Qualitative Shape of Image Formed by Concave Mirror for Triangular Wire

To determine the qualitative shape of the image formed by the concave mirror, we analyze the mapping of each side of the triangular wire frame using the mirror formula and transverse magnification formula.

1. Coordinate System Setup

Let the pole of the concave mirror be at the origin $(0,0)$ on the principal axis. Since the mirror is concave with a focal length $f$, its focal point is located at $x = -f$.

The object is a right-angled triangular wire located in front of the mirror ($x < 0$):

• Vertical leg: Located at $x = -f/2$, extending from $y = 0$ to $y = f/2$.
• Horizontal leg: Lies along the principal axis from $x = -f/2$ to $x = -f$.
• Hypotenuse: Connects the points $\left(-\frac{f}{2}, \frac{f}{2}\right)$ and $(-f, 0)$.

The equation describing the hypotenuse of the triangle in object space is: $y(x) = f + x \quad \text{for } x \in \left[-f, -\frac{f}{2}\right]$

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2. Mirror Formula and Magnification

For a point object located at distance $u = x$ (where $x < 0$), the image position $v$ is given by the mirror formula: $\frac{1}{v} + \frac{1}{u} = \frac{1}{-f}$

Substituting $u = x$: $\frac{1}{v} = -\frac{1}{f} - \frac{1}{x} = \frac{-x - f}{f x}$

$\implies v(x) = \frac{f x}{-x - f} = \frac{f x}{-(x + f)}$

The transverse magnification $m$ is given by: $m(x) = -\frac{v}{u} = -\frac{v(x)}{x} = \frac{f}{x + f}$

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3. Image of Each Segment

A. Image of the Vertical Leg ($x = -f/2$, $y \in [0, f/2]$)

• Position: $v\left(-\frac{f}{2}\right) = \frac{f \left(-f/2\right)}{-\left(-f/2 + f\right)} = \frac{-f^2 / 2}{-f/2} = +f$ The image of the vertical leg is formed behind the mirror (virtual image) at $v = f$.

• Height: $m\left(-\frac{f}{2}\right) = \frac{f}{-f/2 + f} = +2$ Since the height of the vertical leg is $h = f/2$, the height of its image is: $y' = m \cdot y = 2 \times \frac{f}{2} = f$

  Thus, the vertical leg maps to a vertical line segment at $v = f$ extending from $y' = 0$ to $y' = f$.

B. Image of the Horizontal Leg ($y = 0$, $x \in [-f, -f/2]$)

• At $x = -f/2$, $v = +f$.

• As $x \to -f$, $v \to +\infty$.

  Thus, the horizontal leg along the principal axis maps to a ray lying on the principal axis extending from $v = f$ to $v = \infty$.

C. Image of the Hypotenuse ($y(x) = f + x$, $x \in [-f, -f/2]$)

The height $y'$ of any point on the image of the hypotenuse is given by: $y'(x) = m(x) \cdot y(x)$

Substituting $m(x) = \frac{f}{x + f}$ and $y(x) = x + f$: $y'(x) = \left(\frac{f}{x + f}\right) (x + f) = f$

Remarkably, $y'(x) = f$ is constant and independent of $x$ for all points along the hypotenuse!

• At $x = -f/2$, $v = f$ and $y' = f$.
• As $x \to -f$, $v \to +\infty$ and $y' = f$.

Thus, the hypotenuse maps to a horizontal straight line parallel to the principal axis at a height $y' = f$, extending from $v = f$ to $v \to \infty$.

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Conclusion

Combining the images of all three sides:

1. A vertical segment at $v = f$ of height $f$.
2. A horizontal segment on the principal axis from $v = f$ to $\infty$.
3. A horizontal line parallel to the principal axis at height $f$ extending from $v = f$ to $\infty$.

This matches the qualitative shape depicted in Option (D).