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

Q: Why does the image form behind the mirror?

The entire object is placed between the pole and the focus of the concave mirror. For a concave mirror, any object placed within the focal length forms a virtual, erect, and magnified image behind the mirror.

Q: Why does the image of the horizontal side extend to infinity?

The horizontal side extends from $f/2$ up to the focus $f$. As the object distance approaches the focal length, the image distance approaches infinity. Hence, the image of the segment near the focus stretches out to infinity.

Q: How did we know the height of the vertical side is $f/2$?

The triangle has an angle of $45^\circ$ at the focus. The base of this right-angled triangle is the distance from the vertical side to the focus, which is $f - f/2 = f/2$. Using trigonometry, the height is base $\times \tan(45^\circ) = f/2 \times 1 = f/2$.

Q: Why is the image of the hypotenuse a horizontal line?

When we calculate the image height for any point on the hypotenuse, the magnification $m = f/x$ perfectly cancels out the object height $y = x$. This leaves a constant image height of $f$ for all points on the hypotenuse, resulting in a horizontal line.

Q: Why are the lines in the correct option pointing to the right?

In our sign convention, the object is on the left (negative $x$), and the virtual image forms on the right (positive $x$). Since the image distance $v$ goes from $+f$ to $+\infty$, the lines must extend towards the right.

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

The problem of finding the image of an extended object in a spherical mirror is a classic test of your understanding of optics. But when the object is a 2D shape like a triangle, it becomes a beautiful exercise in mapping coordinates!

### Analyzing the Setup

Imagine you are looking at a concave mirror. The principal axis runs right through the middle. We place a wire bent into a right-angled triangle in front of it.

The vertical side of this triangle, let's call it $AB$, is positioned exactly halfway between the pole and the focus, at a distance of $f/2$. The horizontal side lies flat on the principal axis, stretching from $f/2$ all the way to the focus $f$. The hypotenuse connects the top of the vertical side to the focus, making a neat $45^\circ$ angle.

Because the entire triangle is placed within the focal length of the concave mirror, we know right away that the image will be **virtual, erect, and formed behind the mirror**.

### The Master Equation

To find the shape of the image, we need to break the triangle down into its three sides and find the image of each part separately. We will rely on two trusty tools: the mirror formula and the magnification formula.

The mirror formula is:
$$ \frac{1}{v} + \frac{1}{u} = \frac{1}{f} $$

And the transverse magnification is:
$$ m = -\frac{v}{u} $$

Let's start with the vertical side $AB$. It is located at $u = -f/2$. Plugging this into our mirror formula:
$$ \frac{1}{v} + \frac{1}{-f/2} = \frac{1}{-f} $$
$$ \frac{1}{v} - \frac{2}{f} = -\frac{1}{f} $$
$$ \frac{1}{v} = \frac{1}{f} \Rightarrow v = f $$

So, the image of the vertical side forms at a distance $f$ behind the mirror. What about its height? The magnification is $m = -f / (-f/2) = 2$. Since the height of $AB$ is $f/2$ (thanks to that $45^\circ$ angle), the image height is $2 	imes (f/2) = f$. The image of $AB$ is a vertical line of height $f$ at $v = f$.

Next, consider the horizontal side on the principal axis. It starts at $u = -f/2$ and goes up to $u = -f$. We already know the image of the starting point is at $v = f$. As the object point moves towards the focus ($u 	o -f$), its image shoots off to infinity ($v 	o +\infty$). Thus, the image of the horizontal side is a horizontal line extending from $v = f$ to $+\infty$.

### Final Calculation

Now for the grand finale: the hypotenuse! Let's pick a random vertical segment $PQ$ on the hypotenuse, located at a distance $x$ from the focus. Its distance from the pole is $u = -(f - x)$.

Because of the $45^\circ$ angle, the height of this segment is simply $y = x$. Let's find its image position $v$:
$$ \frac{1}{v} + \frac{1}{-(f-x)} = \frac{1}{-f} $$
$$ \frac{1}{v} = \frac{1}{f-x} - \frac{1}{f} = \frac{x}{f(f-x)} $$
$$ v = \frac{f(f-x)}{x} $$

Now, let's calculate the magnification for this specific segment:
$$ m = -\frac{v}{u} = -\frac{\frac{f(f-x)}{x}}{-(f-x)} = \frac{f}{x} $$

To find the height of the image of $PQ$, we multiply the magnification by the object height:
$$ y' = m 	imes y = \left(\frac{f}{x}ight) 	imes x = f $$

Look at that! The $x$ perfectly cancels out. This means that no matter where you are on the hypotenuse, the image height is always exactly $f$. The image of the slanted hypotenuse is a perfectly **horizontal line** at height $f$, extending to infinity!

Combining all three parts, the final image consists of a vertical line at $v = f$, a horizontal line on the axis going to $+\infty$, and a top horizontal line at height $f$ going to $+\infty$. This perfectly matches option (d).

Select Answer:

Visualized Solution

$Analyzing the Setup$

$Image of Vertical Side A B$

$Image of Horizontal Side A C$

$Image of the Hypotenuse$

$Height of the Hypotenuse Image$

$Final Shape of the Image$

The Sigma Insight: Spherical Mirror

Analyzing the Setup

The Master Equation

Final Calculation

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