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? (the figures are not to scale.)

A)

B)

C)

D)

Hint: We need to understand the relation between the shape of the object placed in front of the mirror, the position of the object and its size or orientation with the image that can be formed from the image made of these specifications to solve the problem.

Complete answer:
We are given a concave mirror of focal length f. A wire is bent to a right-angled triangle and kept in front of the mirror. The one tip of the wire is at a distance of ‘f’ from the mirror and the other is at a distance of \[\dfrac{f}{2}\] from the mirror. We can find the image of the both points using the mirror formula as –
\[\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}\]
For, the point F, the image formed will be at –
\[\begin{align}
 & \dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u} \\
 & \Rightarrow \dfrac{1}{-f}=\dfrac{1}{{{v}_{F}}}+\dfrac{1}{-f} \\
 & \therefore {{v}_{F}}=\infty \\
\end{align}\]
For the point E, the image formed will be at –
\[\begin{align}
 & \dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u} \\
 & \Rightarrow \dfrac{1}{-f}=\dfrac{1}{{{v}_{E}}}+\dfrac{1}{\dfrac{-f}{2}} \\
 & \therefore {{v}_{E}}=f \\
\end{align}\]
We can find the magnification of the images formed due to the points E and F as –
\[\begin{align}
 & m=\dfrac{-v}{u} \\
 & \Rightarrow {{m}_{F}}=\dfrac{-{{v}_{F}}}{{{u}_{F}}} \\
 & \Rightarrow {{m}_{F}}=-\infty \\
 & \text{and,} \\
 & {{m}_{E}}=\dfrac{-{{v}_{E}}}{{{u}_{E}}} \\
 & \Rightarrow {{m}_{E}}=\dfrac{-f}{-\dfrac{f}{2}} \\
 & \therefore {{m}_{E}}=2 \\
\end{align}\]
Using this information, we can draw the required image as –

So, the image of the triangular wire appears to be a ‘L’ shaped figure when the image is formed on a concave lens. This is the required solution.

The correct answer is option C.

Note:
We should be careful with sign conventions while solving the optics problems. In this problem we can see that we have considered the conventional signs for the concave mirror such as the negative focal length, but the focal point is on the right side.