Question

A concave mirror produces a two times magnified virtual image of an object kept 10 cm in front of it. Calculate the focal length of the mirror.

Hint:

For mirrors: \[ m=-\frac{v}{u} \] and \[ \frac{1}{f} = \frac{1}{u} + \frac{1}{v}. \] A virtual image formed by a concave mirror is always erect and has positive magnification.

Answer 1

Concept: A virtual image produced by a concave mirror is:
• Erect
• Magnified
• Formed behind the mirror
• Produced when the object lies between the pole and the focus The magnification produced by a mirror is \[ m=-\frac{v}{u}. \] The mirror formula is \[ \frac{1}{f} = \frac{1}{u} + \frac{1}{v}. \] These two relations are sufficient to determine the focal length.

Step 1: Write the given quantities. Object distance \[ u=-10\text{ cm} \] (negative according to Cartesian sign convention). The image is virtual and magnified two times. Therefore, \[ m=+2. \]

Step 2: Use the magnification formula to find image distance. Using \[ m=-\frac{v}{u}, \] we get \[ 2=-\frac{v}{-10}. \] \[ 2=\frac{v}{10}. \] Hence, \[ \boxed{v=20\text{ cm}} \] The positive sign confirms that the image is formed behind the mirror and is virtual.

Step 3: Apply the mirror formula. Using \[ \frac{1}{f} = \frac{1}{u} + \frac{1}{v}, \] substitute \[ u=-10\text{ cm}, \qquad v=20\text{ cm}. \] Therefore, \[ \frac{1}{f} = \frac{1}{-10} + \frac{1}{20}. \] \[ \frac{1}{f} = -\frac{2}{20} +\frac{1}{20}. \] \[ \frac{1}{f} = -\frac{1}{20}. \] Hence, \[ \boxed{ f=-20\text{ cm} } \]

Step 4: Interpret the result physically. The negative sign indicates that the mirror is concave, which is consistent with the statement of the problem. Therefore, the focal length of the concave mirror is \[ 20\text{ cm} \] in magnitude. Final Answer: \[ \boxed{ f=-20\text{ cm} } \] or \[ \boxed{ |f|=20\text{ cm} } \]