Distance Between Two Object Positions for Concave Mirror Double Magnification

To find the distance between the two object positions, we use the optical concepts of a concave mirror, including the mirror formula and the magnification formula.

1. Given Parameters:

• Focal length of the concave mirror, $f = -10\ \text{cm}$ (by sign convention).
• The magnitude of linear magnification, $|m| = 2$.

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2. Relation between Magnification, Focal Length, and Object Distance:

The linear magnification $m$ produced by a spherical mirror in terms of focal length $f$ and object distance $u$ is given by: $m = \frac{f}{f - u}$

Since the size of the image is double that of the object ($|m| = 2$), there are two possible cases for the formed image:

1. Real Image: The image is inverted, so $m = -2$.
2. Virtual Image: The image is erect, so $m = +2$.

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3. Case 1: Real Image ($m = -2$)

Substituting $m = -2$ and $f = -10\ \text{cm}$ into the magnification formula: $-2 = \frac{-10}{-10 - u_1}$

Cross-multiplying: $-2(-10 - u_1) = -10$ $20 + 2u_1 = -10$ $2u_1 = -30$ $u_1 = -15\ \text{cm}$

Thus, the object is placed at a distance of $15\ \text{cm}$ in front of the mirror.

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4. Case 2: Virtual Image ($m = +2$)

Substituting $m = +2$ and $f = -10\ \text{cm}$ into the magnification formula: $2 = \frac{-10}{-10 - u_2}$

Cross-multiplying: $2(-10 - u_2) = -10$ $-20 - 2u_2 = -10$ $-2u_2 = 10$ $u_2 = -5\ \text{cm}$

Thus, the object is placed at a distance of $5\ \text{cm}$ in front of the mirror.

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5. Calculating the Distance Between the Two Positions:

The distance $d$ between the two positions of the object is given by the difference between their absolute positions along the principal axis: $d = |u_1 - u_2| = |-15\ \text{cm} - (-5\ \text{cm})| = 10\ \text{cm}$

Final Answer: The distance between the two positions of the object is 10 $\text{cm}$.