they perfectly cancel each other out, resulting in a constant height.

Question

A wire is bent in the shape of a right angled triangle and is placed in front of a concave mirror of focal length f, as shown in the figure. Which of the figures shown in the four options qualitatively represent(s) the shape of the image of the bent wire ? (These figures are not to scale.) ?

Sigma Percentile · Accepted Answer

### Analyzing the Setup
Imagine you are looking at a concave mirror, and right in front of it, between the pole and the focus, someone has placed a wire bent into a perfect right-angled triangle. Our mission is to figure out what the reflection of this entire shape looks like.

At first glance, you might think a triangle will just reflect as a bigger or smaller triangle. But spherical mirrors distort shapes in fascinating ways depending on where the object is placed. Since the entire wire is placed between the pole and the focus (from $f/2$ to $f$), we know from the basic principles of ray optics that the image formed will be **virtual, erect, and magnified**, appearing behind the mirror.

### The Vertical and Horizontal Wires
Let's break the triangle down into its three constituent wires. We start with the vertical wire, which is positioned at a distance of $f/2$ from the pole. Using the standard mirror formula:

$$\frac{1}{v} + \frac{1}{u} = \frac{1}{F}$$

Substituting our values ($u = -f/2$ and $F = -f$), we find that the image distance $v = +f$. The positive sign confirms it's a virtual image formed behind the mirror. The magnification $m = -v/u$ gives us $+2$. Since the original height of this vertical segment is $f/2$, its image will be stretched to a height of $f$.

Next, consider the horizontal base of the triangle lying directly on the principal axis. It stretches from $u = -f/2$ to $u = -f$. We already know the image of the point at $f/2$ forms at $f$. But what about the point at the focus ($u = -f$)? Rays originating from the focus reflect parallel to the principal axis, meaning their image forms at **infinity**. Thus, the image of the horizontal base is a straight line extending from $v = f$ all the way to $+\infty$.

### The Master Equation for the Hypotenuse
Now comes the most thrilling part of the problem: the hypotenuse. It's a slanted line connecting the top of the vertical wire to the focus. To find its image, we can write the mathematical equation of this line.

The line passes through $(-f, 0)$ and $(-f/2, f/2)$. Calculating the slope gives us $1$, so the equation of the line is simply $y = x + f$. Since our object coordinate $x$ is just $u$, we can write the height of any point on the hypotenuse as:

$$y = u + f$$

Now, we need to find the height of the image, $y'$, for every point on this line. We know that $y' = m \cdot y$. Let's express the magnification $m$ entirely in terms of $u$ and $f$:

$$m = \frac{f}{f - u}$$

Wait, considering our sign convention where $u$ is negative, the magnification formula simplifies to $m = \frac{f}{u + f}$. Now, let's multiply this magnification by the object height $y$:

$$y' = \left(\frac{f}{u + f}\right) \cdot (u + f)$$

Look at that beautiful cancellation! The $(u + f)$ terms completely vanish, leaving us with:

$$y' = f$$

### The Elegant Shortcut
What does $y' = f$ mean physically? It means that no matter which point you pick on the hypotenuse, its image will always form at a constant height of $f$. The slanted line transforms into a perfectly **horizontal line** parallel to the principal axis!

If you love physics, there is an even faster, purely conceptual way to see this without writing a single equation. Look at the hypotenuse again. If you extend that slanted line, it passes exactly through the focus of the mirror.

What is the golden rule of spherical mirrors? **Any incident ray that passes through the focus will reflect parallel to the principal axis.** If we treat the hypotenuse as a collection of light rays originating from the focus, their reflections must all travel parallel to the axis. Hence, the image of the hypotenuse is a horizontal line extending to infinity. This perfectly matches Option (D).

Select Answer:

Visualized Solution

\text{Visualizing the Setup}

\text{Mirror Formula \& Sign Convention}

\text{Image of the Vertical Wire}

\text{Computing Vertical Image}

\text{Image of the Horizontal Base}

\text{Equation of the Hypotenuse}

\text{Image of the Hypotenuse}

\text{The Elegant Shortcut}

\text{Final Shape of the Image}

The Sigma Insight: Spherical Mirror

Analyzing the Setup Imagine you are looking at a concave mirror, and right in front of it, between the pole and the focus, someone has placed a wire bent into a perfect right-angled triangle

The Vertical and Horizontal Wires Let's break the triangle down into its three constituent wires

The Master Equation for the Hypotenuse Now comes the most thrilling part of the problem: the hypotenuse

The Elegant Shortcut What does $y^{'} = f$ mean physically? It means that no matter which point you pick on the hypotenuse, its image will always form at a constant height of $f$

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