Parallax Method and Image Position for Concave Mirror

To determine the range of the object distance $x$, we analyze the concept of parallax and the imaging properties of a concave mirror:

1. Parallax Principle:
   Let an observer view two objects, $O_1$ and $O_2$, aligned along the line of sight, where $O_1$ is closer to the eye than $O_2$.
   When the observer moves their eye in a particular direction (e.g., to the left), the line of sight to the nearer object rotates by a greater angle than that to the farther object. Consequently, relative to the farther object, the nearer object appears to shift in the direction opposite to the displacement of the eye (i.e., towards the right).

2. Application to the Problem:

   • The student looks from a distance at the pin $\text{A}$ and its real, inverted image. The eye is positioned along the principal axis, farther away from the pole $\text{P}$ than both the object pin and its image.
   • When the student shifts their eye towards the left, the image appears to move towards the right relative to the object pin.
   • According to the principle of parallax, the object which shifts opposite to the eye's movement (to the right) must be closer to the eye.
   • Therefore, the image is closer to the student's eye than the object pin $\text{A}$.

3. Relationship between Image Distance and Object Distance:

   • Since both the image and the object pin lie between the mirror pole $\text{P}$ and the eye, being closer to the eye means being farther from the mirror's pole $\text{P}$.
   • Thus, the image distance $v$ is greater than the object distance $u$:
     $v > u = x$

4. Mirror Formula Analysis:
   For a concave mirror forming a real, inverted image:

   • The mirror equation is:
     $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$
   • When $x = 2f$, the image is formed at the center of curvature ($v = 2f = u$), and there is no parallax.
   • When $x > 2f$, the image is formed between $f$ and $2f$, so $v < u$.
   • When $f < x < 2f$, the image is formed beyond the center of curvature ($v > 2f$), which gives:
     $v > u$
   • For $x < f$, the image formed is virtual and erect, which contradicts the observation of an inverted image.

Hence, the condition $v > u$ implies that:
$f < x < 2f$

Therefore, the correct option is B.