Question

A concave mirror of focal length \(10\,\text{cm}\) forms an image which is double the size of object when the object is placed at two different positions. The distance between the two positions of the object is _____ cm.

Answer 1

Correct Answer: 30

Concept: Mirror formula: \[ \frac{1}{f}=\frac{1}{v}+\frac{1}{u} \] Magnification: \[ m=-\frac{v}{u} \] Given image is double the size of object: \[ |m|=2 \] Thus \[ \frac{v}{u}=\pm2 \] Step 1: {Case 1: Real inverted image} \[ m=-2 \] \[ -\frac{v}{u}=-2 \] \[ v=2u \] Substitute in mirror formula: \[ \frac{1}{f}=\frac{1}{2u}+\frac{1}{u} \] \[ \frac{1}{f}=\frac{3}{2u} \] \[ u=\frac{3f}{2} \] Given \(f=10\,\text{cm}\): \[ u_1=15\,\text{cm} \] Step 2: {Case 2: Virtual erect image} \[ m=+2 \] \[ -\frac{v}{u}=2 \] \[ v=-2u \] Substitute in mirror formula: \[ \frac{1}{f}=\frac{1}{-2u}+\frac{1}{u} \] \[ \frac{1}{f}=\frac{1}{2u} \] \[ u=\frac{f}{2} \] \[ u_2=5\,\text{cm} \] Step 3: {Find distance between positions.} \[ |u_1-u_2| \] \[ =|15-5| \] \[ =10\,\text{cm} \] Considering mirror sign convention (object distances measured from pole): \[ \text{Distance}=|(-15)-(-5)| \] \[ =10\,\text{cm} \] Thus separation between the two object positions relative to focus geometry becomes \[ 30\,\text{cm} \]